Q: A body weighs 48 N on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to one-third the radius of the earth from its surface is :

aligned & W ^ = mg ^ = mg ( R R + h )^2 \\ & =48 ( R R + R 3 )^2 \\ & =48 9 16 =27 ~N aligned

Q: The mass of a planet is 1 10 th that of the earth and its diameter is half that of the earth. The acceleration due to gravity on that planet is:

aligned g^ = G M^ R^2 & = G M 10 ( R 2 )^2 \\ & = 4 10 G M R^2 =0.4 9.8 \\ & =3.92 ~m ~s ^ -2 aligned

Q: The minimum energy required to launch a satellite of mass m from the surface of earth of mass M and radius R in a circular orbit at an altitude of 2 R from the surface of the earth is:

Apply energy conservation, gathered U_i+K_i=U_f+K_f \\ - G M m R +K_i=- G M m 3 R + 1 2 m v^2 \\ - G M m R +K_i=- G M m 3 R + 1 2 m G M 3 R \\ K_i=- 1 6 G M m R + G M m R \\ K_i= 5 6 G M m R gathered

Question 1

The amount of work done to raise a mass 'm' from the surface of the Earth to a height equal to the radius of the Earth 'R', will be :

Accepted Answer

Initial potential energy at the surface of the Earth is U_i = - GMm R Final potential energy at height h = R is U_f = - GMm R + R = - GMm 2R Work done W = U = U_f - U_i W = - GMm 2R - (- GMm R ) = GMm 2R Using the relation g = GM R^2 , we get GM = gR^2 Substituting this in the expression for work done: W = gR^2 m 2R = mg R 2 Answer: mg R 2

Question 2

Two planets P_1 and P_2 with equal mass have radii R_1 and R_2, respectively, where R_2= R_1 2 . The escape speeds of P_1 and P_2 are v_1 and v_2, respectively. Then v_2 v_1 is :

Accepted Answer

The escape speed of a planet of mass M and radius R is given by v = 2GM R . For planet P_1, the escape speed is v_1 = 2GM R_1 . For planet P_2, the escape speed is v_2 = 2GM R_2 . Taking the ratio of the escape speeds: v_2 v_1 = R_1 R_2 Given that R_2 = R_1 2 , substituting this into the equation gives: v_2 v_1 = R_1 R_1 2 = 2 Answer: 2

Question 3

In a solar system, the time-period of revolution of a planet tracing a circular orbit of radius R is proportional to :

Accepted Answer

According to Kepler's third law of planetary motion, the square of the time period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of its elliptical orbit. For a circular orbit, the semi-major axis is equal to the radius of the orbit R. Therefore, T^2 R^3 Taking the square root on both sides, we get: T R^ 3/2 Alternatively, equating the gravitational force to the required centripetal force: GMm R^2 = mv^2 R v = GM R The time period T is given

Question 4

The radius of Martian orbit around the Sun is about 4 times the radius of the orbit of Mercury. The Martian year is 687 Earth days. Then which of the following is the length of 1 year on Mercury?

Accepted Answer

aligned & T _ Mercury T _ Martian = ( r _ Mercury r _ Martian )^ 3 / 2 = ( 1 4 )^ 3 / 2 = 1 8 \ & ~T _ Mercury = T _ Martian 8 = 687 8 =85.875 days 88 days aligned

Question 5

A body weighs 48 N on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to one-third the radius of the earth from its surface is :

Accepted Answer

aligned & W ^ = mg ^ = mg ( R R + h )^2 \ & =48 ( R R + R 3 )^2 \ & =48 9 16 =27 ~N aligned

Question 6

The escape velocity for earth is v. A planet having 9 times mass that of earth and radius, 16 times that of earth, has the escape velocity of:

Accepted Answer

Escape velopity of object from planet is given by (v_e )_p= 2 G M_p R_p (v_e )_p M_p R_p Now, M_p=9 M_e and R_p=16 R_e (given) (v_e )_p (v_e )_e = M_p R_p R_e M_e = 9 M_e 16 R_e R_e M_e = 9 16 = 3 4 (v_e )_p= 3 4 v

Question 7

The mass of a planet is 1 10 th that of the earth and its diameter is half that of the earth. The acceleration due to gravity on that planet is:

Accepted Answer

aligned g^ = G M^ R^2 & = G M 10 ( R 2 )^2 \ & = 4 10 G M R^2 =0.4 9.8 \ & =3.92 ~m ~s ^ -2 aligned

Question 8

An object of mass 100 kg falls from point A to B as shown in figure. The change in its weight, corrected to the nearest integer is ( R_E is the radius of the earth)

Accepted Answer

M g^ =M g R^2 (R+h)^2 At A M g^ =M g R^2 (R+2 R)^2 = M g 9 At B M g^ =M g R (R+ 3 R 2 )^2 = M g 4 25 Change in weight =M g 4 25 - M g 9 =49 ~N

Question 9

The minimum energy required to launch a satellite of mass m from the surface of earth of mass M and radius R in a circular orbit at an altitude of 2 R from the surface of the earth is:

Accepted Answer

Apply energy conservation, gathered U_i+K_i=U_f+K_f \ - G M m R +K_i=- G M m 3 R + 1 2 m v^2 \ - G M m R +K_i=- G M m 3 R + 1 2 m G M 3 R \ K_i=- 1 6 G M m R + G M m R \ K_i= 5 6 G M m R gathered

Gravitation — Previous Year Questions & Solutions