Physics – Gravitation Textual Solved Problems

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Example – 01:

  • Show that the escape velocity of a body orbiting very close to the earth’s surface of the earth is √2vc. Where vc is the critical velocity of a body when it is orbiting very close to the earth’s surface.
  • Solution:

For an orbiting satellite, critical velocity is given by

where G = Universal gravitational constant

M = the mass of the earth



R = the radius of the earth

h = height of the satellite above the earth’s surface.

  • For satellite orbiting very close to the earth, h can be neglected as h < < R. Hence it can be neglected. Therefore the critical velocity of the satellite orbiting very close to earth’s surface.

The escape velocity of a satellite oh the surface of the earth is given by



Dividing equation (1) by (2)

  •  Thus the escape velocity of a body from the surface of the planet (earth) is √2  times the critical velocity of the body when it is orbiting close to the planet’s (earth’s) surface.

Example – 02:

  • Show that the escape velocity of a body from the surface of a planet of radius R and mean density ρ is

  • Solution:

The escape velocity of a satellite on the surface of the planet is given by



Where G = Universal gravitational constant

M = the mass of the Planet

R = the radius of the Planet

Let d    = density of the material of the planet

Now, Mass of Planet   =  Volume of Planet x Density



This is an expression for escape velocity in terms of density of the material of the planet.

Example – 03:

  • An artificial satellite has to be set up to revolve around the planet in a circular orbit close to its surface. If r is density and R is the radius of the planet, show that the period of revolution is

  • Solution:

The time period of a satellite orbiting around the earth is given by



If satellite orbiting very close to the earth  (i.e. h < < R)

then h can be neglected. Then R + h = R

This is an expression for the period of a satellite in terms of density of the material of the planet.



Example – 04:

  • Find the acceleration due to gravity on the surface of the moon. Given that the mass of the moon is 1/80 times that of the earth and the diameter of the moon is 1/4 times that of the earth. g = 9.8 m/s2.
  • Given: MM = ME, RM = RE, gE = 9.8 m/s2,
  • To find: gM = ?
  • Solution:
    Ans:The acceleration due to gravity on the surface of the moon is 1.96 m/s2.

Example – 05:

  • At what height above the earth’s surface will the acceleration due to gravity be 25% of the value at the surface of the earth? g= 9.8 m/s2, R= 6400 km.
  • Given: R =6400 km, gh = 25% g = 0.25 g
  • To find: h =?
  • Solution:


Ans: At height of 6400 km, the acceleration due to gravity be 25% of the value at the surface of the earth

Example – 06:

  • Find the acceleration due to gravity at a depth of 2000 km from the surface of the earth, assuming earth to be a homogeneous sphere. R = 6400 km, g = 9.8 m/s2.
  • Solution:
  • Given: d = 2000 km, Radius of earth = R = 6400 km, g = 9.8 m/s2.
  • To find:   acceleration due to gravity =  gd =?

Variation in acceleration due to gravity

Ans: The acceleration at depth of 2000 km below the surface of the earth is 6,738  m/s2

Example – 07:

  • What would be the duration of the year if the distance of the Earth from the sun gets tripled?
  • Given: r1= R, r2= 32R , T1 = 1 year,
  • To find: T2 =?
  • Solution:

By Keppler’s law
Ans: The duration of the year will be 5.196 years.

Example – 08:

  • Calculate the period of revolution of the planet Jupiter around the Sun. The ratio of the radius of Jupiter’s orbit to that of earth’s orbit around the Sun is 5.2.
  • Solution:
  • Given: rJ : r = 5.2 , Time period of the Earth  T1 = 1 year.
  • To Find:  Period of Jupiter TJ =?

By Keppler’s law, we have T2 ∝ r3

Gravitation 10



Ans: The period of revolution of planet Jupiter is 11.86 years.

Example – 09:

  • Calculate the escape velocity of a body from the surface of the planet of diameter 2200 km. Acceleration due to gravity on the surface of the planet is 160 cm/s2.
  • Given diameter of planet = 2200 km, radius of planet = 1100 km= 1.1 x 106 m, Acceleration due to gravity = g = 160 cm/s2 = 1.6 m/s2.
  • To Find: Escape velocity = Ve =?

Ans: The escape velocity on the surface of the planet is 1.876 km/s

Example – 10:

  • What are the (1) KE (2) PE (3) total energy and (4) binding energy of an artificial satellite of mass 1000 kg orbiting at a height of 3600 km above the surface of the earth? Mass of the earth = 6 x 1024 kg; Radius of earth = 6400 km, G = 6.67 × 10-11 N m2/kg2,
  • Given: Mass of satellite = m = 1000 kg, G = 6.67 × 10-11 N m2/kg2, radius of earth = R = 6400 km; M = 6 x 1024 kg. h = 3600 km, r = R + h = 6400 + 3600 = 10000 km = 10 x 106 m;
  • To Find: K.E. = ?, P.E. = ?, T.E. = ?, B.E. = ?,
  • Solution:



Now Total energy = – B.E. = – 2.0 x 109 J
Now Potential Energy = – 2 x B.E. = – 2 x 2.0 x 109 J = 4.0 x 109 J
Kinetic energy = B.E. = 2.0 x 109 J
Ans: The kinetic energy of satellite = 2.0 x 109 J
Potential energy of satellite = – 4.0 x 109 J
Total energy of satellite = – 2.0 x 109 J
The binding energy of satellite = 2.0 x 109 J

Example – 11:

  • A body weights  4.5 kg wt on the surface of the earth. How much will it weigh on the surface of a planet whose mass is 1/9 that of earth and whose radius is half that of earth
  • Solution:
  • Given: , Weight of body on earth = WE = 4.5 kg wt, mass of planet = 1/9 mass of earth i.e MP = 1/9 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 – 12:

  • The mass of mars is 1/9 that of earth and the radius is 1/2 that of earth. What is the weight of the body having mass of 1000 kg on earth.
  • Solution:
  • Given: , mass of body on earth = mE =1000 kg, mass of mars = 1/9 mass of earth i.e MP = 1/9 ME, radius of mars = 1/2 radius of earth i.e.  RP = 1/2 RE.
  • To find: Weight of body on the mars = WP = ?

Weight on earth = 1000 x 9.8 = 9800 N

Mass of the body remains constant



Ans: The weight of the body on the surface of the mars is 4356 N

Example – 13:

  • What is the decrease in the weight of a body of mass 500 kg when it is taken into a mine of depth 1000 m? R = 6400 km, g = 9.8 m/s2.
  • Solution:
  • Given: Mass of body = m = 500 kg, depth d = 1000 m = 1 km, Radius of earth = R = 6400 km, g = 9.8 m/s2.
  • To find:   decrease in weight = W – Wd =?

Ans: The decrease in the weight of the body is 1 N.

Example – 14:

  • Find the weight of a body of mass 100 kg on the earth at a) equator b) pole c) latitude of 30°. R = 6400 km, g = 9.8 m/s2
  • Solution:
  • Given: m =  100  kg,  R = 6400  km = 6.4 × 106 m,
  • To find: WE = ?, WP = ? WΦ = ?

Time period of earth = 24 hours = 24 x 60 x 60 s

The acceleration due to gravity at latitude Φ is given by

gΦ = g – Rω2cos2Φ

The Weight of body at latitude Φ is given by

WΦ = mgΦ = mg – mRω2cos2Φ

At equator Φ = 0°

WE = 100 x 9.8  – 100 x 6.4 × 10x (7.273 x 10-5)2cos20

WE = 980  – 100 x 6.4 × 10x (7.273 x 10-5)x (1)2

WE = 980  – 3.386 = 976.6 N

At poles Φ =90°

WE = 100 x 9.8  – 100 x 6.4 × 10x (7.273 x 10-5)2cos290

WE = 980  – 100 x 6.4 × 10x (7.273 x 10-5)x (0)2

WE = 980  – 0 = 980 N

At latitude Φ = 30°

WE = 100 x 9.8  – 100 x 6.4 × 10x (7.273 x 10-5)2cos230

WE = 980  – 100 x 6.4 × 10x (7.273 x 10-5)x (0.866)2

WE = 980  – 2.539 = 977.5 N

Ans: Weight of the body on equator, on ploe and on latitude 30° are

976.6 N, 980 N and 977.5 N respectively.

Example – 15:

  • Taking G = 6.67 × 10-11 N m2/kg2, the radius of the earth as 6400 km and mean density of earth as 5478 kg/m3, calculate g at the surface of the earth.
  • Solution:
  • Given: Radius of the Earth = R = 6400 km = 6.4 × 10m,  Density of material of earth = ρ = 5478 kg/m3, G = 6.67 × 10-11 N m2/kg2
  • .To find: Acceleeartion due to gravityg = ?

Ans: Acceleration due to gravity = 9.8 m/s2.

Example – 16:

  • A playful astronaut releases a bowling ball of mass 500 g into circular orbit about an altitude of 600 km. What is the mechanical energy of the ball in its orbit. radius of the earth = 6400 km, mass of earth = 6 x 1024 kg.
  • Given: mass of a ball = 500 g = 0.5 kg, height of satellite above the surface of the earth = 600 km, Radius of the earth = 6400 km, radius of orbit = 6400 + 600 = 7000 km = 7 x 106 m, mass of earth = 6 x 1024 kg.
  • To Find: Mechanical energy of ball = ET = ?
  • Solution:

Now total energy of satellite = T.E. = – B.E. = – 1.43 x 107 J

Ans: The mechanical energy of the ball in its orbit is – 1.43 x 107 J

Example – 17:

  • Find the escape velocity from the earth for a 1000 kg spacecraft and find the kinetic energy it must have at the surface of the earth in order to escape the earth’s gravitational field.
  • Given: Mass of spacecraft = m = 1000 kg,
  • To Find: Escape velocity ve = ?, kinetic energy = K.E. = ?
  • Solution:

The escape velocity on the surface of the earth is given by

K.E. = ½ mv2 = ½ (1000)(11.2 x 103)2

K.E. ½  x 103 x 11.22 x 106

K.E.= 6.261 x 1010 J

Ans: Escape velocity is 11.2 km/s and kinetic energy required is  6.261 x 1010 J

Science > Physics > Gravitation > You are Here
Physics Chemistry Biology Mathematics

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