Physics · Textbook solutions

Gravitation

Every solved example, exercise, and miscellaneous question — in the order the textbook teaches them. · 58 questions

5. Gravitation — worked examples

11 q

Solved Examples

Worked · 11
  1. Solved Ex.5.1
    What would be the average duration of year if the distance between the Sun and the Earth becomes (A) thrice the present distance. (B) twice the present distance.
  2. Solved Ex.5.2
    The gravitational force between two bodies is 1 N. If distance between them is doubled, what will be the gravitational force between them?
  3. Solved Ex.5.3
    Three particles A, B, and C each having mass mm are kept along a straight line with AB = BC = ll. A fourth particle D is kept on the perpendicular bisecter of AC at a distance ll from B. Determine the gravitational force on D.
  4. Solved Ex.5.4
    Calculate mass of the Earth from given data, Acceleration due to gravity g=9.81g = 9.81 m/s2^{2} Radius of the Earth RE=6.37×106R_{E} = 6.37 \times 10^{6} m G=6.67×10−11G = 6.67 \times 10^{-11} N m2^{2}/kg2^{2}
  5. Solved Ex.5.5
    Calculate the acceleration due to gravity on the surface of moon if mass of the moon is 1/80 times that of the Earth and diameter of the moon is 1/4 times that of the Earth (g=9.8g = 9.8 m/s2^{2})
  6. Solved Ex.5.6
    Find the acceleration due to gravity on a planet that is 10 times as massive as the Earth and with radius 20 times of the radius of the Earth (g=9.8g = 9.8 m/s2^{2}).
  7. Solved Ex.5.7
    At what distance above the surface of Earth the acceleration due to gravity decreases by 10% of its value at the surface? (Radius of Earth = 6400 km). Assume the distance above the surface to be small compared to the radius of the Earth.
  8. Solved Ex.5.8
    What will be the change in potential energy of a body of mass mm when it is raised from height RER_{E} above the Earth's surface to 52RE\dfrac{5}{2}R_{E} above the Earth's surface? RER_{E} and MEM_{E} are the radius and mass of the Earth respectively.
  9. Solved Ex.5.9
    Show that the critical velocity of a body revolving in a circular orbit very close to the surface of a planet of radius RR and mean density ρ\rho is 2RGπρ32R\sqrt{\dfrac{G\pi\rho}{3}}.
  10. Solved Ex.5.9b
    Calculate the period of revolution of a polar satellite orbiting close to the surface of the Earth. Given R=6400R = 6400 km, g=9.8g = 9.8 m/s2^{2}.
  11. Solved Ex.5.10
    An artificial satellite revolves around a planet in circular orbit close to its surface. Obtain the formula for period of the satellite in terms of density ρ\rho and radius RR of planet.

Exercises

47 q

Choose the correct option

Practice · 4
  1. Choose the correct option.
    Ex Q.1 (i)
    The value of acceleration due to gravity is maximum at
    1. A.
      the equator of the Earth .
    2. B.
      the centre of the Earth.
    3. C.
      the pole of the Earth.
    4. D.
      slightly above the surface of the Earth.
  2. Ex Q.1 (ii)
    The weight of a particle at the centre of the Earth is
    1. A.
      infinite.
    2. B.
      zero.
    3. C.
      same as that at other places.
    4. D.
      greater than at the poles.
  3. Ex Q.1 (iii)
    The gravitational potential due to the Earth is minimum at
    1. A.
      the centre of the Earth.
    2. B.
      the surface of the Earth.
    3. C.
      a points inside the Earth but not at its centre.
    4. D.
      infinite distance.
  4. Ex Q.1 (iv)
    The binding energy of a satellite revolving around planet in a circular orbit is 3×1093 \times 10^{9} J. Its kinetic energy is
    1. A.
      6×1096 \times 10^{9} J
    2. B.
      −3×109-3 \times 10^{9} J
    3. C.
      −6×10+9-6 \times 10^{+9} J
    4. D.
      3×10+93 \times 10^{+9} J

Answer the following questions

Practice · 15
  1. Answer the following questions.
    Ex Q.2 (i)
    State Kepler's law equal of area.
  2. Ex Q.2 (ii)
    State Kepler's law of period.
  3. Ex Q.2 (iii)
    What are the dimensions of the universal gravitational constant?
  4. Ex Q.2 (iv)
    Define binding energy of a satellite.
  5. Ex Q.2 (v)
    What do you mean by geostationary satellite?
  6. Ex Q.2 (vi)
    State Newton's law of gravitation.
  7. Ex Q.2 (vii)
    Define escape velocity of a satellite.
  8. Ex Q.2 (viii)
    What is the variation in acceleration due to gravity with altitude?
  9. Ex Q.2 (ix)
    On which factors does the escape speed of a body from the surface of Earth depend?
  10. Ex Q.2 (x)
    As we go from one planet to another planet, how will the mass and weight of a body change?
  11. Ex Q.2 (xi)
    What is periodic time of a geostationary satellite?
  12. Ex Q.2 (xii)
    State Newton's law of gravitation and express it in vector form.
  13. Ex Q.2 (xiii)
    What do you mean by gravitational constant? State its SI units.
  14. Ex Q.2 (xiv)
    Why is a minimum two stage rocket necessary for launching of a satellite?
  15. Ex Q.2 (xv)
    State the conditions for various possible orbits of a satellite depending upon the tangential speed of projection.

Answer the following questions in detail

Practice · 16
  1. Answer the following questions in detail.
    Ex Q.3 (i)
    Derive an expression for critical velocity of a satellite.
  2. Ex Q.3 (ii)
    State any four applications of a communication satellite.
  3. Ex Q.3 (iii)
    Show that acceleration due to gravity at height hh above the Earth's surface is gh=g(RR+h)2g_{h} = g\left(\dfrac{R}{R+h}\right)^{2}
  4. Ex Q.3 (iv)
    Draw a labelled diagram to show different trajectories of a satellite depending upon the tangential projection speed.
  5. Ex Q.3 (v)
    Derive an expression for binding energy of a body at rest on the Earth's surface.
  6. Ex Q.3 (vi)
    Why do astronauts in an orbiting satellite have a feeling of weightlessness?
  7. Ex Q.3 (vii)
    Draw a graph showing the variation of gravitational acceleration due to the depth and altitude from the Earth's surface.
  8. Ex Q.3 (viii)
    At which place on the Earth's surface is the gravitational acceleration maximum? Why?
  9. Ex Q.3 (ix)
    At which place on the Earth surface the gravitational acceleration minimum? Why?
  10. Ex Q.3 (x)
    Define the binding energy of a satellite. Obtain an expression for binding energy of a satellite revolving around the Earth at certain attitude.
  11. Ex Q.3 (xi)
    Obtain the formula for acceleration due to gravity at the depth 'd' below the Earth's surface.
  12. Ex Q.3 (xii)
    State Kepler's three laws of planetary motion.
  13. Ex Q.3 (xiii)
    State the formula for acceleration due to gravity at depth 'd' and altitude 'h' Hence show that their ratio is equal to (R−dR−2h)\left(\dfrac{R-d}{R-2h}\right) by assuming that the altitude is very small as compared to the radius of the Earth.
  14. Ex Q.3 (xiv)
    What is critical velocity? Obtain an expression for critical velocity of an orbiting satellite. On what factors does it depend?
  15. Ex Q.3 (xv)
    Define escape speed. Derive an expression for the escape speed of an object from the surface of the each.
  16. Ex Q.3 (xvi)
    Describe how an artificial satellite using two stage rocket is launched in an orbit around the Earth.

Solve the following problems

Practice · 12
  1. Solve the following problems. Where a value is not stated in the question, take G=6.67×10−11G = 6.67 \times 10^{-11} N m2^{2}/kg2^{2} and, for the Earth, escape speed ve=11.2v_{e} = 11.2 km/s — the values stated in Sec. 5.3 and Sec. 5.7.4 of this chapter. (The book's Exercises print no constants preamble; these are supplied so that each question can be attempted on its own.)
    Ex Q.4 (i)
    At what distance below the surface of the Earth, the acceleration due to gravity decreases by 10% of its value at the surface, given radius of Earth is 6400 km.
  2. Ex Q.4 (ii)
    If the Earth were made of wood, the mass of wooden Earth would have been 10% as much as it is now (without change in its diameter). Calculate escape speed from the surface of this Earth.
  3. Ex Q.4 (iii)
    Calculate the kinetic energy, potential energy, total energy and binding energy of an artificial satellite of mass 2000 kg orbiting at a height of 3600 km above the surface of the Earth. Given:- G=6.67×10−11G = 6.67 \times 10^{-11} Nm2^{2}/kg2^{2} R=6400R = 6400 km M=6×1024M = 6 \times 10^{24} kg
  4. Ex Q.4 (iv)
    Two satellites A and B are revolving around a planet. Their periods of revolution are 1 hour and 8 hours respectively. The radius of orbit of satellite B is 4×1044 \times 10^{4} km. find radius of orbit of satellite A .
  5. Ex Q.4 (v)
    Find the gravitational force between the Sun and the Earth. Given Mass of the Sun =1.99×1030= 1.99 \times 10^{30} kg Mass of the Earth =5.98×1024= 5.98 \times 10^{24} kg The average distance between the Earth and the Sun =1.5×1011= 1.5 \times 10^{11} m.
  6. Ex Q.4 (vi)
    Calculate the acceleration due to gravity at a height of 300 km from the surface of the Earth. (M=5.98×1024M = 5.98 \times 10^{24} kg, R=6400R = 6400 km).
  7. Ex Q.4 (vii)
    Calculate the speed of a satellite in an orbit at a height of 1000 km from the Earth's surface. ME=5.98×1024M_{E} = 5.98 \times 10^{24} kg, R=6.4×106R = 6.4 \times 10^{6} m.
  8. Ex Q.4 (viii)
    Calculate the value of acceleration due to gravity on the surface of Mars if the radius of Mars =3.4×103= 3.4 \times 10^{3} km and its mass is 6.4×10236.4 \times 10^{23} kg.
  9. Ex Q.4 (ix)
    A planet has mass 6.4×10246.4 \times 10^{24} kg and radius 3.4×1063.4 \times 10^{6} m. Calculate energy required to remove on object of mass 800 kg from the surface of the planet to infinity.
  10. Ex Q.4 (x)
    Calculate the value of the universal gravitational constant from the given data. Mass of the Earth =6×1024= 6 \times 10^{24} kg, Radius of the Earth =6400= 6400 km and the acceleration due to gravity on the surface =9.8= 9.8 m/s2^{2}
  11. Ex Q.4 (xi)
    A body weighs 5.6 kg wt on the surface of the Earth. How much will be its weight on a planet whose mass is 7 times the mass of the Earth and radius twice that of the Earth's radius.
  12. Ex Q.4 (xii)
    What is the gravitational potential due to the Earth at a point which is at a height of 2RE2R_{E} above the surface of the Earth, Mass of the Earth is 6×10246 \times 10^{24} kg, radius of the Earth =6400= 6400 km and G=6.67×10−11G = 6.67 \times 10^{-11} Nm2^{2} kg−2^{-2}.