Gravitation — Page 4 | NEET Physics

Q31

The radius of a planet is twice the radius of earth. Both have almost equal average mass densities. V P and V E are escape velocities of the planet and the earth, respectively, then

A V P = 1.5 V E

B V P = 2 V E

C V E = 3 V P

D V E = 1.5 V P

Correct Answer

Option B

Solution

Here, R P = 2R E ,

{\rho _E} = {\rho _P}

Escape velocity of the earth,

{V_E} = \sqrt {{{2G{M_E}} \over {{R_E}}}} = \sqrt {{{2G} \over {{R_E}}}\left( {{4 \over 3}\pi R_E^3{\rho _E}} \right)}

= {R_E}\sqrt {{8 \over 3}\pi G{\rho _E}}

...(i) Escape velocity of the planet

{V_P} = \sqrt {{{2G{M_P}} \over {{R_P}}}} = \sqrt {{{2G} \over {{R_P}}}\left( {{4 \over 3}\pi R_P^3{\rho _P}} \right)}

= {R_P}\sqrt {{8 \over 3}\pi G{\rho _P}}

...(ii) Divide (i) by (ii), we get

{{{V_E}} \over {{V_P}}} = {{{R_E}} \over {{R_P}}}\sqrt {{{{\rho _E}} \over {{\rho _P}}}} = {{{R_E}} \over {2{R_E}}}\sqrt {{{{\rho _E}} \over {{\rho _E}}}} = {1 \over 2}

\Rightarrow {V_P} = 2{V_E}

Q32

A particle of mass 'm' is kept at rest at a height 3R from the surface of earth, where 'R' is radius of earth and 'M' is mass of earth. The minimum speed with which it should be projected , so that it does not return back, is (g is acceleration due to gravity on the surface of earth)

A

{\left( {{{GM} \over {2R}}} \right)^{1/2}}

B

{\left( {{{gR} \over 4}} \right)^{1/2}}

C

{\left( {{{2g} \over R}} \right)^{1/2}}

D

{\left( {{{GM} \over R}} \right)^{1/2}}

Correct Answer

Option A

Solution

The minimum speed with which the particle should be projected from the surface of the earth so that it does not return back is known as escape speed and it is given by

{v_e} = \sqrt {{{2GM} \over {\left( {R + h} \right)}}}

Here, h = 3R $\therefore$

{v_e} = \sqrt {{{2GM} \over {\left( {R + 3R} \right)}}} = \sqrt {{{2GM} \over {4R}}} = \sqrt {{{GM} \over {2R}}}

= \sqrt {{{gR} \over 2}}

\left( \because {g = {{GM} \over {{R^2}}}} \right)

Q33

Infinite number of bodies, each of mass 2 kg are situated on x-axis at distance 1 m, 2 m, 4 m, 8 m, . . . , respectively, from the origin. The resulting gravitational potential due to this system at the origin will be

A

- {4 \over 3}

G

B

-

4G

C

-

G

D

- {8 \over 3}

G

Correct Answer

Option B

Solution

Gravitational potential V =

{{ - Gm} \over r}

{V_0} = - {{G \times 2} \over 1} - {{G \times 2} \over 2} - {{G \times 2} \over 4} - {{G \times 2} \over 8}

=

- 2G\left[ {1 + {1 \over 2} + {1 \over 4} + {1 \over 8} + ...\infty } \right]

= - 2G \times {1 \over {1 - {1 \over 2}}}

= - 2G \times {1 \over {{1 \over 2}}} = - 4G

Q34

A body of mass 'm' is taken from the earth's surface to the height equal to twice the radius (R) of the earth. The change in potential energy of body will be

A 3mgR

B

{1 \over 3}

mgR

C mg2R

D

{2 \over 3}

mgR

Correct Answer

Option D

Solution

Gravitational potential energy at any point at a distance r from the centre of the earth is

U = - {{GMm} \over r}

where M and m be masses of the earth and the body respectively. At the surface of the earth, r = R

{U_i} = - {{GMm} \over R}

At a height h from the surface, r = R + h = R + 2R = 3R ( h = 2R (Given)) $\therefore$

{U_f} = - {{GMm} \over {3R}}

Change in potential energy,

\Delta U = {U_f} - {U_i}

= - {{GMm} \over {3R}} - \left( { - {{GMm} \over R}} \right) = {{GMm} \over R}\left( {1 - {1 \over 3}} \right)

= {2 \over 3}{{GMm} \over R} = {2 \over 3}mgR

\left( \because{g = {{GM} \over {{R^2}}}} \right)

Q35

If

{v_e}

is escape velocity and

{v_o}

is orbital velocity of a satellite for orbit close to the earth's surface, then these are related by

A

{v_o} = \sqrt {2{v_e}}

B v o

=

v e

C

{v_e} = \sqrt {2{v_o}}

D

{v_e} = \sqrt 2 {v_o}

Correct Answer

Option D

Solution

Escapevelocity,

{v_e} = \sqrt {{{2GM} \over R}}

...(i) where M and R where M and R be the mass and radius of the earth respectively.

The orbital velocity of a satellite close to the earth’s surface is

{v_0} = \sqrt {{{GM} \over R}}

...(ii) From (i) and (ii), we get

{v_e} = \sqrt 2 {v_0}

Q36

A spherical planet has a mass M P and diameter D P . A particle of mass m falling freely near the surface of this planet will experience an acceleration due to gravity, equal to

A

{{4G{M_P}} \over {D_P^2}}

B

{{G{M_P}m} \over {D_P^2}}

C

{{G{M_P}} \over {D_P^2}}

D

{{4G{M_P}m} \over {D_P^2}}

Correct Answer

Option A

Solution

Gravitational attraction force on particle B,

{F_g} = {{G{M_P}m} \over {{{\left( {{D_P}/2} \right)}^2}}}

Acceleration of particle due to gravity

a = {{{F_g}} \over m} = {{4G{M_P}} \over {D_P^2}}

Q37

The height at which the weight of a body becomes

{\left( {{1 \over {16}}} \right)^{th}}

, its weight on the surface of earth (radius R), is

A 5R

B 15R

C 3R

D 4R

Correct Answer

Option C

Solution

Let at h height, the weight of a body becomes 1/16th of its weight on the surface.

{W_h} = 16{W_s}

...(i) $\Rightarrow$

16{W_s} = {W_s}

...(ii)

{W_h} = mg = {{GMm} \over {{{\left( {{R_e} + h} \right)}^2}}}

g' = {{GM} \over {{{\left( {{R_e} + h} \right)}^2}}}

Similarly,

g = {{GM} \over {R_e^2}}

{{g'} \over g} = {{R_e^2} \over {{{\left( {{R_e} + h} \right)}^2}}} \Rightarrow g' = g{\left( {1 + {h \over R}} \right)^{ - 2}}

\Rightarrow {W_h} = 16{W_s}

\Rightarrow g' = {g \over {16}}

{1 \over {16}} = {\left( {1 + {h \over R}} \right)^{ - 2}} \Rightarrow 4 = 1 + {h \over R}

h = 3R

Q38

A geostationary satellite is orbiting the earth at a height of 5R above the surface of the earth, R being the radius of the earth. The time period of another satellite in hours at a height of 2R from the surface of the earth is

A 5

B 10

C 6

\sqrt 2

D

{6 \over {\sqrt 2 }}

Correct Answer

Option C

Solution

According to Kelpner’s law of period T 2 $\propto$ R 3

{{T_1^2} \over {T_2^2}} = {{R_1^3} \over {R_2^3}} = {{{{\left( {6R} \right)}^3}} \over {{{\left( {3R} \right)}^3}}} = 8

{{24 \times 24} \over {T_2^2}} = 8

T_2^2 = {{24 \times 24} \over 8} = 72 = 36 \times 2

$\therefore$

{T_2} = 6\sqrt 2

Q39

A particle of mass M is situated at the centre of a spherical shell of same mass and radius a. The magnitude of the gravitational potential at a point sutuated at a/2 distance from the centre, will be :

A

{{GM} \over a}

B

{{2GM} \over a}

C

{{3GM} \over a}

D

{{4GM} \over a}

Correct Answer

Option C

Solution

Here, Mass of a particle = M Mass of a spherical shell = M Radius of a spherical shell = a Let O be centre of a spherical shell.

Gravitational potential at point P due to particle at O is

{V_1} = - {{GM} \over {a/2}}

Gravitational potential at point P due to spherical shell is

{V_2} = - {{GM} \over a}

Hence, total gravitational potential at point P is V = V 1 + V 2

= - {{GM} \over {a/2}} + \left( { - {{GM} \over a}} \right) = - {{2GM} \over a} - {{GM} \over a} = - {{3GM} \over a}

\left| V \right| = {{3GM} \over a}

Q40

A particle of mass m is thrown upwards from the surface of the earth, with a velocity u. The mass and the radius of the earth are, respectively, M and R. G is gravitational constant and g is acceleration due to gravity on the surface of the earth. The minimum value of u so that the particle does not return back to earth, is

A

\sqrt {{{2GM} \over {{R^2}}}}

B

\sqrt {{{2GM} \over R}}

C

\sqrt {{{2gM} \over {{R^2}}}}

D

\sqrt {2g{R^2}}

Correct Answer

Option B

Solution

According to law of conservation of mechanical energy

{1 \over 2}m{u^2} - {{GMm} \over R} = 0

$\Rightarrow$

{u^2} = {{2GM} \over R}

$\therefore$

u = \sqrt {{{2GM} \over R}} = \sqrt {2gR}

\left(\because {g = {{GM} \over {{R^2}}}} \right)