{"id":11289,"date":"2019-12-11T17:37:23","date_gmt":"2019-12-11T12:07:23","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=11289"},"modified":"2021-07-02T15:42:36","modified_gmt":"2021-07-02T10:12:36","slug":"1st-puc-physics-question-bank-chapter-8","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-physics-question-bank-chapter-8\/","title":{"rendered":"1st PUC Physics Question Bank Chapter 8 Gravitation"},"content":{"rendered":"

You can Download Chapter 8 Gravitation Questions and Answers, Notes, 1st PUC Physics Question Bank with Answers<\/a> Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 1st PUC Physics Question Bank Chapter 8 Gravitation<\/h2>\n

1st PUC Physics Gravitation Textbook Questions and Answers<\/h3>\n

Question 1.
\nAnswer the following:<\/p>\n

1. You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?<\/p>\n

2. An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity?<\/p>\n

3. If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun\u2019s pull is greater than the moon\u2019s pull, (you can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the moon\u2019s pull is greater than the tidal effect of the sun. Why?<\/p>\n

Answer:<\/p>\n

1. No, It is impossible to screen the influence of gravitational force. This is mainly because unlike other forces the gravitational force acts on a body independent of the material medium or the status of the body on which it is acting or that of the body due to which it is present.<\/p>\n

2. Yes, If the mass of the space station is large enough then the astronaut can detect gravity acting on him.<\/p>\n

3. The tidal effect is inversely proportional to the cube of the distance while the gravitational force is inversely proportional to the square of the distance. Since the distance between sun and earth is much greater than that between the moon and earth, the effect observed due to the moon is more that due to the sun.<\/p>\n

Question 2.
\nChoose the correct alternative:<\/p>\n

1. Acceleration due to gravity increases\/decreases with increasing altitude.<\/p>\n

2. Acceleration due to gravity increases\/decreases with increasing depth (assume the earth to be a sphere of uniform density).<\/p>\n

3. Acceleration due to gravity is independent of the mass of the earth\/ mass of the body.<\/p>\n

4. The formula – G Mm(1\/r2<\/sub> – 1\/r1<\/sub>) is more\/less accurate than the formula mg(r2<\/sub> – r1<\/sub>) for the difference of potential energy between two points r2<\/sub> and r1<\/sub> distance away from the centre of the earth.<\/p>\n

Answer:
\n1. Decreases.
\nThe value of acceleration due to gravity at a height \u2018h\u2019 above the ground \u2018gh<\/sub>\u2019 is given by,
\ngh<\/sub> = \\(g\\left[1-\\frac{2 h}{R_{e}}\\right]\\), where Re<\/sub> is radius of Earth. Therefore gh<\/sub> decreases with an increase in height.<\/p>\n

2. Decreases.
\nThe value of acceleration due to gravity at a depth \u2018d\u2019 below the ground (gd<\/sub>), is given by, gd<\/sub> = \\(g\\left[1-\\frac{d}{R_{e}}\\right]\\), where Re<\/sub> is the radius of earth. This shows that with an increase in the value of \u2018d\u2019 i the value of gd<\/sub> decreases.<\/p>\n

3. Mass of the body
\nThe acceleration due to gravity of any planet of mass \u2018M\u2019 and radius \u2018R\u2019 is given by, g = \\(\\frac{G M}{R^{2}}\\), where, G is the universal gravitational constant. So, the value of \u2018g\u2019 is dependent on the mass of the earth but independent of mass of the body.<\/p>\n

4. More accurate
\nThe gravitational potential energy of any point at a distance \u2018r\u2019 from the centre of the earth is
\nV (r) = \\(-\\frac{G m M}{r}\\)
\nSo, Potential Energy difference
\n= V(r2<\/sub>) – V(r1<\/sub>)
\n\"1st
\n= gm (r2<\/sub> – r1<\/sub>)
\n= mgr2<\/sub> – mgr1<\/sub>
\n(Assuming r2<\/sub> – r1<\/sub> \u2248 R2<\/sup>)
\nThus -GmM \\(\\left(\\frac{1}{r_{2}}-\\frac{1}{r_{1}}\\right)\\) is a more accurate relation.<\/p>\n

Question 3.
\nSuppose, there existed a planet that went around the sun twice as fast as the earth. What would be its orbital size as compared to that of the earth?
\nAnswer:
\nLet TE<\/sub> be the time taken for one revolution by earth and Tx<\/sub> be that by the planet. Similarly, let RE<\/sub> be the orbital radius of earth and Rx<\/sub> be that of the planet.
\nFrom Kepler\u2019s 3rd<\/sup> Law of planetary motion, R3<\/sup> \u221d T2<\/sup>
\n\u21d2 \\(\\frac{R_{x}^{3}}{R_{\\varepsilon}^{3}}=\\frac{T_{x}^{2}}{T_{E}^{2}}\\)
\nRx<\/sub> = RE<\/sub>\\(\\left(\\frac{T_{x}}{T_{E}}\\right)^{2 \/ 3}\\)
\n\u21d2 Rx = (0.5)2\/3<\/sup> RE<\/sub>
\n[\u2235 give that TE<\/sub> = 2TX<\/sub>]
\n\u21d2 Rx<\/sub> = (0.63) RE<\/sub>
\nSo, the orbital size of the planet is lesser by a factor of 0.63.<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nOne of the satellites of Jupiter has an orbital period of 1.769 days and the radius of the orbit is 4.22 \u00d7 108<\/sup>m. Show that the mass of Jupiter is about one-thousandth that of the sun.
\nAnswer:
\nFor the satellite around Jupiter
\nT10<\/sub> = 1.769 days
\nR10<\/sub> = 4.22 \u00d7 108<\/sup>m
\nFor earth around sun
\nTe<\/sub> = 365.25 days
\nRe<\/sub> = 1.496 \u00d7 1011<\/sup>m
\nWe know that,
\nM = \\(\\frac{4 \\pi \\mathrm{R}^{3}}{\\mathrm{GT}^{2}}\\)
\n\u21d2 \\(\\frac{M_{S}}{M_{J}}=\\left(\\frac{1.496 \\times 10^{11}}{4.22 \\times 10^{8}}\\right)^{3} \\times\\left(\\frac{1.769}{365.25}\\right)^{2}\\)
\n= 1045.039
\n= 1000
\nThus mass of Sun is about thousand times Mass of Jupiter.<\/p>\n

Question 5.
\nLet us assume that our galaxy consists of 2.5 \u00d7 1011<\/sup> stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution? Take the diameter of the Milky Way to be 105<\/sup> ly.
\nAnswer:
\nNumber of stars in our galaxy = 2.5 \u00d7 1011<\/sup>
\nMass of each star = 1 solar mass
\n= 2 \u00d7 1030<\/sup> kg
\n\u21d2 Mass of our galaxy
\n= 2.5 \u00d7 1011<\/sup> \u00d7 2 \u00d7 1036<\/sup> kg
\n= 5 \u00d7 1041<\/sup> kg
\nDiameter of milky way = 105<\/sup> ly
\nd = 105<\/sup> \u00d7 3 \u00d7 108<\/sup> ms-1<\/sup> \u00d7 1yr
\nThe distance of the star = radius of its orbit = 5 \u00d7 104<\/sup> ly.
\nTherefore, the time period of the revolving star is given by,
\n\"1st
\nT2<\/sup> = 1.259 \u00d7 1017<\/sup> yr2<\/sup>
\nT = 3.549 \u00d7 108<\/sup> yrs
\nIt takes about 3.55 \u00d7 108 <\/sup>years to complete one revolution for a star around the milky way.<\/p>\n

Question 6.
\nChoose the correct alternative:<\/p>\n

    \n
  1. If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of Its kinetic\/ potential energy.<\/li>\n
  2. The energy required to launch an orbiting satellite out of earth\u2019s gravitational influence is more\/less than the energy required to project a stationary object at the same height (as the satellite) out of earth\u2019s influence.<\/li>\n<\/ol>\n

    Answer:
    \n1. Kinetic energy
    \nThe potential energy of a satellite rotating in its orbit is zero. The total energy of a system is the sum of its kinetic energy (+ve) and potential energy. Since the earth satellite system is a bound system. The satellite has a negative total energy. So, the energy of the satellite is the negative of its kinetic energy.<\/p>\n

    2. Less
    \nAn orbiting satellite has more energy than a stationary object at the same height. This additional energy is provided by the orbit. It requires lesser energy to make it move out of the earth\u2019s influence than a stationary object.<\/p>\n

    Question 7.
    \nDoes the escape speed of a body from the earth depend on?<\/p>\n

      \n
    1. the mass of the body,<\/li>\n
    2. the location from where it is projected,<\/li>\n
    3. the direction of projection,<\/li>\n
    4. the height of the location from where the body is launched?<\/li>\n<\/ol>\n

      Answer:
      \nEscape velocity of a body from earth is given by v = \\(\\sqrt{2 g R}\\).
      \ng = Acceleration due to Gravity.
      \nR = Radius of earth.
      \n1. No.
      \nEscape velocity is independent of mass of the object.<\/p>\n

      2. No.
      \nAssuming the value of g is the same in all the locations, it is independent of location of projection.<\/p>\n

      3. No.
      \nThe direction of projection is also immaterial as long as it is projected with the velocity above ground.<\/p>\n

      4. Yes.
      \nIt depends on the gravitational potential energy which depends on the height. Since g changes with height. The escape velocity also depends on height.<\/p>\n

      Question 8.
      \nA comet orbits the sun in a highly elliptical orbit. Does the comet have a constant?<\/p>\n

        \n
      1. linear speed,<\/li>\n
      2. angular speed,<\/li>\n
      3. angular momentum,<\/li>\n
      4. kinetic energy,<\/li>\n
      5. potential energy,<\/li>\n
      6. total energy throughout its orbit?<\/li>\n<\/ol>\n

        Neglect any mass loss of the comet when it comes very close to the Sun.
        \nAnswer:<\/p>\n

          \n
        1. No, linear speed varies from point to point in the orbit.<\/li>\n
        2. No, angular speed varies from point to point in the orbit.<\/li>\n
        3. Yes, angular momentum is conserved at any point<\/li>\n
        4. No, kinetic energy varies as the speed varies.<\/li>\n
        5. No, potential energy varies as distance, from Sun varies.<\/li>\n
        6. Yes, the total energy of the system is always constant.<\/li>\n<\/ol>\n

          \"KSEEB<\/p>\n

          Question 9.
          \nWhich of the following symptoms is likely to afflict an astronaut in space<\/p>\n

            \n
          1. swollen feet,<\/li>\n
          2. swollen face,<\/li>\n
          3. headache,<\/li>\n
          4. orientational problem.<\/li>\n<\/ol>\n

            In the following two exercises, choose the correct answer from among the given ones:
            \nAnswer:<\/p>\n

              \n
            1. No, swollen feet will not affect the astronaut as the astronaut feels weightlessness and legs do not support the weight.<\/li>\n
            2. Yes, since most of the sense organs are on the face astronaut will feel its effect.<\/li>\n
            3. Yes, headaches are not affected by gravity.<\/li>\n
            4. Yes, orientation is necessary for doing any work, & without gravity also is necessary.<\/li>\n<\/ol>\n

              Question 10.
              \nThe gravitational intensity at the centre of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see the following figure).
              \n1. a,
              \n2. b,
              \n3. c,
              \n4. 0
              \n\"1st
              \nAnswer:
              \nThe answer is (3) c.
              \nThe gravitational intensity at any point is defined as the negative of the potential gradient i.e., \\(\\left(-\\frac{\\mathrm{dv}}{\\mathrm{dr}}\\right)\\)
              \nAt any point inside the spherical shell, the value of gravitational potential is constant implying the intensity is zero.
              \n\"1st
              \nSo, intensity at all points inside is zero. Since the upper half is cut off there is no attractive force on the upper portion to pull if upwards. So effectively the net force is downwards along c.<\/p>\n

              Question 11.
              \nFor the above problem, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow.<\/p>\n

                \n
              1. d,<\/li>\n
              2. e,<\/li>\n
              3. \u00a0f,<\/li>\n
              4. g<\/li>\n<\/ol>\n

                Answer:
                \nSince, the gravitational intensity is defined as the negative of the gradient of gravitational potential i.e., \\(\\left(-\\frac{\\mathrm{dv}}{\\mathrm{dr}}\\right)\\) and the value of gravitational potential inside the spherical shell is constant, the value of gravitational intensity is zero.
                \n\"1st
                \nConsider any point p as shown, inside a sphere then the net gravitational intensity at that point is zero. If the upper half is cut off, the pull in the left and right side cancel each other just like the full sphere base but the pull upwards is missing. Thus the effective force is downwards along e.<\/p>\n

                Question 12.
                \nA rocket is fired from the earth towards the sun. At what distance from the earth\u2019s centre is the gravitational force on the rocket zero? Mass of the sun = 2 \u00d7 1030<\/sup> kg, mass of the earth = 6 \u00d7 1024<\/sup> kg. Neglect the effect of other planets etc. (orbital radius = 1.5 \u00d7 1011<\/sup> m).
                \nAnswer:
                \nLet the distance from the earth to the point where gravitational force is to zero be x
                \n\"1st
                \nGiven that,
                \nMass of sum = Ms<\/sub> = 2 \u00d7 1030<\/sup> kg
                \nMass of Earth = Me<\/sub> = 6 \u00d7 1024<\/sup>kg
                \nOrbital radius = r = 1.5 \u00d7 1011<\/sup> m
                \nNow,
                \nWe know that, gravitational force \u221d \\(\\frac{M}{R^{2}}\\)
                \nSince, gravitational force of Earth = Gravitational force of Sun
                \n\"1st
                \nx = 2.594 \u00d7 108<\/sup> m
                \nAt a distance of about 2.6 \u00d7 105<\/sup> km gravitational force is zero.<\/p>\n

                Question 13.
                \nHow will you \u2018weigh the sun\u2019, that is estimate its mass? The mean orbital radius of the earth around the sun is 1.5 \u00d7 108<\/sup> km.
                \nAnswer:
                \nGiven that
                \nOrbital radius, r = 1.5 \u00d7 108<\/sup> km
                \n= 1.5 \u00d7 1011<\/sup> m
                \nWe know that, Time for one revolution = 1 year
                \n= 365.25 \u00d7 24 \u00d7 60 \u00d7 60s
                \nNow,
                \nM = \\(\\frac{4 \\pi^{2} r^{3}}{G T^{2}}\\)
                \nG is universal gravitational constant.
                \n\u21d2 M = \\(\\frac{4 \\pi^{2} \\times\\left(1.5 \\times 10^{8}\\right)^{3}}{6.67 \\times 10^{-11} \\times\\left(3.156 \\times 10^{7}\\right)^{2}}\\)
                \n= 2.005 \u00d7 1030<\/sup> kg
                \n\u2248 2 \u00d7 1030<\/sup> kg
                \nThe estimate mass of sun is 2 \u00d7 1030<\/sup> kg<\/p>\n

                Question 14.
                \nA Saturn year is 29.5 times the earth year. How far is the Saturn from the sun if the earth is 1.50 \u00d7 108<\/sup> km away from the sun?
                \nAnswer:
                \nDistance of earth from sun is the Earth\u2019s orbital radius;
                \nre<\/sub> = 1.5 \u00d7 108<\/sup> km
                \n= 1.5 \u00d7 1011<\/sup> m
                \nTime period of Earth = Te<\/sub>
                \nTime period of Saturn = (29.5 \u00d7 Te<\/sub>)
                \nFrom Kepler\u2019s 3rd<\/sup> Law of planetary motion
                \nT2<\/sup> \u221d R3<\/sup>
                \n\u21d2 \\(\\frac{T_{s}^{2}}{T_{e}^{2}}=\\frac{r_{s}^{3}}{r_{e}^{2}}\\)
                \n\u21d2 rs<\/sub>3<\/sup> = \\(\\left(\\frac{T_{s}}{T_{e}}\\right)^{2}\\) re3<\/sup>
                \n= (29.5)2<\/sup> \u00d7 (1.5 \u00d7 1011<\/sup>)3<\/sup>
                \nrs<\/sub> = (29.5)2\/3<\/sup> \u00d7 1.5 \u00d7 1011<\/sup>
                \n= 1.432 \u00d7 1012<\/sup> m.<\/p>\n

                \"KSEEB<\/p>\n

                Question 15.
                \nA body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?
                \nAnswer:
                \nWeight of the body = 63 N = mg
                \nNow at a certain height h, let the acceleration due to gravity be gn.<\/sub>
                \nSo, New weight = mgn<\/sub>
                \nWe know that gn<\/sub> = \\(\\frac{g}{\\left(1+\\frac{h}{R_{e}}\\right)^{2}}\\)
                \nat a height of Re<\/sub>\/2
                \ngn<\/sub> = \\(\\frac{g}{\\left(1+\\frac{R_{e} \/ 2}{R_{e}}\\right)^{2}}\\)
                \n= \\(\\frac{g}{(3 \/ 2)^{2}}\\)
                \ngn<\/sub> = \\(\\frac{4}{9}\\) g
                \nNew weight = mgn<\/sub>= (mg) \u00d7 \\(\\frac{4}{9}\\)
                \n= 63 \u00d7 \\(\\frac{4}{9}\\)
                \n= 28 N<\/p>\n

                Question 16.
                \nAssuming the earth to be a sphere of uniform mass density, how much would a body weigh halfway down to the centre of the earth if it weighed 250 N on the surface?
                \nAnswer:
                \nWeight of body = 250 N
                \nNow, at a depth d, let acceleration due to gravity be gd<\/sub>,
                \n\"1st
                \ngd<\/sub> = g(1-1\/2)
                \n= g\/2
                \nNew weight = mgd<\/sub>
                \n= mg\/2
                \n= \\(\\frac{250 \\mathrm{N}}{2}\\)
                \n= 125 N<\/p>\n

                Question 17.
                \nA rocket is fired vertically with a speed of 5 km s-1<\/sup> from the earth\u2019s surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth = 6.0 \u00d7 1024<\/sup> kg; mean radius of the earth = 6.4 \u00d7 106<\/sup> m; G= 6.67 \u00d7 10-11<\/sup> N m2<\/sup> kg-2<\/sup>.
                \nAnswer:
                \nInitial velocity of the rocket,
                \nVi<\/sub> = 5 \u00d7 103<\/sup> m s-1<\/sup>
                \nMass of the Earth, M = 6.0 \u00d7 1024<\/sup> kg
                \nRadius of the Earth, R = 6.4 \u00d7 106<\/sup>\u00a0m
                \nLet the rocket be of mass, \u2018m\u2019 and reach an height \u2018h\u2019
                \nNow, The initial energy of the rocket is given by,
                \nTotal initial energy (TEi<\/sub>) = Initial Kinetic Energy (KEi<\/sub>) + Initial Potential Energy (PEi<\/sub>)
                \nKEi<\/sub> = \\(\\frac{1}{2}\\) \u00d7 m \u00d7 Vi<\/sub>2<\/sup>
                \n= \\(\\frac{m\\left(5 \\times 10^{3}\\right)^{2}}{2}\\) = 1.25 \u00d7 m \u00d7 107<\/sup>
                \nPEi<\/sub> = \\(-\\frac{G M m}{R}\\)
                \n\u21d2 TEi<\/sub> = 1.25 \u00d7 m \u00d7 107<\/sup> – \\(\\frac{\\mathrm{GM}}{\\mathrm{R}}\\) \u00d7 m
                \nAt the heighest point, the kinetic energy is 0. Total Energy at height \u2018h\u2019 (TEn<\/sub>) = PEn<\/sub>
                \n\u21d2 TEn<\/sub> = \\(-\\frac{\\mathrm{GM}}{(\\mathrm{R}+\\mathrm{h})}\\) \u00d7 m
                \nSince the total energy of system is conserved TEi<\/sub> = TEn<\/sub>
                \n\"1st
                \n\"1st
                \n0.2 (R+h) = h
                \n0.2 R = 0.8 h
                \nh = (R\/4) = 1.6 \u00d7 106<\/sup>m
                \nThe rocket goes 1.6 \u00d7 106<\/sup> m from the surface pf the Earth and (6.4 + 1.6) \u00d7 106<\/sup> = 8.0 \u00d7 106<\/sup> m from the center of the Earth.<\/p>\n

                Question 18.
                \nThe escape speed of a projectile on the earth\u2019s surface is 11.2 km s-1<\/sup>. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets.
                \nAnswer:
                \nEscape velocity of projectile = 11.2 km s-1<\/sup>
                \nVesc<\/sub> =1.12 \u00d7 104<\/sup> m s-1<\/sup>
                \nInitial velocity of the projectile,
                \nVi<\/sub> =3 Vesc<\/sub>
                \n= 3.36 \u00d7 104<\/sup> m s-1<\/sup>
                \nLet m be the mass of the projectile and Vf<\/sub> be the final velocity after it has escaped from the Earth\u2019s gravitational field.
                \nInitial Energy = KEi<\/sub> + PEi<\/sub>
                \nKEi<\/sub> = \\(\\frac{1}{2}\\) mVi<\/sub>2<\/sup>
                \nBy definition of Vesc<\/sub>, PEi<\/sub> = \\(-\\frac{1}{2} m V_{e s c}^{2}\\)
                \nFinal Energy is = \\(\\frac{1}{2} m V_{f}^{2}\\)
                \nSo, By conservation of energy
                \n\\(\\left(-\\frac{1}{2} m V_{e s c}^{2}\\right)+\\left(\\frac{1}{2} m V_{1}^{2}\\right)=\\left(\\frac{1}{2} m V_{f}^{2}\\right)\\)
                \n\u21d2 Vf<\/sub>2<\/sup> = Vi<\/sub>2<\/sup> – V2<\/sup>esc<\/sub>
                \n\"1st
                \n\u2248 3.17 km s-1<\/sup><\/p>\n

                Question 19.
                \nA satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth\u2019s gravitational influence? Mass of the satellite = 200 kg; mass of the earth = 6.0 \u00d7 1024<\/sup> kg; radius of the earth = 6.4 \u00d7 106<\/sup> m; G = 6.67 \u00d7 10-11<\/sup> N m2<\/sup> kg-2<\/sup>.
                \nAnswer:
                \nMass of the satellite = 200 kg.
                \nMass of the Earth = 6.0 \u00d7 1024<\/sup> kg
                \nRadius of Earth = 6.4 \u00d7 106<\/sup> m
                \nHeight of the satellite = 400 km
                \nWe knew that,
                \nOrbital velocity of the satellite,
                \nVo<\/sub> = \\(\\sqrt{\\frac{\\mathrm{GM}}{(\\mathrm{R}+\\mathrm{h})}}\\)
                \nSo, Initial KE = \\(\\frac{1}{2} m V_{0}^{2}\\)
                \n\"1st
                \nNow we need to apply (- TEi<\/sub>) energy to force it out of the Earth\u2019s influence.
                \nSo, Energy needed = \\(\\frac{\\mathrm{GMm}}{2(\\mathrm{R}+\\mathrm{h})}\\)
                \n\"1st
                \n= 5.9 \u00d7 109<\/sup> J<\/p>\n

                Question 20.
                \nTwo stars each of one solar mass (2 \u00d7 1030<\/sup> kg) are approaching each other for a head-on collision. When they are a distance 109<\/sup> km, their speeds are negligible. What is the speed with which they collide? The radius of each star is 104<\/sup> km. Assume the stars to remain undistorted until they collide. (Use the known value of G).
                \nAnswer:
                \nMass of the Stars, M = 2 \u00d7 1030<\/sup>kg
                \nRadius of the Stars, R = 104<\/sup> km
                \n= 107<\/sup> m
                \nDistance between Stars, d = 109<\/sup> km
                \n= 1012<\/sup> m
                \nInitial velocity almost 0, So KEi<\/sub> of each is 0. So, Initial potential energy
                \n\\(=-\\frac{\\mathrm{GMM}}{(\\mathrm{d}+2 \\mathrm{R})}=-\\frac{\\mathrm{GMM}}{\\mathrm{d}}\\)
                \n[\u2235 d>>>R]
                \nSo, Initial Total Energy TEi<\/sub> = \\(-\\frac{\\mathrm{GM}}{\\mathrm{d}}\\)
                \nWhen the two Stars are about to collide let their velocity be V and separaion is 2R.
                \nSo, TE
                \n\"1st
                \nBy conservation of energy;
                \n\"1st
                \n\u2248 \\(\\sqrt{6.67 \\times 10^{12}}\\)
                \nv = \\(\\sqrt{6.67}\\) \u00d7 106<\/sup> m s-1<\/sup><\/p>\n

                Question 21.
                \nTwo heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the midpoint of the line joining the centres of the spheres ? is an object placed at that point in equilibrium? if so, is the equilibrium stable or unstable?
                \nAnswer:
                \nMass of each sphere = 100 kg
                \nRadius of each sphere = 0.1 m
                \n\"1st
                \nAt the midpoint M, the gravitational pull by each sphere is exactly the same since they are identical spheres.
                \nSo, The gravitational force is zero.
                \nThe potential at M = – \\(\\frac{\\mathrm{GM}}{(0.5)}-\\frac{\\mathrm{GM}}{(0.5)}\\)
                \n= – 4GM
                \n= – 4 \u00d7 6.67 \u00d7 10-11<\/sup> \u00d7 100
                \n= – 2.668 \u00d7 10-8<\/sup> J\/kg
                \nSo, the potential at \u2018M\u2019 is – 2.67 \u00d7 10-8<\/sup> J\/kg
                \nAt object at M will be in equilibrium. Since mt force on it is zero. The equilibrium is unstable as a slight ctange in position of the mass, the net foce becomes non zero and it will move.<\/p>\n

                \"KSEEB<\/p>\n

                Question 22.
                \nIt’s you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of the earth. What is the potential due to earth\u2019s gravity at the sit* of this satellite? (Take the potential energy at infinity to be zero). Mas of the earth = 6.0 \u00d7 1024<\/sup> kg, radius = 6400 km.
                \nAnswer:
                \nMass of the Earth = 6.0\u00d7 1024<\/sup> kg
                \nRadius of Earth = 6.4 \u00d7 106<\/sup>m.
                \nHeight of the satellite from surface of Earth is 36000 km = 3.6 \u00d7 107<\/sup>m
                \nDistance of the satellite from centre of Earth is d = (3.6 \u00d7 107<\/sup>m + 6.4 \u00d7 106<\/sup> m)
                \n= 4.24 \u00d7 107<\/sup> m
                \nGravitational potential due to Earth on the satellite is v = \\(-\\frac{G M}{d}\\)
                \n\\(-\\frac{6.67 \\times 10^{-11} \\times 6 \\times 10^{24}}{4.24 \\times 10^{7}}\\)
                \n= – 9.438 \u00d7 107<\/sup> J\/kg
                \n= – 9.4 \u00d7 107<\/sup>J\/kg.<\/p>\n

                Question 23.
                \nA star 2.5 times the mass of the sun and collapsed to a size of 42 km rotates with a speed of 1.2 rev. per second. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objects called pulsars belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity? (mass of the sun = 2 \u00d7 1030<\/sup> kg).
                \nAnswer:
                \nSince the Star is rotating, there is a force pushing it outward (centrifugal force) and a force pulling it inwards, (gravitational force).
                \nIf gravitational force (FG<\/sub>) > Centrifugal force (Fc<\/sub>)
                \nThen the object on the equator remains stuck on the star.
                \nMass of Star, M = 2.5 (MSun<\/sub>)
                \n= 2.5 \u00d7 2 \u00d7 1030<\/sup> kg
                \n= 5 \u00d7 1030<\/sup> kg
                \nRadius of Star = 12 km = 1.2 \u00d7 104<\/sup> m
                \n\u21d2 FG<\/sub> = \\(\\frac{\\mathrm{GMm}}{\\mathrm{R}^{2}}=\\frac{6.67 \\times 10^{-11} \\times 5 \\times 10^{30}}{\\left(1.2 \\times 10^{4}\\right)^{2}}\\)
                \n= \\(\\frac{6.67 \\times 5}{1.44}\\) \u00d7 1011<\/sup> \u00d7 m
                \n= 2.316 \u00d7 1012<\/sup> \u00d7 m N
                \nFc<\/sub> = mr \u03c92<\/sup>
                \nWhere, r = radius of Star = 12 km
                \n\u03c9 = angular speed = 2 \u00d7 \u03c0 \u00d7 (1.2)
                \n\u21d2 Fc<\/sub> = m \u00d7 1.2 \u00d7 104<\/sup> \u00d7 (2.4\u03c0)2<\/sup>
                \n= 6.82 \u00d7 105<\/sup> \u00d7 m N
                \nSince Fc<\/sub> < FG<\/sub> the object is stuck to Star\u2019s surface.<\/p>\n

                Question 24.
                \nA spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system? Mass of the space ship = 1000 kg; mass of the sun = 2 \u00d7 1030<\/sup> kg; mass of mars = 6.4 \u00d7 1023<\/sup> kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28 \u00d7 108<\/sup> km; G = 6.6 7 \u00d7 10-11<\/sup> N m2<\/sup> kg-2<\/sup>.
                \nAnswer:
                \nMass of Spaceship, ms<\/sub> = 1000 kg
                \nMass of mars Ms<\/sub> = 6.4 \u00d7 1023<\/sup> kg;
                \nRadius of mars rm<\/sub> = 3.395\u00d7 106<\/sup> m
                \nRadius of the orbit of Mars
                \n= 2.28 \u00d7 108<\/sup> km = 2.28 \u00d7 1011<\/sup> m
                \nInitial velocity of spaceship is = 0
                \nSo, the Total Initial Energy is only potential energy.
                \n\"1st
                \n= – 6.67 \u00d7 10-8<\/sup>[1.88 \u00d7 1017<\/sup> + 8.772 \u00d7 1018<\/sup>]
                \n= – 6.67 \u00d7 8.96 \u00d7 1010<\/sup>
                \n= – 5.976 \u00d7 1011<\/sup> = – 6 \u00d7 1011<\/sup> J<\/p>\n

                Question 25.
                \nA rocket is fired \u2018vertically\u2019 from the surface of mars with a speed of 2 km s-1<\/sup>. If 20% of its initial energy is lost due to martian atmospheric resistance, how far will the rocket go from the surface of Mars before returning to it? Mass of mars = 6.4 \u00d7 1023 <\/sup>kg; radius of mars = 3395 km; G = 6.67 \u00d7 10-11<\/sup> Nm2<\/sup>kg-2<\/sup>.
                \nAnswer:
                \nMass of Mars = 6.4 \u00d7 1023<\/sup>kg
                \nRadius of Mars, R = 3.395 \u00d7 106<\/sup> m.
                \nInitial velocity of the rocket = 2 km s-1<\/sup>
                \n= 2 \u00d7 103<\/sup> ms-1<\/sup>
                \nLet the mass of the rocket be m, then the initial kinetic energy is, \\(\\frac{1}{2}\\) mv2<\/sup>
                \n= \\(\\frac{1}{2}\\) \u00d7 m \u00d7 (2 \u00d7 103)2<\/sup>
                \n= 2 m \u00d7 106<\/sup> J
                \nSince 20% of the energy is lost due to atmost resistance, only 80% of energy is to be considered. Now,
                \n\"1st
                \nLet the rocket reach a height ‘h\u2019, at height h the velocity is 0. So, KEf<\/sub> = 0
                \nand PEf<\/sub> = \\(-\\frac{\\mathrm{GMm}}{(\\mathrm{R}+\\mathrm{h})}\\)
                \nTEf<\/sub> = \\(-\\frac{\\mathrm{GMm}}{\\mathrm{R}+\\mathrm{h}}\\) ….(2)
                \nBy conservation of Energy , (1) = (2)
                \n\"1st
                \n\"1st
                \n0.1872 (R + h)= h
                \n0.1272 R = 0.878 h
                \n\u21d2 h = \\(\\frac{0.1272 \\times 3.395}{0.8728}\\) \u00d7 106<\/sup>
                \nh = 0.495 \u00d7 106<\/sup> m
                \nh = 495 km
                \nSo, the satellite reaches a height of 495 km.<\/p>\n

                1st PUC Physics Gravitation One Mark Questions and Answers<\/h3>\n

                Question 1.
                \nWhat is acceleration due to gravity?
                \nAnswer:
                \nThe acceleration produced on a body falling freely under gravity is known as acceleration due to gravity.<\/p>\n

                Question 2.
                \nWrite the dimensional formula of G
                \nAnswer:
                \nM-1<\/sup>L3<\/sup>T-2<\/sup><\/p>\n

                \"KSEEB<\/p>\n

                Question 3.
                \nWhat is the unit of G?
                \nAnswer:
                \nNm2<\/sup>kg-2<\/sup><\/p>\n

                Question 4.
                \nWhat are the factors affecting the value of G?
                \nAnswer:
                \nIt is a universal constant and is not affected by any factors.<\/p>\n

                Question 5.
                \nGive the SI unit and dimensional formula of \u2018g\u2019.
                \nAnswer:
                \nSI unit – m s-2<\/sup>, Dimensional formula LT-2<\/sup><\/p>\n

                Question 6.
                \nWrite the expression for acceleration due to gravity at a height (altitude) \u2018h\u2019.
                \nAnswer:
                \nThe acceleration due to gravity at a height \u2018h\u2019 from the surface of earth is given by, g1<\/sup>= g \\(\\left[1-\\frac{2 h}{R}\\right]\\) where g is the acceleration due to gravity on the surface of earth and R is the radius of the earth.<\/p>\n

                Question 7.
                \nHow does g vary with height above the surface of earth?
                \nAnswer:
                \ng decreases with an increase in height.<\/p>\n

                Question 8.
                \nWrite the expression for acceleration due to gravity at a depth h below Earth\u2019s surface.
                \nAnswer:
                \n\\(g^{\\prime}=g\\left[1-\\frac{d}{R}\\right]\\), where g is the acceleration due to gravity and R is the radius of the Earth.<\/p>\n

                \"KSEEB<\/p>\n

                Question 9.
                \nWhere is the acceleration due to gravity on earth maximum?
                \nAnswer:
                \nAcceleration due to gravity is maximum at the poles.<\/p>\n

                Question 10.
                \nWhere is acceleration due to gravity on earth minimum?
                \nAnswer:
                \nAcceleration due to gravity is minimum at the equator.<\/p>\n

                Question 11.
                \nWhat is the orbital velocity of a satellite?
                \nAnswer:
                \nThe velocity with which a satellite revolves in its closed orbit is called orbital velocity.<\/p>\n

                Question 12.
                \nWrite the expression for the orbital velocity of the satellite.
                \nAnswer:
                \nThe expression for orbital velocity of a satellite revolving round the earth in a closed orbit of radius \u2018r\u2019 is given by,
                \nv = \\(\\sqrt{\\frac{\\mathrm{GM}}{\\mathrm{r}}}\\)
                \nwhere G is the gravitational constant and M is the mass of the earth.<\/p>\n

                Question 13.
                \nWhat is the escape velocity of a body?
                \nAnswer:
                \nThe minimum velocity with which a body must be projected such that it escapes from the earth\u2019s gravitational pull is called escape velocity.<\/p>\n

                Question 14.
                \nGive the expression for the escape velocity.
                \nAnswer:
                \nEscape velocity is given by,
                \nVe<\/sub> = \\(\\sqrt{\\frac{2 G M}{R}}\\)
                \nwhere G is the gravitational constant, M is the mass of the earth and R is the radius of the earth.<\/p>\n

                \"KSEEB<\/p>\n

                Question 15.
                \nWhat is a satellite?
                \nAnswer:
                \nSatellite is an object moving round a planet in a closed and stable orbit.<\/p>\n

                Question 16.
                \nWhat is a geostationary satellite?
                \nAnswer:
                \nA satellite which revolves round the earth with the same angular speed and the same direction as that of earth around its axis is a geostationary satellite.<\/p>\n

                Question 17.
                \nWhat is the period of a geostationary satellite?
                \nAnswer:
                \n24 hours<\/p>\n

                Question 18.
                \nWhat is meant by remote sensing?
                \nAnswer:
                \nObtaining data about a material from a distance using sensors is called remote sensing.<\/p>\n

                Question 19.
                \nMention the relation between escape velocity and Orbiting velocity.
                \nAnswer:
                \nEscape velocity ve<\/sub> = \\(\\sqrt{2}\\) vo<\/sub><\/p>\n

                Question 20.
                \nAssume a hole is made through the centre of the Earth along the diameter. What type of motion does the mass execute if it is dropped into the hole?
                \nAnswer:
                \nSimple Harmonic Motion.<\/p>\n

                Question 21.
                \nIs the potential energy of an object on the surface of Earth zero?
                \nAnswer:
                \nNo, It is given by \\(-\\frac{\\mathrm{GM}}{\\mathrm{R}}\\).<\/p>\n

                Question 22.
                \nWhat is basis for Kepler\u2019s 2nd<\/sup> Law of Planetary Motion?
                \nAnswer:
                \nThe law of conservation of angular momentum.<\/p>\n

                \"KSEEB<\/p>\n

                Question 23.
                \nIf the Earth stops rotating axis. What will happen to the value of
                \nAnswer:
                \nSince the Earth stopped rotation the centrifugal force pushing the objects outside becomes zero. So the value of g increases.<\/p>\n

                Question 24
                \nWrite one most important application of geostationary satellite.
                \nAnswer:
                \nLong-distance communication.<\/p>\n

                Question 25.
                \nWhat is the effect of Non-sphericity of Earth on the value of \u2018g\u2019?
                \nAnswer:
                \nThe value of g keeps varying with different places on Earth. It is maximum at poles and minimum at the equator.<\/p>\n

                Question 26.
                \nWhat is the condition for a uniform sphere to be black hole?
                \nAnswer:
                \nCondition is, \\(\\sqrt{\\frac{2 \\mathrm{GM}}{\\mathrm{R}}}\\) = c,
                \nWhere M is mass of the sphere,
                \nR is radius of sphere,
                \nC is velocity of light.<\/p>\n

                Question 27.
                \nHow does the Orbital velocity of a satellite depends on the mass of the satellite?
                \nAnswer:
                \nIt is Independent of mass of satellite.<\/p>\n

                Question 28.
                \nWhat does a spring balance measure mass or weight?
                \nAnswer:
                \nWeight.<\/p>\n

                Question 29.
                \nThe Gravitational Potential Energy of a body at a point in a gravitational field of another body is \\(-\\frac{G M m}{r}\\) What does -ve sign show?
                \nAnswer:
                \nThe negative sign implies that the body is bound to the body of mass M.<\/p>\n

                1st PUC Physics Gravitation Two Marks Questions and Answers<\/h3>\n

                Question 1.
                \nDefine the potential energy of a body. Give an expression for the gravitational potential energy of a body.
                \nAnswer:
                \nThe energy possessed by a body due to its position is called potential energy. Gravitational potential energy is = mg h.<\/p>\n

                Question 2.
                \nState and explain Newton\u2019s law of gravitation.
                \nAnswer:
                \nEverybody in this universe attracts every other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. If m1<\/sub> and m2<\/sub> are the masses of two bodies separated by a distance \u2018r\u2019, then the gravitational force F between them is,
                \n\\(\\mathrm{F} \\propto \\frac{\\mathrm{m}_{1} \\mathrm{m}_{2}}{\\mathrm{r}^{2}}\\) or
                \n\\(\\mathrm{F}=\\frac{\\mathrm{Gm}_{1} \\mathrm{m}_{2}}{\\mathrm{r}^{2}}\\)
                \nWhere \u2018G\u2019 is.called the gravitational constant.<\/p>\n

                Question 3.
                \nDefine universal gravitational constant G. What is the dimensional formula of G?
                \nAnswer:
                \nGravitational constant is numerically equal to the force of attraction between two bodies of unit masses separated by unit distance.
                \nDimensional formula of G is [M-1<\/sup>L3<\/sup> T-2<\/sup>]<\/p>\n

                Question 4.
                \nDerive the relation between acceleration due to gravity (g) and the gravitational constant (G)
                \nAnswer:
                \nConsider a body of mass \u2018m\u2019 on the surface of the earth of mass M. Let R be the radius of the Earth. The gravitational force exerted on the body according to Newton\u2019s second law is,
                \nF1<\/sub> = mg ……(1)
                \nwhere \u2018g\u2019 is acceleration due to gravity. According to Newton\u2019s law of gravitation, the force between the body close to the surface of the earth and the earth is,
                \nF2<\/sub> = G. \\(\\frac{m M}{R^{2}}\\) ………(2)
                \nBut F1<\/sub> = F2<\/sub>
                \n\u2234 mg = \\(\\frac{G m M}{R^{2}}\\)
                \ng = \\(\\frac{G M}{R^{2}}\\).<\/p>\n

                \"KSEEB<\/p>\n

                Question 5.
                \nExplain the variation of \u2018g\u2019 due to the shape of the earth?
                \nAnswer:
                \nOn the surface of the earth g = \\(\\frac{G M}{R^{2}}\\)
                \nSince G and M are constants g \\(\\propto \\frac{1}{R^{2}}\\)
                \nDue to the ellipsoidal shape of the earth equatorial radius is greater than the polar radius. So \u2018g\u2019 is maximum at the poles than at the equator.<\/p>\n

                Question 6.
                \nMention the factors on which the orbital velocity and escape velocity depends.
                \nAnswer:
                \nThe orbital velocity and escape velocity depends on mass of the planet and its radius.<\/p>\n

                Question 7.
                \nState and explain Kepler\u2019s law for planetary motion.
                \nAnswer:
                \nI Law :
                \nEvery planet moves in an elliptical orbit around the sun with sun at one of the focus.<\/p>\n

                II Law:
                \nThe line joining the planet to the sun sweeps out equal areas in equal interval of time.<\/p>\n

                III Law:
                \nThe square of the period of any planet about the sun is proportional to the cube of the semi-major axis of the ellipse.<\/p>\n

                Question 8.
                \nExplain what is meant by weightlessness?
                \nAnswer:
                \nThe weight of a body is measured by the reaction of the surface which supports the body. When a person is standing on the lift the person the presses the surface with a force equal to his weight. The equal and opposite reaction of the surface on the person gives the sensation of weight.
                \nWhen the lifts starts moving downwards with an acceleration, the reaction on the person becomes less and he feels his weight to be less.<\/p>\n

                When the lifts falls freely, both the person and the lift are moving downwards with equal acceleration. Thus the surface does not exert any force on the body. Such a state is called weightlessness. Similarly, the astronauts in a spacecraft orbiting round the earth experience weightlessness.<\/p>\n

                \"KSEEB<\/p>\n

                Question 9.
                \nGive any two applications of remote sensing.
                \nAnswer:<\/p>\n

                  \n
                1. remote sensing is used for mapping of forests.<\/li>\n
                2. remote sensing is used for groundwater exploration.<\/li>\n<\/ol>\n

                  Question 10.
                  \nHow is a satellite launched?
                  \nAnswer:
                  \nIn order to launch a satellite, it is taken vertically upwards to the required orbit and then appropriate horizontal velocity is imparted to the satellite so that it revolves round the earth.<\/p>\n

                  Question 11.
                  \nDistinguish between inertial mass and gravitational mass.
                  \nAnswer:
                  \nInertial mass of a body is a measure of its inertia. It is determined using Newton\u2019s second law of motion. Gravitational mass is a measure of the gravitational pull acting on it. It is determined using Newton\u2019s law of gravitation.<\/p>\n

                  Question 12.
                  \nFind the expression for the weight of the body at centre of the Earth.
                  \nAnswer:
                  \nWe know that, the value of acceleration due to gravity at depth d is given by \\(\\mathrm{g}_{\\mathrm{d}}=\\mathrm{g}\\left[1-\\frac{\\mathrm{d}}{\\mathrm{R}}\\right]\\) where R is radius of Earth.
                  \nAt depth = R, gd<\/sub> = g[1 – R\/R] = 0
                  \nSo, weight of the body at centre of Earth is 0.<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 13.
                  \nIs it possible to keep a satellite, so that it is always over Kashmir? Why?
                  \nAnswer:
                  \nNo, It is possible to place a geostationary satellite on an equatorial plane. Since Kashmir is not on the equator it is not possible to place a satellite above it.<\/p>\n

                  Question 14.
                  \nFind the height at which g is the same as that at depth R\/2.
                  \nAnswer:
                  \nThe acceleration due to- gravity at depth R\/2 is
                  \ngd<\/sub> = g[1 – d\/R] = g \\(\\left[1-\\frac{\\mathrm{R} \/ 2}{\\mathrm{R}}\\right]\\)
                  \n= g\/2
                  \nAt height h
                  \n\"1st
                  \nh = R\/4<\/p>\n

                  Question 15.
                  \nThe mass of 2 bodies are tripled and distance is halved, how will the gravitational force change?
                  \nAnswer:
                  \nWe know that Fg<\/sub> = \\(\\frac{\\mathrm{Gm}_{1} \\mathrm{m}_{2}}{\\mathrm{r}^{2}}\\)
                  \nNow, m1<\/sub> and m2<\/sub> are made 3 m1<\/sub> and 3 m2<\/sub> respectively and r is made (r\/2).
                  \nG (3m1)(3m2)
                  \nSo, \\(F_{g}^{\\prime}=\\frac{G\\left(3 m_{1}\\right)\\left(3 m_{2}\\right)}{(r \/ 2)^{2}}\\) = 36 Fg<\/sub>
                  \nGravitational force increases 36 times.<\/p>\n

                  Question 16.
                  \nDefine gravitational field strength. Find the field at a point distance \u2018x\u2019 form a mass ‘m1<\/sub>‘.
                  \nAnswer:
                  \nThe gravitational field is defined as the gravitation force experienced by an object of unit mass at any point. The Gravitational force on an object of mass m due to mass \u2018m1<\/sub>‘ at a distance x.
                  \nF = \\(\\frac{\\mathrm{Gm}_{1} \\mathrm{m}}{\\mathrm{x}^{2}}\\)
                  \nField = \\(\\frac{F}{m}=\\frac{G m_{1}}{x^{2}}\\)<\/p>\n

                  Question 17.
                  \nA planet\u2019s radius is reduced by 5%, with its mass unchanged. What is the percentage change in \u2018g\u2019?
                  \nAnswer:
                  \nWe know that g \\(\\propto \\frac{1}{R^{2}}\\)
                  \nSo, \\(\\frac{\\Delta g}{g}\\) = – 2 \\(\\frac{\\Delta R}{R}\\)
                  \n\u21d2 \\(\\frac{\\Delta g}{g}\\) = – 2(5%) = – 10%<\/p>\n

                  Question 18.
                  \nWhat is the gravitational force on a body inside a spherical shell? Why?
                  \nAnswer:
                  \nInside the shell, the net gravitational force is zero. This is because there is no mass inside, the gravitational field is zero, thereby force is zero.<\/p>\n

                  Question 19.
                  \nThe value of \u2018g\u2019 on the moon is 1\/6th<\/sup> of that of Earth and diametre of moon is 1\/4th<\/sup> that of Earth. Find the ratio of
                  \nescape velocities.
                  \nAnswer:
                  \nWe know that,
                  \n\"1st
                  \nThe escape velocity of moon is about 4.9 times that of Earth.<\/p>\n

                  Question 20.
                  \nState universal law of gravitation. Establish the relation Me<\/sub> = \\(\\frac{\\mathrm{g} \\mathrm{R}_{\\mathrm{e}}^{2}}{\\mathrm{G}}\\), where Me<\/sub> and Re<\/sub> are the mass and radius of Earth respectively.
                  \nAnswer:
                  \nThe universal law of gravitation states that everybody in the universe attracts every other body with a force which is directly proportional to their masses and inversely proportional to the square of distance between them.
                  \nF \\(\\propto \\frac{m_{1} m_{2}}{r^{2}}\\)
                  \n\u21d2 F = \\(\\frac{\\mathrm{Gm}_{1} \\mathrm{m}_{2}}{\\mathrm{r}^{2}}\\)
                  \nFor the body of mass \u2018m\u2019 on earth.
                  \nF = mg = \\(\\frac{\\mathrm{GM}_{\\mathrm{e}} \\mathrm{m}}{\\mathrm{R}_{\\mathrm{e}}^{2}}\\)
                  \n\u21d2 Me<\/sub> = \\(\\frac{R_{e}^{2} g}{G}\\)<\/p>\n

                  Question 21.
                  \nThe radii of 2 planets are R and 2R respectively and their elements e and e\/2 respectively. What is the ratio of their acceleration due to gravity at their surface?
                  \nAnswer:
                  \nWe know that,
                  \ng \u221d M and g \u221d 1\/R2<\/sup>
                  \n\u21d2 g \u221d \\(\\frac{\\rho \\mathrm{V}}{\\mathrm{R}^{2}}\\) (\u2235 M = \u03c1V)
                  \nSince V \u221d R3<\/sup>
                  \ng \u221d \\(\\frac{\\rho \\mathrm{R}^{3}}{\\mathrm{R}^{2}}\\)
                  \ni.e.; g \u221d \u03c1R
                  \nGiven that, \u03c11<\/sub> = \u03c1
                  \n\u03c12<\/sub> = \u03c1\/2
                  \nR1<\/sub> = R
                  \nR1<\/sub> = 2R
                  \n\u21d2 \\(\\frac{\\mathrm{g}_{1}}{\\mathrm{g}_{2}}=\\frac{\\rho_{1} \\mathrm{R}_{1}}{\\rho_{2} \\mathrm{R}_{2}}=\\frac{\\rho \\mathrm{R}}{(\\rho \/ 2)(2 \\mathrm{R})}=1\\).<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 22.
                  \nThe change In the value of \u2018g\u2019 at a height \u2018h\u2019 and depth \u2018d\u2019 is same. Assuming h and d are both very small compared to radius of Earth. Find ratio of h with respect to d.
                  \nAnswer:
                  \nWe know that \u2018g\u2019 at a height \u2018h\u2019 is given by \\(g^{\\prime}=g\\left[1-\\frac{2h}{R}\\right]\\) …….(1)
                  \ng at a depth \u2018d\u2019 is given by,
                  \n\\(g^{\\prime}=g\\left[1-\\frac{d}{R}\\right]\\) ……..(2)
                  \nComparing (1) and (2), we can write
                  \n2h = d
                  \n\u21d2 \\(\\frac{h}{d}=\\frac{1}{2}\\)<\/p>\n

                  Question 23.
                  \nIf the Earth\u2019s mass is 9 times and radius is 4 times that of Mars. Find the escape velocity of Mars. If that of Earth is 11.2 km s-1<\/sup>
                  \nAnswer:
                  \nWe know that, escape velocity is given by,
                  \nVesc<\/sub> = \\(\\sqrt{\\frac{2 \\mathrm{GM}}{\\mathrm{R}}}\\)
                  \nLet Vm<\/sub>, Ve<\/sub> be escape velocity of mars and Earth then, Vm<\/sub> = \\(\\sqrt{\\frac{2 \\mathrm{GM}_{\\mathrm{m}}}{\\mathrm{R}_{\\mathrm{m}}}}\\)
                  \n\\(\\sqrt{\\frac{2 \\mathrm{G}\\left(\\mathrm{M}_{\\mathrm{e}} \/ 9\\right)}{\\left(\\mathrm{R}_{\\mathrm{e}} \/ 4\\right)}}\\)
                  \n\"1st
                  \n= 7.47 kms-1<\/sup><\/p>\n

                  Question 24.
                  \nA black hole is a body from whose surface nothing may ever escape. What is the condition for a uniform spherical mass M to be a black hole? What should be radius of earth if it has to be a black hole?
                  \nAnswer:
                  \nFor a body to be a black hole, even light should not escape. So limiting escape velocity is 3 \u00d7 108<\/sup> ms-1<\/sup>
                  \nSo, for body of Mass M, condition is,
                  \n\\(\\sqrt{\\frac{2 \\mathrm{GM}}{\\mathrm{R}}}\\) \u2265 3 \u00d7 108<\/sup> ms-1<\/sup>
                  \nFor Earth, M = 6 \u00d7 1024<\/sup> kg
                  \n\\(\\frac{2 \\times 6.67 \\times 10^{-11} \\times 6 \\times 10^{24}}{R}\\) = (3 \u00d7 108<\/sup>)2<\/sup> R
                  \n= 9 \u00d7 10-2<\/sup> m
                  \nR = 9 cm.<\/p>\n

                  1st PUC Physics Gravitation Three Marks Questions and Answers<\/h3>\n

                  Question 1.
                  \nThe mass and diameter of a planet are 4 times those of Earth. What will be the time period of oscillation of the seconds pendulum on that planet?
                  \nAnswer:
                  \nWe know that the time period of oscillation of the pendulum is given by,
                  \n\\(\\mathrm{T}=\\sqrt[2 \\pi]{\\frac{1}{g}}\\)
                  \nLet the value of \u2018g\u2019 on the planet be gp<\/sub> and on earth be ge<\/sub>.
                  \nWe know that, g \u221d \\(\\frac{M}{R^{2}}\\)
                  \n\"1st
                  \n\"1st
                  \n\u21d2 Tp<\/sub> = 2 Ts<\/sub>
                  \nSince it is a seconds pendulum Te<\/sub> = 1s
                  \n\u21d2 Tp<\/sub> = 2s.<\/p>\n

                  Question 2.
                  \nThree equal masses are placed on three corners of equilateral triangles. What is work done in doing this?
                  \nAnswer:
                  \nWhen ma<\/sub> is brought no work is done when mb<\/sub> is brought work done is, \\(-\\frac{G m_{a} m_{b}}{x}\\)
                  \n\"1st
                  \nWhen mc<\/sub> is brought to point C from infinity.
                  \nThe work done is \\(-\\frac{G m_{a} m_{c}}{x}\\)\\(-\\frac{G m_{b} m_{c}}{x}\\)
                  \nTotal work done
                  \n\"1st
                  \nSince all masses are equal, total work done is = \\(-\\frac{3 G m^{2}}{x}\\)<\/p>\n

                  Question 3.
                  \nShow that the gravitational potential at point of distant \u2018a’ from the mass M1<\/sub> is given by, V = \\(-\\frac{\\mathrm{GM}_{1}}{\\mathrm{a}}\\).
                  \nAnswer:
                  \nConsider a mass \u2018m\u2019 placed at a distance a from a mass M, then force is given by,
                  \n\\(\\mathrm{F}=\\frac{\\mathrm{GmM}_{1}}{\\mathrm{a}^{2}}\\)
                  \nLet the work done in moving the body through a distance dx, if it originally at a distance x, the work done be dw.
                  \ndw = F.dx
                  \n\"1st
                  \nThe potential energy is the amount of work done in bringing the object from \u221e to point \u2018a\u2019. PE, U = \\(-\\frac{\\mathrm{GM}_{1} \\mathrm{m}}{\\mathrm{a}}\\)
                  \nGravitational potential is PE per unit mass
                  \n\u21d2 V = \\(\\frac{\\mathrm{u}}{\\mathrm{m}}=-\\frac{\\mathrm{GM}_{1}}{\\mathrm{a}}\\)<\/p>\n

                  Question 4.
                  \nWhat happens to a body when it’s projected up with a velocity of 11.2 km s-1<\/sup>? Why? Compare escape speeds of 2 planets of mass M and 4 M and radii 2 R and R respectively.
                  \nAnswer:
                  \nWhen the object is throw up with 11.2 km s-1<\/sup>, which is equal to the escape velocity. The object doesn\u2019t come back and escapes from Earth\u2019s gravitational field. The escape velocity is given by,
                  \n\"1st
                  \n\\(\\frac{V_{\\mathrm{esc}_{1}}}{V_{\\mathrm{esc}_{2}}}=\\frac{1}{2 \\sqrt{2}}\\)
                  \nThe escape velocity of the 2nd<\/sup> planet is greater by a factor of 2\\(\\sqrt{2}\\).<\/p>\n

                  Question 5.
                  \nWhat are geostationary satellite? Calculate the height of the orbit above surface of Earth in which a satellite, if placed, will appear stationary.
                  \nAnswer:
                  \nA geostationary satellite is a satellite which revolves around the Earth with exact same angular speed and direction as the Earth. Thus it appears stationary from Earth.
                  \nT = 24 \u00d7 3600 s = 8.64 \u00d7 104<\/sup>s
                  \nLet h be the height of orbit. Then,
                  \n\"1st
                  \n\u2245 (4.24 \u00d7 107<\/sup>) – (6.4 \u00d7 106<\/sup>)m
                  \n\u2248 3.6 \u00d7 107<\/sup> m or 36,000 km from the surface of Earth.<\/p>\n

                  Question 6.
                  \nExplain the variation of \u2018g\u2019 with<\/p>\n

                    \n
                  1. height<\/li>\n
                  2. depth<\/li>\n<\/ol>\n

                    Answer:
                    \nConsider M to be mass of Earth and R to be radius of Earth.
                    \n1. At a point \u2018h\u2019 above the ground
                    \nNew g is gn<\/sub> = \\(\\frac{\\mathrm{GM}}{(\\mathrm{R}+\\mathrm{h})^{2}}\\)
                    \n\\(=\\frac{\\mathrm{GM}}{\\mathrm{R}^{2}\\left[1+\\frac{\\mathrm{h}}{\\mathrm{R}}\\right]^{2}}\\)
                    \ngn<\/sub> = g[1 + \\(\\frac{h}{R}\\)]-2<\/sup>
                    \nif h << R,
                    \ngn<\/sub> = g[1 – \\(\\frac{2h}{R}\\)]
                    \nSo, g decreases with increase in height,<\/p>\n

                    2. At a depth \u2018d\u2019 below the ground. The new g, gd is given by
                    \ndd<\/sub> = g[1 – \\(\\frac{d}{R}\\)]
                    \nSo, g also decreases with depth.<\/p>\n

                    \"KSEEB<\/p>\n

                    Question 7.
                    \nIf the earth is one half of its present distance from the sun, How many days will the present one year on the surface will change?
                    \nAnswer:
                    \nWe know that, T2<\/sup> \u221d R3<\/sup>
                    \n\\(\\left(\\frac{T_{1}}{T_{2}}\\right)^{2}=\\left(\\frac{R_{1}}{R_{2}}\\right)^{3}\\)
                    \nSince, R is halved
                    \nT1<\/sub> = (1\/2)3\/2<\/sup>T
                    \n= \\(\\frac{T}{2 \\sqrt{2}}\\)
                    \nT1<\/sub> = 129 days
                    \nThe year of the new Earth will have 236 days lesser.<\/p>\n

                    Question 8.
                    \nDefine gravitational potential and field intensity. Give their units. Also, write relation between them.
                    \nAnswer:
                    \nGravitational potential at point is defined as the work done in bringing a body of unit mass from infinity to that point. Its unit is J\/kg, denoted by V Gravitational field intensity is defined as the force experienced by a unit mass at that point. It is expressed in N\/kg, denoted by I.
                    \nI = \\(-\\frac{\\mathrm{V}}{\\mathrm{R}}\\)<\/p>\n

                    Question 9.
                    \nA body weighs 90 kg f on the surface of Earth. What is its weight on a planet whose mass ln \\(\\frac{1}{9}\\) and radius is \\(\\frac{1}{2}\\) of Earth ?
                    \nAnswer:
                    \nWe know that g \u221d \\(\\frac{M}{R^{2}}\\)
                    \n\"1st
                    \nLet m be the mass of the object, then its weight is mg\u03c1<\/sub> = m [\\(\\frac{4}{9}\\) ge<\/sub>]
                    \n\\(\\frac{4}{9}\\)(mg\u03c1<\/sub>)
                    \n\\(\\frac{4}{9}\\) (90kgf)
                    \n= 40kgf<\/p>\n

                    Question 10.
                    \nThe escape velocity v of a body depends on<\/p>\n

                      \n
                    1. the acceleration due to gravity \u2018g\u2019 and<\/li>\n
                    2. \u00a0the radius of planet \u2018R\u2019<\/li>\n<\/ol>\n

                      Find the relation between them using dimension analysis.
                      \nAnswer:
                      \nWe know that,
                      \nEscape velocity = V = [LT-1<\/sup>]
                      \nAcceleration due to gravity = g = [LT-2<\/sup>]
                      \nand Radius = R = [L]
                      \n\u21d2 Let V = kga<\/sup>Rb<\/sup>
                      \n\u21d2 [LT-1<\/sup>] = [La<\/sup>T-2a<\/sup>] [Lb<\/sup>]
                      \na + b = 1 and -2a = -1
                      \n\u21d2 a = 1\/2 and b =1\/2
                      \nSo, V = k \\(\\sqrt{\\mathrm{Rg}}\\).<\/p>\n

                      \"KSEEB<\/p>\n

                      Question 11.
                      \nState Newton\u2019s Law of Gravitation. Find a % decrease in weight of the body when taken to a height of 16 km above surface of Earth. Radius of Earth a 6400 km.
                      \nAnswer:
                      \nAccording to Newton\u2019s Law of Gravitation, everybody in the universe attracts every other body with a force proportional to their masses and inversely proportional to the square of distance between them.
                      \n\\(\\mathrm{F} \\propto \\frac{\\mathrm{m}_{1} \\mathrm{m}_{2}}{\\mathrm{r}^{2}}\\)
                      \n\u21d2 F = \\(\\frac{\\mathrm{Gm}_{1} \\mathrm{m}^{2}}{\\mathrm{r}^{2}}\\), where
                      \nG \u2192 Universal Gravitational Constant
                      \nm1<\/sub> and m2<\/sub> \u2192 masses of the 2 bodies
                      \nr \u2192 distance between the bodies.
                      \nThe % change is weight is given by
                      \n\\(\\frac{\\Delta m g}{m g}\\) \u00d7 100 = \\(\\frac{\\Delta g}{ g}\\) \u00d7 100 = \\(\\frac{\\mathrm{g}-\\mathrm{g}_{n}}{\\mathrm{g}}\\) \u00d7 100
                      \nWe know that, gn<\/sub> = g \\(g\\left(1-\\frac{2 h}{R}\\right)\\)
                      \n\"1st
                      \n= 0.005
                      \n% change = \\(\\frac{\\mathrm{g}-\\mathrm{g}_{n}}{\\mathrm{g}}\\) \u00d7 100 = 0.5%<\/p>\n

                      Question 12.
                      \nDefine gravitational field strength. Which plane in the solar system has the greatest gravitational field strength? What is the gravitational field strength of a planet where a 60 kg astronaut weighs 300 N?
                      \nAnswer:
                      \nThe gravitational field strength at a point in a field is defined as the force experienced by a body of unit mass placed at that point.
                      \nI = \\(\\frac{G M}{R^{2}}\\) \u21d2 I = g
                      \nThe value of I is maximum for Saturn.
                      \nI = \\(\\frac{300 N}{60 \\mathrm{kg}}\\)
                      \nI = 5 ms-2<\/sup><\/p>\n

                      1st PUC Physics Gravitation Five Marks Questions and Answers<\/h3>\n

                      Question 1.
                      \nWhat is a geostationary satellite? Write any four applications of artificial satellites.
                      \nAnswer:
                      \n1 Part:
                      \nA satellite which revolves round the earth with the same angular speed and the same direction as that of earth around its axis is a geostationary satellite.
                      \nUses of artificial satellites:<\/p>\n

                        \n
                      1. for mapping of forests (or to study deforestation)<\/li>\n
                      2. for mapping of rivers, reservoirs<\/li>\n
                      3. for pollution monitoring.<\/li>\n
                      4. for weather forecasting, prediction, and warning of cyclones, mapping of flood-affected areas<\/li>\n
                      5. geostationary satellites are used for telephones, mobiles, broadcasting of radio and television.<\/li>\n<\/ol>\n

                        Question 2.
                        \nWhat are orbital and escape velocity? Write relation between them. Mention any two characteristics of geostationary satellite.
                        \nAnswer:
                        \nThe velocity with which a satellite revolves in its closed orbit is called orbital velocity. The minimum velocity with which body must be projected so that it escapes form the gravitational force of attraction of earth is called escape velocity.
                        \nEscape velocity ve<\/sub> = \\(\\sqrt{2}\\)Vo<\/sub> Where Vo<\/sub> is the orbital velocity.
                        \nTwo properties of geostationary satellites:<\/p>\n

                          \n
                        1. These satellites are appears to be stationary with respect to the earth\u2019s surface.<\/li>\n
                        2. Their height form earth\u2019s surface is about 36,000km and they revolve around the earth from west to east in the equatorial plane.<\/li>\n<\/ol>\n

                          \"KSEEB<\/p>\n

                          Question 3.
                          \nState and explain Newton\u2019s Law of gravitation. Hence define universal gravitational constant and find the dimensional formula for it.
                          \nAnswer:
                          \nThe gravitational force of attraction between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. If two masses m1<\/sub> and m2<\/sub> are separated by a distance r then \\(\\mathrm{F} \\propto \\frac{\\mathrm{m}_{\\mathrm{m}_{2}}}{\\mathrm{r}^{2}}\\) Or \\(\\mathrm{F}=\\mathrm{G} \\frac{\\mathrm{m}_{\\mathrm{1}} \\mathrm{m}_{2}}{\\mathrm{r}^{2}}\\) Where G is the proportionality constant known as universal gravitational constant.
                          \nFrom the above equation, G = \\(\\frac{\\mathrm{Fr}^{2}}{\\mathrm{m}_{1} \\mathrm{m}_{2}}\\)
                          \nIf m1<\/sub> = m2<\/sub> = 1 and r = 1 then G = \\(=\\frac{F(1)^{2}}{1 \\times 1}\\) = F
                          \nGravitational constant is defined as the gravitational force attraction between two bodies of unit masses separated by unit distance. Dimensional formula for [G] = [M-1<\/sup> L3<\/sup> T-2<\/sup> ].<\/p>\n

                          Question 4.
                          \nWhat is the difference between gravitational potential energy and gravitation potential? Derive an expression for gravitational potential energy.
                          \nAnswer:
                          \nThe gravitational potential is the gravitational potential energy per unit mass. So it is independent of the mass considered. The gravitational potential energy is defined as the work done in bringing a mass \u2018m\u2019 from infinity to the given point.
                          \nWe know that, F = \\(\\frac{\\mathrm{GMm}}{\\mathrm{x}^{2}}\\), where x is the distance between M and m.
                          \nWork done in moving the mass by dx from
                          \n\"1st
                          \nThe gravitational potential V= \\(\\frac{\\mathrm{U}}{\\mathrm{m}}\\)
                          \n= \\(-\\frac{\\mathrm{GM}}{\\mathrm{r}}\\)<\/p>\n

                          Question 5.
                          \nDefine the term orbital velocity. Establish a relation for the orbital velocity of a satellite very close to the surface of the earth. Find the ratio of orbital velocity and escape velocity.
                          \nAnswer:
                          \nOrbital velocity is the minimum velocity that is needed to put the satellite into a given orbit around Earth. Consider a satellite of mass \u2018m\u2019 moving around in an orbit at height \u2018h\u2019 above the ground. Let M be mass of Earth, R be the radius of earth and the orbital velocity be Vo<\/sub>.
                          \nWe know that, Fg<\/sub> = \\(\\frac{\\mathrm{GMm}}{(\\mathrm{R}+\\mathrm{h})^{2}}\\)
                          \nand centrifugal force FC<\/sub> = \\(\\frac{m V_{0}^{2}}{(R+h)}\\)
                          \nIn equilibrium i.e., while rotating around the orbit
                          \n\"1st
                          \n(Since, h << R ; R + h = R)
                          \n\\(\\frac{V_{0}}{V_{e s c}}=\\frac{\\sqrt{g R}}{\\sqrt{2 g R}}=\\frac{1}{\\sqrt{2}}\\)<\/p>\n

                          Question 6.
                          \nObtain an expression for escape velocity of an object of mass \u2018m\u2019 from the surface of planet of mass M and radius R. If the escape velocity of planet is know to be 11.2 km s-1<\/sup>. How fast will the object move If its velocity of launch is 22.4 km s-1<\/sup> from the surface?
                          \nAnswer:
                          \nThe minimum velocity required to escape from the gravitational force of the planet is known as the escape velocity of the planet. Consider the object \u2018m\u2019 on the surface of Earth.
                          \nInitial PE = \\(-\\frac{\\mathrm{GMm}}{\\mathrm{R}}\\)
                          \nInitial KE = \\(\\frac{1}{2} m\\left(V_{\\text {esc }}\\right)^{2}\\)
                          \nTotal energy = \\(\\frac{1}{2} m\\left(V_{\\text {esc }}\\right)^{2}-\\frac{G M m}{R}\\)
                          \nTo escape, the KE should greater than or equal to its PE.
                          \n\"1st
                          \nInitial velocity of object = 22.4 km s-1<\/sup>
                          \n= 2 Vesc<\/sub>
                          \nTotal Initial Energy
                          \n= \\(\\frac{1}{2}\\) m(2Vesc<\/sub>)2<\/sup> – \\(\\frac{\\mathrm{GMm}}{\\mathrm{R}}\\)
                          \n= 2 m \\(V_{\\text {esc }}^{2}\\) – \\(\\frac{1}{2}\\) m \\(V_{\\text {esc }}^{2}\\) \u00a0[From(1)]
                          \n= \\(\\frac{3}{2}\\) m \\(V_{\\text {esc }}^{2}\\)\u00a0 \u00a0……(2)
                          \nTotal Final energy = PE (at \u221e) + \\(\\frac{1}{2}\\) mVf<\/sub>2<\/sup>
                          \n= 0+ \\(\\frac{1}{2}\\) mVf<\/sub>2<\/sup>
                          \n= \\(\\frac{1}{2}\\) mVf<\/sub>2<\/sup> ………(3)
                          \nBy conservation of energy, (2) = (3)
                          \nVf<\/sub> = \\(\\sqrt{3}\\) Vesc<\/sub>
                          \n= 19.4 km s-1<\/sup><\/p>\n

                          1st PUC Physics Gravitation Numerical Problems Questions and Answers<\/strong><\/p>\n

                          Question 1.
                          \nFind the gravitational force between two masses 15 kg each separated by a distance of 2m.
                          \n(G = 6.67 \u00d7 10-11<\/sup>Nm2<\/sup>kg-2<\/sup>)
                          \nSolution:
                          \nGiven, m1<\/sub> = m2<\/sub> = 15 kg
                          \nd = 2 m
                          \nG = 6.67 \u00d7 10-11<\/sup>Nm2<\/sup>kg-2<\/sup>
                          \n\u2234 F = \\(\\frac{m_{1} m_{2}}{d^{2}}\\)
                          \n= \\(\\frac{6.67 \\times 10^{-11} \\times 15 \\times 15}{2^{2}}\\)
                          \n= 3.75 \u00d7 10-9<\/sup> N.<\/p>\n

                          Question 2.
                          \nFind the height at which the acceleration due to gravity is 1% less than its value on the surface of the earth. Radius of the earth 6400 km.
                          \nSolution:
                          \nLet h be the height where the acceleration due to gravity is 1 % less than that on the surface of the earth. Then,
                          \n\\(g^{\\prime}=g\\left(1-\\frac{2 h}{R}\\right)\\)
                          \nHere, g’ = 1 % less than g.
                          \n\"1st
                          \n= 32 km.<\/p>\n

                          Question 3.
                          \nThe mass of a sphere is twice the mass of another sphere. These spheres attract each other with a force of 2 \u00d7 10-4<\/sup> N. When they are moved 0.3 m apart, the force reduces to 5 \u00d7 10-5<\/sup>N. Find<\/p>\n

                            \n
                          1. the mass of the spheres.<\/li>\n
                          2. Initial separation between the spheres.<\/li>\n<\/ol>\n

                            Solution:
                            \nLet m be the mass of one sphere. Then the mass of second sphere is 2 m. If \u2018d’ is the initial separation between the two spheres,
                            \nF1<\/sub> = \\(\\mathrm{G} \\frac{\\mathrm{m}_{1} \\mathrm{m}_{2}}{\\mathrm{d}^{2}}\\)
                            \n= G \\(\\frac{\\mathrm{m} \\times 2 \\mathrm{m}}{\\mathrm{d}^{2}}\\) ……(1)
                            \nWhen they are moved 0.3 m further apart,
                            \nF2<\/sub> = G \\(\\frac{\\mathrm{m} \\times 2 \\mathrm{m}}{(\\mathrm{d}+0.3)^{2}}\\) ………(2)
                            \nDividing (1) by (2),
                            \n\"1st
                            \nTaking square root,
                            \n2 = \\(\\frac{(\\mathrm{d} + 0.3)}{\\mathrm{d}}\\) or d = 0. 3 m
                            \n\u2234 Initial separation between the spheres is 0. 3 m
                            \nTo find the mass, substituting in equation (1)
                            \n\"1st
                            \n\u2234 m = 367.3 kg
                            \n2m = 734.61 kg<\/p>\n

                            Question 4.
                            \nCalculate the height at which the acceleration due to gravity becomes\\(\\left(\\frac{1}{16}\\right)^{t h}\\) of its value on the surface.(Radius of the earth 6400 km).
                            \nSolution:
                            \nAcceleration due to gravity at a height h is given by,
                            \n\"1st
                            \n\\(\\frac{h}{R}\\) = 3
                            \nh = 3R = 3 \u00d7 6400
                            \n= 19,200 km.<\/p>\n

                            Question 5.
                            \nAn artificial satellite revolves in a circular orbit at height 600 km above the earths surface. What is the orbital velocity and period of revolution of the satellite? ( Radius of the earth = 8400 km, g = 9.8 ms-2<\/sup>)
                            \nSolution:
                            \nOrbital velocity is given by,
                            \nv = \\(\\sqrt{\\frac{\\mathrm{GM}}{\\mathrm{r}}}\\)
                            \nwhere G is the gravitational constant, M is the mass of the earth.
                            \nWe have, g = \\(\\frac{G M}{R^{2}}\\) and GM = gR2<\/sup>
                            \nv = \\(\\sqrt{\\frac{\\mathrm{gR}^{2}}{\\mathrm{r}}}\\)
                            \n\\(=\\sqrt{\\frac{g R^{2}}{R+h}}\\)
                            \nwhere ‘h\u2018 is the height of the satellite above the earth\u2019s surface.
                            \n\u2234 v = \\(\\sqrt{\\frac{9.8 \\times\\left(6400 \\times 10^{3}\\right)^{2}}{6400 \\times 10^{3}+600 \\times 10^{3}}}\\)
                            \n= 7.57 \u00d7 103<\/sup> ms-1<\/sup>
                            \nPeriod of revolution, T = \\(\\frac{2 \\pi r}{v}\\)
                            \n\"1st
                            \n= 5810 seconds<\/p>\n

                            Question 6.
                            \nThe acceleration due to gravity on moon is 1.7 ms-2<\/sup> and its radius is 0.27 times the radius of the earth. Calculate the ratio of mass of the earth to the mass of the moon. The acceleration due to gravity on the earths surface is 9.8 ms-2<\/sup>
                            \nSolution:
                            \nAcceleration due to gravity on moon, gm<\/sub> = 1.7 ms-2<\/sup>
                            \nAcceleration due to gravity on earths surface gE<\/sub> = 9.8 ms-2<\/sup>
                            \nGiven, Rm<\/sub> = 0.27 RE<\/sub> where Rm<\/sub> and RE<\/sub> are the radii of moon and the earth respectively. Acceleration due to gravity on moon is
                            \ngm<\/sub> = \\(\\frac{\\mathrm{GM}_{\\mathrm{m}}}{\\mathrm{R}_{\\mathrm{m}}^{2}}\\) …..(1)
                            \nAcceleration due to gravity on earth is
                            \ngE<\/sub> = \\(\\frac{\\mathrm{GM}_{\\mathrm{E}}}{\\mathrm{R}_{\\mathrm{E}}^{2}}\\) ……(2)
                            \nwhere Mm<\/sub> and ME<\/sub> are the masses of moon and earth respectively.
                            \n\"1st
                            \n= 79
                            \nMass of the earth is 79 times the mass of the moon.<\/p>\n

                            Question 7.
                            \nA Saturn year is 30 times the earth year. How far is the Saturn from the sun if the earth is 1.50 \u00d7 108<\/sup> km away from the sun?
                            \nSolution:
                            \nAccording to Kepler’s third law, T2<\/sup> \u221d R3<\/sup>
                            \n\u2234\\(\\frac{\\mathrm{T}_{\\mathrm{S}}^{2}}{\\mathrm{T}_{\\mathrm{E}}^{2}}=\\left(\\frac{\\mathrm{R}_{\\mathrm{S}}}{\\mathrm{R}_{\\mathrm{E}}}\\right)^{3}\\) …..(1)
                            \nwhere TS<\/sub> and TE<\/sub> are the period of revolution of Saturn and the earth and RS<\/sub> and RE<\/sub> are the distances of Saturn and the earth from the sun respectively.
                            \n\"1st
                            \nGiven RE<\/sub> = 1.50 \u00d7 1011<\/sup>m
                            \nTS<\/sub> = 30 \u00d7 TE<\/sub>
                            \n\u2234 RS<\/sub> = 1.50 \u00d7 1011<\/sup> \u00d7 \\(\\left(\\frac{30 T_{E}}{T_{E}}\\right)^{2 \/ 3}\\)
                            \n= 1.50 \u00d7 1011<\/sup> x 302\/3<\/sup>
                            \n= 1.448 \u00d7 1012<\/sup> m
                            \n= 1.448 \u00d7 109<\/sup> km.<\/p>\n

                            Question 8.
                            \nDetermine the escape velocity of a body from the moon assuming moon to be a uniform sphere of radius 1.74 \u00d7 106<\/sup>m and mass 7.36 \u00d7 1022<\/sup>kg. Based on the answer explain why moon has no atmosphere.
                            \n(G = 6.67 \u00d7 10-11<\/sup> Nm2<\/sup>kg-2<\/sup>).
                            \nSolution:
                            \nEscape velocity is given by,
                            \n\"1st
                            \n= 2.38 \u00d7 103<\/sup> ms-1<\/sup>
                            \n= 2.38 km s-1<\/sup>
                            \nThe escape velocity from the moon is 2.38kms-1<\/sup> which is less than the r.m.s. value of velocity of gas molecule. Hence all gases which might have been on the surface of the moon had escaped from moon. Hence moon has no atmosphere.<\/p>\n

                            Question 9.
                            \nMass of Jupiter is 320 times that of earth and its radius is 11 times the earth’s radius. Calculate the escape velocity of a body from Jupiter’s surface, given that escape velocity of earth’s surface is 11.2 km s-1<\/sup>
                            \nSolution:
                            \nThe escape velocity from Jupiter\u2019s surface is given by,
                            \nVJ<\/sub> = \\(\\sqrt{\\frac{2 \\mathrm{GM}_{\\mathrm{J}}}{\\mathrm{R}_{\\mathrm{J}}}}\\)
                            \nwhere MJ<\/sub> and RJ<\/sub> are the mass and radius of Jupiter respectively.
                            \nGiven that, MJ<\/sub> = 320 ME<\/sub> and RJ<\/sub> = 11 RE<\/sub>
                            \n\"1st
                            \n= \\(\\sqrt{\\frac{320}{11}}\\) \u00d7 11.2 \u00d7 103<\/sup>
                            \n= 5.394 \u00d7 11.2 \u00d7 103<\/sup> ms-1<\/sup>
                            \n= 60.41 kms-1<\/sup>.<\/p>\n

                            \"KSEEB<\/p>\n

                            Question 10.
                            \nDetermine the angular velocity with which earth would rotate on its axis so that a person on the equator would weigh 3\/5 of his present weight. Radius of earth = 6400 km, g = 9.8 ms-2<\/sup>
                            \nSolution:
                            \nReal weight of the person W = mg
                            \nApparent weight of the person due to the rotation of the earth,
                            \nW’ = mg’
                            \nBut W’ = \\(\\frac{3}{5}\\)W
                            \nmg’ = \\(\\frac{3}{5}\\) mg or
                            \ng’ = \\(\\frac{3}{5}\\) g
                            \nVariation of acceleration due to gravity on the equator is given by,
                            \ng’ = g – \u03c92<\/sup>R
                            \ni.e. \\(\\frac{3}{5}\\) g = g – \u03c92<\/sup>R
                            \n\u2234 \u03c92<\/sup>R = \\(\\frac{2}{5}\\) g
                            \n\u03c9 = \\(\\sqrt{\\frac{2 g}{5 R}}\\)
                            \n\\(=\\sqrt{\\frac{2 \\times 9.8}{5 \\times 6400 \\times 10^{3}}}\\)
                            \n= 7.8 \u00d7 10-4<\/sup> rads-1<\/sup>.<\/p>\n

                            Question 11.
                            \nA satellite circling round the earth has a period of 10 hours. Calculate its height above the surface of the earth., g = 9.8ms-2<\/sup>
                            \nSolution:
                            \nPeriod of revolution of the satellite, T = 10hrs
                            \n= 10 \u00d7 60 \u00d7 60 = 36,000 sec
                            \ng = 9.8ms-2<\/sup>.
                            \nFrom, the equation
                            \n\"1st
                            \nR = 321.6 \u00d7 106<\/sup>m
                            \norbital velocity vo <\/sub>= \\(\\sqrt{\\mathrm{gR}}\\)
                            \n\\(=\\sqrt{9.8 \\times 321.6 \\times 10^{6}}\\)
                            \n= 56.14 \u00d7 103<\/sup>ms-1<\/sup>
                            \n\u2234 from the equation T = \\(\\frac{2 \\pi(\\mathrm{R}+\\mathrm{h})}{\\mathrm{v}_{0}}\\)
                            \n3600 = \\(\\frac{2 \\times 3.142\\left(321.6 \\times 10^{6}+h\\right)}{56.14 \\times 10^{3}}\\)
                            \n2021040 = 2020.9 \u00d7 106<\/sup> + 6.284h
                            \nh = \\(\\frac{0.14 \\times 10^{6}}{6.284}\\)
                            \n= 222.79km.<\/p>\n

                            Question 12.
                            \nCalculate the acceleration due to gravity<\/p>\n

                              \n
                            1. at a height 16km above Earth\u2019s surface and<\/li>\n
                            2. at a depth 2.8km below the surface of the earth. (Radius of the earth is 6400 km and acceleration due to gravity on the surface of the earth = 9.8m\/s-2<\/sup>).<\/li>\n<\/ol>\n

                              Solution:
                              \n1. Acceleration due to gravity,
                              \ng=9.8ms-2<\/sup>, Height h = 16 \u00d7 103<\/sup>m Radius of the earth, R = 6400 \u00d7 103<\/sup> m From the equation g1<\/sup> = g \\(\\left(\\frac{\\mathrm{R}}{\\mathrm{R}+\\mathrm{h}}\\right)^{2}\\)
                              \n= 9.8 \\(\\left(\\frac{6400 \\times 10^{3}}{6400 \\times 10^{3}+16 \\times 10^{3}}\\right)^{2}\\)
                              \n= 9.75 ms-2<\/sup><\/p>\n

                              2.\u00a0 depth, h = 2.8 \u00d7 103<\/sup>m
                              \nFrom the equation g1<\/sup> = g \\(\\left[1-\\frac{h}{R}\\right]\\)
                              \n= 9.8 \\(\\left[1-\\frac{2.8 \\times 10^{3}}{6400 \\times 10^{3}}\\right]\\)
                              \n= 9.796 ms-2<\/sup><\/p>\n

                              \"KSEEB<\/p>\n

                              Question 13.
                              \nThe gravitational force between two bodies. A and B is 6.4 \u00d7 10-7<\/sup> N. When the distance between them is decreased by 0.2m. the force increases to 14.4 \u00d7 10-7<\/sup> N. Find their original distance apart. If the mass of A is twice that of B, find the mass of A and B. Given G = 6.67 \u00d7 10-11<\/sup>Nm2<\/sup>kg-2<\/sup>
                              \nSolution:
                              \n1. Let the mass of body A = m1<\/sub>
                              \nLet the mass of body B = m2<\/sub>
                              \nLet \u2018 \u03b1 \u2019 be the distance
                              \nforce F = 6.4 \u00d7 10-7<\/sup>N
                              \nF = \\(\\frac{\\mathrm{G} \\mathrm{m}_{1} \\mathrm{m}_{2}}{\\mathrm{d}^{2}}\\)
                              \n6.4 \u00d7 10-7<\/sup> = \\(\\frac{\\mathrm{G} \\mathrm{m}_{1} \\mathrm{m}_{2}}{\\mathrm{d}^{2}}\\) (1)<\/p>\n

                              2. Now the distance, d = d – 0.2 force, F= 14.4 \u00d7 10-7<\/sup>N
                              \n14.4 \u00d7 10-7<\/sup> = \\(\\frac{\\mathrm{G}_{\\mathrm{m}_{1} \\mathrm{m}_{2}}}{(\\mathrm{d}-0.2)^{2}}\\) (2)
                              \n\"1st
                              \nTaking Square root on both sides we have 1.5 = \\(\\frac{d}{d-0.2}\\)
                              \nd = 1.5d – 0.3
                              \nd = 0.6m<\/p>\n

                              3. If the mass, of A is twice that of B then m1<\/sub>\u00a0= 2m2<\/sub>, force F = 14.4 \u00d7 10-7<\/sup>N
                              \n\u2234 14.4 \u00d7 10-7<\/sup> = \\(\\frac{6.67 \\times 10^{-11} \\times 2 m_{2} \\cdot m_{2}}{0.6}\\)
                              \nm2<\/sub>2<\/sup> = 0.6477 \u00d7 104<\/sup>
                              \nm2<\/sub> = 0.8048 \u00d7 102<\/sup>kg = 80.48kg
                              \nm1<\/sub> = 2m2<\/sub> = 1.61 \u00d7 102<\/sup>kg
                              \n= 161kg.<\/p>\n

                              Question 14.
                              \nThe mass and diameter of a planet are 3 times that of the earth, what is the acceleration due to gravity on the surface of planet (given \u2018g\u2019 on the surface of earth is 9.8m\/s2<\/sup>).
                              \nSolution:
                              \nMass of the planet M1<\/sup> = 3Mearth<\/sub>
                              \nDiameter of the planet, R1<\/sup> = 3Rearth<\/sub>
                              \nLet g1<\/sup> be the acceleration due to gravity on the surface of the earth whose mass Mearth<\/sub> & Radius Rearth<\/sub> Then,
                              \ng = \\(\\frac{\\mathrm{GM}_{\\mathrm{earth}}}{\\mathrm{R}_{\\mathrm{earth}}^{2}}\\) (1)
                              \nLet g1<\/sup> be the accleration due to grav-ity. on the surface of the planet whose mass is M1<\/sup> & Radius is R1<\/sup> then
                              \ng1<\/sup> = \\(\\frac{\\mathrm{GM}^{1}}{\\left(\\mathrm{R}^{1}\\right)^{2}}\\)
                              \ng1<\/sup> = \\(\\frac{\\text { G3M earth }}{3 \\mathrm{R} \\text { earth }}=\\frac{\\text { GMearth }}{3 \\mathrm{R}_{\\text {earth }}^{2}}\\) (2)
                              \nDivide equ (2) by equ (1)
                              \n\"1st
                              \n= 3.267 m s-2<\/sup><\/p>\n

                              Question 15.
                              \nAn artificial satellite revolves round the earth at a height of 1200km, the radius of the earth is 6400km. What is its orbital velocity and time period of revolution?
                              \nAnswer:
                              \nOrbital velocity, vo<\/sub> = \\(\\sqrt{\\frac{\\mathrm{GM}}{\\mathrm{r}}}\\)
                              \nBut we have, g = \\(\\frac{\\mathrm{GM}}{\\mathrm{R}^{2}}\\) \u21d2 GM = gR2<\/sup>
                              \n\u2234 vo<\/sub> = \\(\\sqrt{\\frac{9 R^{2}}{r}}=\\sqrt{\\frac{g R^{2}}{R+h}}\\)
                              \nwhere \u2018h\u2019 is the height of the satellite above the earths surface.
                              \n\u2234 vo<\/sub> = \\(\\sqrt{\\frac{9.8 \\times\\left(6400 \\times 10^{3}\\right)^{2}}{6400 \\times 10^{3}+1200 \\times 10^{3}}}\\)
                              \n= 7.27 x 103<\/sup>ms-1<\/sup>
                              \nPeriod of revolution, T = \\(\\frac{2 \\pi r}{v_{0}}\\)
                              \n\"1st
                              \n= 6569 seconds<\/p>\n

                              Question 16.
                              \nCalculate the mass and density of the earth from the following data. Gravitational constant = 6.67 \u00d7 10-11<\/sup>Nm2<\/sup>Kg-2<\/sup> Radius of the earth = 6.4 \u00d7 106<\/sup>m. Acceleration due to gravity = 9.8ms-2<\/sup>.
                              \nAnswer:
                              \nG = 6.67 \u00d7 10-11<\/sup>Nm2<<\/sup>Kg-2<\/sup>,
                              \nR = 6.4 \u00d7 106<\/sup>m , g = 9.8 m\/s-2<\/sup>, M = ? and \u03c1 =?
                              \ng = \\(\\frac{\\mathrm{GM}}{\\mathrm{R}^{2}}\\) or M = \\(\\frac{\\mathrm{g} \\mathrm{R}^{2}}{\\mathrm{G}}\\)
                              \n= \\(\\frac{9.8 \\times\\left(6.4 \\times 10^{6}\\right)^{2}}{6.67 \\times 10^{-11}}\\)
                              \n= 60.18 \u00d7 1023<\/sup>kg
                              \n= 6 \u00d7 1024<\/sup> kg.
                              \nDensity
                              \n\u03c1 = \\(\\frac{3 g}{4 \\pi R G}\\)
                              \n= \\(\\frac{3 \\times 9.8}{4 \\times 3.14 \\times 6.4 \\times 10^{6} \\times 6.67 \\times 10^{-11}}\\)
                              \n= 5.483 \u00d7 103<\/sup> kg\/m3<\/sup>.<\/p>\n

                              Question 17.
                              \nA person can jump 1.2 m on Earth. How high can he jump on a planet whose density is 1\/3rd<\/sup> and radius is 4 times that of Earth, assuming he uses same energy.
                              \nAnswer:
                              \nWe know that,
                              \ng \u221d \\(\\frac{M}{R^{2}}\\)
                              \n\"1st
                              \nNow, since same energy is spent
                              \nmgp<\/sub>hp<\/sub> = mge<\/sub>he<\/sub>
                              \n\u21d2 hp<\/sub> = \\(\\left(\\frac{g_{e}}{g_{p}}\\right)\\) he<\/sub>
                              \n\\(\\frac{3}{4}\\) \u00d7 1.2m
                              \nhp<\/sub> = 0.9m.<\/p>\n

                              Question 18.
                              \nFind ratio of the accelerations due to gravity of 2 planets if their radii ratio is 2 : 3. Under condition<\/p>\n

                                \n
                              1. Mass remains constant<\/li>\n
                              2. Material remains constant.<\/li>\n<\/ol>\n

                                Answer:
                                \n1. Here since mass remains constant, g \u221d \\(\\frac{M}{R^{2}}\\) \u21d2 g \u221d 1\/R2<\/sup>
                                \n\u21d2 Ratio of gravitation is,
                                \n\\(\\frac{g_{1}}{g_{2}}=\\left(\\frac{R_{2}}{R_{1}}\\right)^{2}=\\frac{9}{4}\\)<\/p>\n

                                2. Here since material is same its density is same \u21d2 g \u221d \u03c1R \u21d2 g \u221d R
                                \n\u2234 \\(\\frac{g_{1}}{g_{2}}=\\left(\\frac{R_{1}}{R_{2}}\\right)=\\frac{2}{3}\\)<\/p>\n

                                Question 19.
                                \nFind the percentage change in velocity of moon to escape the Earth\u2019s gravity.
                                \nAnswer:
                                \nInitial Velocity of moon = Orbital
                                \nVelocity, Vo<\/sub> = \\(\\sqrt{\\frac{\\mathrm{GM}}{\\mathrm{r}}}\\)
                                \nFor the moon to escape Earth\u2019s influence,
                                \nits velocity must become \\(\\sqrt{2 \\frac{\\mathrm{GM}}{\\mathrm{r}}}\\)
                                \nSo, increase in velocity required
                                \n\"1st
                                \nPercentage change
                                \n\"1st
                                \n= 41.4 %<\/p>\n

                                Question 20.
                                \nA rocket is launched with 0.8 times escape velocity. How high will it go before it starts falling?
                                \nAnswer:
                                \nInitial KE = \\(\\frac{1}{2}\\) mVr<\/sub>2<\/sup>
                                \n\\(\\frac{1}{2}\\) m (0.8)2<\/sup> (Vesc<\/sub>)2<\/sup>
                                \n= (0.64) [\\(\\frac{1}{2}\\) m \\(V_{\\text {esc }}^{2}\\)]
                                \n= 0.64 \\(\\frac{\\mathrm{GMm}}{\\mathrm{R}}\\)
                                \nInitial PE = \\(-\\frac{\\mathrm{GMm}}{\\mathrm{R}}\\)
                                \nAt the height \u2018h\u2019, its velocity is 0.
                                \nSo Final Potential Energy = \\(-\\frac{\\mathrm{GMm}}{\\mathrm{R}+\\mathrm{h}}\\)
                                \nBy Law of Conservation of Energy
                                \n(0.64)\\(\\frac{\\mathrm{GMm}}{\\mathrm{R}}-\\frac{\\mathrm{GMm}}{\\mathrm{R}}=\\frac{\\mathrm{GMm}}{(\\mathrm{R}+\\mathrm{h})}\\)
                                \n0.36(R + h) = R
                                \nh = \\(\\frac{0.64}{0.36}\\) R = \\(\\frac{16}{9}\\) R
                                \nSo, the rocket reaches a height of \\(\\frac{16}{9}\\) R,
                                \nwhere R is radius of the Earth.<\/p>\n

                                \"KSEEB<\/p>\n

                                Question 21.
                                \nA mass m is left to fall freely to fall from a height R\/3 from the surface of Earth. What is Its velocity when it hits the surface of Earth.
                                \nAnswer:
                                \nInitial KE = 0,
                                \nInitial PE = \\(\\frac{-G M m}{(R+R \/ 3)}\\) = \\(-\\frac{3}{4} \\frac{\\mathrm{GMm}}{\\mathrm{R}}\\)
                                \nTotal Energy = \\(-\\frac{3}{4} \\frac{\\mathrm{GMm}}{\\mathrm{R}}\\) …..(1)
                                \nat the surface of the Earth, let its velocity be Vf<\/sub> then
                                \nTE = \\(\\frac{1}{2}\\) m(Vf<\/sub>)2<\/sup> – \\(\\frac{\\mathrm{GMm}}{\\mathrm{R}}\\) ….(2)
                                \nBy conservation of energy (1) = (2)
                                \n\"1st
                                \n\\(\\frac{1}{2}\\) Vesc<\/sub>
                                \n= 5.6 km s-1<\/sup><\/p>\n

                                Question 22.
                                \nCalculate the increase in potential energy when a body is moved from the surface of Earth to a height 2 R from the surface, where R is the radius of Earth.
                                \nAnswer:
                                \nThe Potential Energy of object on the surface of Earth is U l<\/sub> = \\(-\\frac{\\mathrm{GMm}}{\\mathrm{R}}\\)
                                \nThe Potential Energy at the height of 2 R from surface is Uf<\/sub> = \\(-\\frac{G M m}{(R+2 R)}\\)
                                \n\"1st<\/p>\n

                                Question 23.
                                \nShow that velocity of a body released from a height r from centre of Earth (r > R) is
                                \nR [2g \\(\\left(\\frac{1}{\\mathbf{R}}-\\frac{1}{\\mathbf{r}}\\right)\\)]1\/2<\/sup> when it reaches the Earth\u2019s surface is, where R is radius of Earth.
                                \nAnswer:
                                \nFrom law of conservation of Energy we can write \\(-\\frac{\\mathrm{GMm}}{\\mathrm{r}}=\\frac{1}{2} \\mathrm{mV}^{2}-\\frac{\\mathrm{GMm}}{\\mathrm{R}}\\)
                                \n\"1st
                                \nV = \\(\\mathrm{R}\\left(2 \\mathrm{g}\\left(\\frac{1}{\\mathrm{R}}-\\frac{1}{\\mathrm{r}}\\right)\\right)^{1 \/ 2}\\)<\/p>\n

                                Question 24.
                                \nDetermine the speed with which the Earth has to rotate so that the weight of a person on the equator is 2\/3rd<\/sup> of the current weight. Assume radius of Earth is 6.4 \u00d7 106<\/sup> m.
                                \nAnswer:
                                \nLet m be the mass of the person, then,
                                \nOriginal weight = mg
                                \nand new weight = mg’
                                \nGiven that \\(\\frac{\\mathrm{mg}^{\\prime}}{\\mathrm{mg}}=\\frac{2}{3}\\)
                                \n\u21d2 g’ = \\(\\frac{2}{3}\\) g
                                \nWe know that,
                                \ng’ = g – R\u03c92<\/sup> (for equatorial plane)
                                \n\\(\\frac{2}{3}\\) g = g – R\u03c92<\/sup>
                                \n\u21d2 \u03c9 = \\(\\sqrt{\\frac{\\mathrm{g}}{3 \\mathrm{R}}}\\)
                                \n\u03c9 = \\(\\sqrt{\\frac{9.8}{3 \\times 6.4 \\times 10^{6}}}\\)
                                \n\u03c9 = 5.104 \u00d7 10-4<\/sup>rad s10-1<\/sup>.<\/p>\n

                                Question 25.
                                \nTwo stars of mass Mx<\/sub> and My<\/sub> of radii Rx<\/sub> and Ry<\/sub> are d distance apart. Now if there is a particle of mass m at the midway, find the minimum speed with which it needs to be projected to escape to infinity.
                                \nAnswer:
                                \nThe initial potential energy of the mass is, given by
                                \n\"1st
                                \nLet the initial velocity be Vi<\/sub>
                                \nthen KEi<\/sub> = \\(\\frac{1}{2}\\) m(Vi<\/sub>)2<\/sup>
                                \nSo, Total Energy
                                \n\\(\\frac{1}{2}\\) mVi<\/sub>2<\/sup> – \\(\\frac{2 G m}{d}\\) (Mx<\/sub> + My<\/sub>)
                                \nTotal Final Energy = 0
                                \nBy conservation of Energy,
                                \n\"1st<\/p>\n

                                Question 26.
                                \nThe mass of Jupiter is 1.9 \u00d7 1027<\/sup> kg and that of Sun is 2 \u00d7 1030<\/sup>kg. The mean distance between them is 7.8 \u00d7 1011<\/sup>m. Find the gravitational force exerted by Sun. Also, calculate the speed of Jupiter assuming circular orbit around the Sun.
                                \nAnswer:
                                \nMass of Jupiter = 1.9 \u00d7 1027<\/sup>kg
                                \nMass of Sun = 2 \u00d7 1030<\/sup>kg
                                \nDistance, r = 7.8 \u00d7 1011<\/sup>m
                                \nSo, gravitational Force, Fg<\/sub> = \\(-\\frac{G M_{1} M_{2}}{r^{2}}\\)
                                \n= \\(\\frac{6.67 \\times 10^{-11} \\times 1.9 \\times 2 \\times 10^{57}}{(7.8)^{2} \\times 10^{22}}\\)
                                \n= 4.16 \u00d7 1023<\/sup>N
                                \nNow, the Fg<\/sub> is equal the centrifugal force of Jupiter.
                                \n\"1st
                                \n= 1.3 \u00d7 104<\/sup> m s-1<\/sup> or 13 km s-1<\/sup>.<\/p>\n

                                Question 27.
                                \nA 500 kg satellite is in a circular orbit of radius 2 RE<\/sub> about the Earth. How much energy is required to move it to another orbit of radius 4 Re<\/sub>? What are changes in kinetic and potential energies?
                                \nAnswer:
                                \nMass of satellite, ms<\/sub> = 500 kg
                                \nLet ME<\/sub> be mass of Earth and RE<\/sub> be radius of earth.
                                \nThe PE1<\/sub> = \\(-\\frac{G M_{E} m_{s}}{2 R_{E}}\\)
                                \nand KE1<\/sub> = \\(\\frac{1}{2}\\) mvo<\/sub>2<\/sup> = \\(\\frac{1}{2}\\) ms<\/sub> \\(\\left(\\frac{\\mathrm{GM}_{\\mathrm{E}}}{2 \\mathrm{R}_{\\mathrm{E}}}\\right)\\)
                                \n= \\(\\frac{\\mathrm{GM}_{\\mathrm{E}} \\mathrm{m}_{\\mathrm{s}}}{4 \\mathrm{R}_{\\mathrm{E}}}\\)
                                \nTotal Energy, TE1<\/sub>
                                \n\"1st
                                \nSimilarly for new orbit
                                \n\"1st
                                \nDifference in Energy = TE2<\/sub> – TE1<\/sub>
                                \n\"1st
                                \n= 3.92 \u00d7 109<\/sup> J
                                \nAn energy of 3.92 GJ needs to be supplied to the satellite to change its orbit.
                                \nChange in PE = \\(\\frac{G M_{E} m_{s}}{2 R_{E}}-\\frac{G M_{E} m_{s}}{4 R_{E}}\\)
                                \n= \\(\\frac{\\mathrm{GM}_{\\mathrm{E}} \\mathrm{m}_{\\mathrm{s}}}{4 \\mathrm{R}_{\\mathrm{E}}}\\)
                                \n= 2 \u00d7 3.92 \u00d7 109<\/sup> J
                                \n= 7.84 \u00d7 109<\/sup> J
                                \nChange in KE = \\(\\frac{G M_{E} m_{s}}{8 R_{E}}-\\frac{G M_{E} m_{s}}{4 R_{E}}\\)
                                \n= – 3.92 \u00d7 109<\/sup> J.<\/p>\n

                                \"KSEEB<\/p>\n

                                Question 28.
                                \nA solid sphere of R\/2 is cut and removed as shown. If \u03c1 is the density of material. A mass \u2018m\u2019 is place at a distance R from centre of sphere. Find the gravitational force of attraction the mass.
                                \nAnswer:
                                \n\"1st
                                \nMass of the Sphere is,
                                \nM1<\/sub> = \u03c1 \u00d7 \\(\\frac{4}{3}\\)R3<\/sup>
                                \nForce due to the sphere.
                                \nF1<\/sub> = \\(\\frac{\\mathrm{GM}_{1} \\mathrm{m}}{(3 \\mathrm{R})^{2}}\\)
                                \nThe Mass of missing portion
                                \nM2<\/sub> = \u03c1 \\(\\frac{4}{3}\\) \u03c0 (R\/2)3<\/sup> = \u03c1 \u00d7 \\(\\frac{4}{3}\\) \u00d7 \\(\\frac{\\pi \\mathrm{R}^{3}}{8}\\)
                                \n= \\(\\frac{M_{1}}{8}\\)
                                \nForce on mass from only the missing portion
                                \n\"1st
                                \nThe total effective force on the mass is, F = F1<\/sub> – F2<\/sub>
                                \n\"1st
                                \n= 0.38 \u00d7 G \u00d7 R \u00d7 \u03c1 \u00d7 m.<\/p>\n

                                Question 29.
                                \nInfinite number of masses are placed at x = 1, 2, 4, 8….. If each of these masses are equal to \u2018m’. Find the potential energy on a mass \u2018m\u2019 at origin.
                                \nAnswer:
                                \nThe gravitational potential energy at a point distance x from a mass m is given
                                \nby \\(-\\frac{\\mathrm{Gmm}}{\\mathrm{x}}\\)
                                \nThe total potential energy is given by,
                                \n\"1st
                                \n[sum of infinity GP]
                                \n= – 2Gm2<\/sup><\/p>\n

                                Question 30.
                                \nTwo masses m1<\/sub> and m2<\/sub> are separated by a distance r. Find the expression for gravitational potential at a point where the net force is zero.
                                \nAnswer:
                                \nLet x be the distance of the point where net force is zero from m.
                                \n\"1st
                                \n\"1st
                                \nThe potential at the point x is
                                \n\"1st<\/p>\n","protected":false},"excerpt":{"rendered":"

                                You can Download Chapter 8 Gravitation Questions and Answers, Notes, 1st PUC Physics Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 1st PUC Physics Question Bank Chapter 8 Gravitation 1st PUC Physics Gravitation Textbook Questions and Answers Question 1. Answer the …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/11289"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=11289"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/11289\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=11289"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=11289"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=11289"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}