Physics – Gravitation Textual Unsolved Problems

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Example – 01:

  • Show that the critical velocity of a body revolving in a circular orbit very close to the surface of the planet of radius R and mean density ρ is

  • Solution:

The critical velocity of a satellite orbiting very close  to the earth’s surface is given by

  • This is the expression for the critical velocity of a satellite orbiting very close to the earth’s surface in terms of density of the material of the planet.

Example – 02:

  • What would be the duration of the year if the distance between the earth and sun gets doubled the present distance?
  • Solution:
  • Given: r2 = 2  r1, old period T1 = 365 days
  • To Find:  New period  T2 =?

By Keppler’s law, we have T2 ∝ r3



Gravitation 09

Ans: Thus the duration of the year would be 1032 days

Example – 03:

  • Calculate the height of communication satellite above the surface of the earth? Given G = 6.67 x 10-11 N m2/kg2; radius of earth = 6400 km, Mass of the earth = 6 x 1024 Kg
  • Given: For communication satellite, T = 24 hr = 24 x 60 x 60 s, Universal gravitational Constant = G = 6.67 x 10-11 N m2/kg2; radius of earth = R = 6400 km = 6.4 x 106 m, Mass of the earth = M = 6 x 1024 Kg
  • To find: height of communication satellite above the surface of the earth = h = ?

The time period of satellite is given by

Now, r = R + h
Hence, h = r – R = – 6.37 x 106
= 35.95 x 106 m
= 35.95 x 103 x 103 m
= 35950 km
Ans: The height of satellite above the surface of the earth is 35950 km.



Example – 04:

  • The distances of two planets from the sun are 1013 m 1012 m respectively. Find the ratio of their periods and orbital speeds.
  • Given: r1 = 1013 m ,  r2 = 1012 m
  • To find: T1 : T2 = ?, vc1 : vc2 = ? ,

orbital speeds

Ans : The ratio of their period is 31.62:1

The ratio of their critical velocities is 0.3162:1

Example – 0.5:

  • A body weights 3.5 kg wt, on the surface of the earth. How much will it weigh on the surface of a planet whose mass is 1/7 that of earth and whose radius is half that of earth
  • Solution:
  • Given: , Weight of body on earth = WE = 1.8 kg, mass of planet = 1/9 mass of earth i.e MP = 1/7 ME, radius of planet = 1/2 radius of earth i.e.  RP = 1/2 RE.
  • To find: Weight of body on the planet = WP = ?

Ans: The weight of the body on the surface of the planet is 2 kg wt.



Example – 06:

  • The radii of the orbits of two satellites revolving around the earth are in the ratio 3:8. Compare their i) critical speed and ii) periods.
  • Given: ratio of radii of orbits r1 :  r =  3:8
  • To find: vc1 : vc2  =? , T1 : T2 = ?

Ans : The ratio of their critical velocities is 1.633:1

The ratio of their period is 0.2296:1

Example – 07:

  • Calculate the escape velocity of a body from the surface of the earth, Given: G = 6.67 x 10-11 N m2/kg2, Radius of the earth = 6400 km, and g =9.8 m/s2.
  • Solution:
  • Given: radius of earth = R = 6400 km = 6.4 x 106 m, Acceleration due to gravity = 9.8 m/s2. Universal gravitational Constant = G = 6.67 x 10-11 N m2/kg2,
  • To Find: ve =?

Ans: Thus the escape velocity of a body from the surface of the earth is 11.2 km/s.



Example – 08:

  • Find the binding energy of a body of mass 50 kg at rest on the surface of the earth. Given: G = 6.67 x 10-11 N m2/kg2, R = 6400 km = 6.4 x 106 m; M = 6 x 1024 kg.
  • Given: the mass of body = 50 kg, Universal gravitational constant = G = 6.67 x 10-11 N m2/kg2, R = 6400 km = 6.4 x 106 m; M = 6 x 1024 kg.
  • To find: binding energy of satellite = B.E. =?
  • Solution:

The binding energy of the body at rest on the surface of earth is given by

Ans: Binding energy of the body is  3.127 x 109 J

Example – 09:

  • What are the total energy and binding energy of an artificial satellite of mass 1000 kg orbiting at a height of 1600 km above the surface of the earth? Mass of the earth = 6 x 1024 kg; Radius of earth = 6400 km, G = 6.67 x 10-11 N m2/kg2,
  • Given: m = 100 kg, Universal gravitational Constant = G = 6.67 x 10-11 N m2/kg2, R = 6400 km = 6.4 x 106 m; M = 6 x 1024 kg. h = 1600 km, r = R + h = 6400 + 1600 = 8000 km = 8 x 106 m;
  • To Find: T.E. = ?, B.E. = ?,
  • Solution:

Now Total energy = – B.E. = – 2.5 x 1010 J
Ans: Total energy of satellite = – 2.5 x 1010 J and
The binding energy of satellite =  2.5 x 1010 J

Example – 10:

  • What would be the duration of the year if the distance between the earth and sun gets halved the present distance?
  • Solution:
  • Given: r2 = 1/2  r1, old period T1 = 365 days
  • To Find:  New period  T2 =?

By Keppler’s law, we have T2 ∝ r3



Gravitation 08

Ans: Thus duration of the year would be 129 days

Example – 11:

  • Two bodies of masses 5 kg and 6 x 1024 kg are placed with their centres 6.4 x 106 m apart. Calculate the force of attraction between the two masses. Also, find the initial acceleration of two masses assuming no other forces act on them.
  • Solution:
  • Given: Mass of first body = m1 = 5 kg, mass of second body = m2 = 6 x 1024 kg, Distance between masses = r =  6.4 x 106 m, Universal gravitational Constant = G = 6.67 x 10-11 N m2/kg2 .
  • To Find: Force of attraction between two masses = F = ?, Initial accelerations of the two masses =?

By Newton’s law of gravitation



  • Initial acceleration of body of mass 5 kg

By Newton’s second law of motion  F = ma

Thus a = F/m = 48.85 / 5 = 9.77 m/s2

  • Initial acceleration of body of mass 6 x 1024 kg

By Newton’s second law of motion  F = ma

Thus a = F/m = 48.85 / 6 x 1024  = 8.142 x 10-24 m/s2

Ans: The force of attarction between the two masses = 48.85 N

The Initial acceleartion of body of mass 5 kg is 9.77 m/sand



That of body of mass 6 x 1024 kg is 8.14 x 10-24 m/s2.

Example – 12:

  • Find the value of Universal gravitational Constant G from the following data: M = 6 x 1024 kg, R = 6400 km, g = 9.774 m/s2,
    Given: M = 6 x 1024 kg, R = 6400 km = 6.4 x 106 m, g = 9.774 m/s2,
    To find: G = ?
  • Solution:
    Universal gravitational constantAns: The value of G is 6.672 x 10-11 N m2/kg2.

Example – 13: 

  • Assuming the earth to be homogeneous sphere, find the density of material of the earth from the following data.g = 9.8 m/s2, G = 6.673 x 10-11 Nm2/kg2 ,    R = 6400 km,
  • Given: g = 9.8 m/s2, Universal gravitational Constant = G = 6.673 x 10-11 Nm2/kg2 ,R = 6400 km = 6.4 x 106 m,
  • To find:  Density = r = ?
  • Solution:

Ans: Density of material of the earth = 5483 kg/m3

Example – 14:

  • The mass of the body on the surface of the earth is 100 kg. What will be its mass and weight at an altitude of 1000 km.
  • Solution:
  • Given: , Mass of body = 100 kg, R = 6400 km  and altitude = h = 1000 km
  • To find: Mass and weight at altitude of 1000 km = ?

r = R + h = 6400 + 1000 = 7400 km



Now weight of body at altitude 100 km = Wh = m gh = 100 x 7.33 = 733 N

The mass is always constant, hence mass at an altitude of 100 km = 100 kg

Ans:  At an altitude of 1000 km the mass of the body is 100 kg and its weight is 733 N.

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