Gravitation — Page 3 | JEE Physics

Q21

Suppose the gravitational force varies inversely as the nth power of distance. Then the time period of a planet in circular orbit of radius

R

around the sun will be proportional to

A

{R^n}

B

{R^{\left( {{{n - 1} \over 2}} \right)}}

C

{R^{\left( {{{n + 1} \over 2}} \right)}}

D

{R^{\left( {{{n - 2} \over 2}} \right)}}

Correct Answer

Option C

Solution

For moving a planet around the sun in the circular orbit, The necessary centripetal force = Gravitational force exerted on it $\therefore$

{{m{v^2}} \over r} = {{GMm} \over {{R^n}}}

$\Rightarrow$

v = \sqrt {{{GM} \over {{R^{n - 1}}}}}

We know,

T = {{2\pi R} \over v}

= 2\pi R \times \sqrt {{{{R^{n - 1}}} \over {GM}}}

=

2\pi \times \sqrt {{{{R^2} \times {R^{n - 1}}} \over {GM}}}

=

2\pi \times {{{R^{{{n + 1} \over 2}}}} \over {\sqrt {GM} }}

$\therefore$

T \propto {R^{{{ \left( {n + 1} \right)} \over 2}}}

Q22

A particle of mass

10

g

is kept on the surface of a uniform sphere of mass

100

kg

and radius

10

cm.

Find the work to be done against the gravitational force between them to take the particle far away from the sphere (you may take

G

= 6.67 \times {10^{ - 11}}\,\,N{m^2}/k{g^2}

)

A

3.33 \times {10^{ - 10}}\,J

B

13.34 \times {10^{ - 10}}\,J

C

6.67 \times {10^{ - 10}}\,J

D

6.67 \times {10^{ - 9}}\,J

Correct Answer

Option C

Solution

We know, Work done = Difference in potential energy $\therefore$

W = \Delta U = {U_f} - {U_i} = 0 - \left[ {{{ - GMm} \over R}} \right]

$\Rightarrow$

W = {{6.67 \times {{10}^{ - 11}} \times 100} \over {0.1}} \times {{10} \over {1000}}

= 6.67 \times {10^{ - 10}}J

Q23

The change in the value of

g

at a height

h

above the surface of the earth is the same as at a depth

d

below the surface of earth. When both

d

and

h

are much smaller than the radius of earth, then which one of the following is correct?

A

d = {{3h} \over 2}

B

d = {h \over 2}

C

d = h

D

d = 2\,h

Correct Answer

Option D

Solution

At height h acceleration due to gravity,

{g_h} = g\left[ {1 - {{2h} \over R}} \right];

At depth d acceleration due to gravity,

{g_d} = g\left[ {1 - {d \over R}} \right]

According to the question,

{g_h} = {g_d}

$\therefore$

g\left[ {1 - {{2h} \over R}} \right]

=

g\left[ {1 - {d \over R}} \right]

\Rightarrow {{2hg} \over R} = {{dg} \over R}

$\Rightarrow$

d=2h

Q24

Average density of the earth

A is a complex function of

g

B does not depend on

g

C is inversely proportional to

g

D is directly proportional to

g

Correct Answer

Option D

Solution

Mass of earth = Volume $\times$ Density of earth( $\rho$ ) $\therefore$ M =

{{4 \over 3}\pi {R^3}}

$\times$ $\rho$ We know,

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

\Rightarrow g = {{G \times \rho \times {4 \over 3}\pi {R^3}} \over {{R^2}}}

$\Rightarrow$

g = {4 \over 3}\rho \pi GR

$\therefore$

g \propto \rho

or

\rho \propto g

Q25

If

{g_E}

and

{g_M}

are the accelerations due to gravity on the surfaces of the earth and the moon respectively and if Millikan's oil drop experiment could be performed on the two surfaces, one will find the ratio

{{electro\,\,ch\arg e\,\,on\,\,the\,\,moon} \over {electronic\,\,ch\arg e\,\,on\,\,the\,\,earth}}\,\,to\,be

A

{g_M}/{g_E}

B

1

C

0

D

{g_E}/{g_M}

Correct Answer

Option B

Solution

electronic charge does does not depend on acceleration due to gravity as it is a universal constant.

So, electronic charge on earth $=$ electronic charge on moon $\therefore$ Required ratio

=1.

Q26

A planet in a distant solar system is

10

times more massive than the earth and its radius is

10

times smaller. Given that the escape velocity from the earth is

11\,\,km\,{s^{ - 1}},

the escape velocity from the surface of the planet would be

A

1.1\,\,km\,{s^{ - 1}}

B

100\,\,km\,{s^{ - 1}}

C

110\,\,km\,{s^{ - 1}}

D

0.11\,\,km\,{s^{ - 1}}

Correct Answer

Option C

Solution

Let Me is mass of earth then mass of planet Mp = 10Me. And let Re is radius of earth then radius of planet Rp =

{{{R_e}} \over {10}}

Escape velocity of earth,

{v_e} = \sqrt {{{2G{M_e}} \over {{R_e}}}}

Escape velocity of planet,

{v_p} = \sqrt {{{2G{M_p}} \over {{R_p}}}}

$\therefore$

{{{{ {{v_p}}}}} \over {{{ {{v_e}} }}}} = {{\sqrt {{{2G{M_p}} \over {{R_p}}}} } \over {\sqrt {{{2G{M_e}} \over {{R_e}}}} }}

= \sqrt {{{{M_p}} \over {{M_e}}} \times {{{{\mathop{\rm R}\nolimits} _e}} \over {{R_p}}}}

= \sqrt {{{10{M_e}} \over {{M_e}}} \times {{{{\mathop{\rm R}\nolimits} _e}} \over {{{\mathop{\rm R}\nolimits} _e}/10}}} = 10

$\therefore$

{{v_p}} = 10 \times {{v_e}}

= 10 \times 11 = 110\,km/s

Q27

This question contains Statement -

1

and Statement -

2

. of the four choices given after the statements, choose the one that best describes the two statements. Statement -

1

: For a mass

M

kept at the center of a cube of side

'a'

, the flux of gravitational field passing through its sides

4\,\pi \,GM.

Statement - 2: If the direction of a field due to a point source is radial and its dependence on the distance

'r'

from the source is given as

{1 \over {{r^2}}},

its flux through a closed surface depends only on the strength of the source enclosed by the surface and not on the size or shape of the surface.

A Statement -

1

is false, Statement -

2

is true

B Statement -

1

is true, Statement -

2

is true; Statement -

2

is a correct explanation for Statement -

1

C Statement -

1

is true, Statement -

2

is true; Statement -

2

is not a correct explanation for Statement -

1

D Statement -

1

is true, Statement -

2

is false

Correct Answer

Option B

Solution

Gravitational field

\overrightarrow g

=

- {{GM} \over {{r^2}}}

where,

M=

mass enclosed in the closed surface Gravitational flux through a closed surface is given by

{\left| {\overrightarrow g .d\overrightarrow S } \right|}

=

4\pi {r^2}.{{GM} \over {{r^2}}}

=

4\pi GM

So Statement - 1 is correct.

Statement - 2 is also correct because when the shape of the earth is spherical, area of the Gaussian surface is

4\pi {r^2}

. This proves inverse square law.

Q28

Two bodies of masses

m

and

4

m

are placed at a distance

r.

The gravitational potential at a point on the line joining them where the gravitational field is zero is:

A

- {{4Gm} \over r}

B

- {{6Gm} \over r}

C

- {{9Gm} \over r}

D zero

Correct Answer

Option C

Solution

Let the gravitational field at

P,

distant

x

from mass

m,

be zero. $\therefore$

{{Gm} \over {{x^2}}} = {{4Gm} \over {{{\left( {r - x} \right)}^2}}}

\Rightarrow 4{x^2} = {\left( {r - x} \right)^2}

\Rightarrow 2x = r - x

\Rightarrow x = {r \over 3}

Gravitational potential at point

P,

V = - {{Gm} \over {{r \over 3}}} - {{4Gm} \over {{{2r} \over 3}}} = -{{9Gm} \over r}

Q29

The mass of a spaceship is

1000

kg.

It is to be launched from the earth's surface out into free space. The value of

g

and

R

(radius of earth ) are

10\,m/{s^2}

and

6400

km

respectively. The required energy for this work will be:

A

6.4 \times {10^{11}}\,

Joules

B

6.4 \times {10^8}\,

Joules

C

6.4 \times {10^9}\,

Joules

D

6.4 \times {10^{10}}\,

Joules

Correct Answer

Option D

Solution

Potential energy at earth surface =

- {{GMm} \over R}

and at free space potential energy = 0 Work done for this =

0 - \left( { - {{GMm} \over R}} \right)

=

{{{GMm} \over R}}

So the required energy for this work is =

{{GMm} \over R}

=

{{{g{R^2}m} \over R}}

[ as

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

] =

mgR

= 1000 \times 10 \times 6400 \times {10^3}

= 6.4 \times {10^{10}}\,\,

Joules

Q30

What is the minimum energy required to launch a satellite of mass

m

from the surface of a planet of mass

M

and radius

R

in a circular orbit at an altitude of

2R

?

A

{{5GmM} \over {6R}}

B

{{2GmM} \over {3R}}

C

{{GmM} \over {2R}}

D

{{GmM} \over {3R}}

Correct Answer

Option A

Solution

Energy of the satellite on the surface of the planet Ei = K.E + P.E = 0 +

\left( { - {{GMm} \over R}} \right)

=

{ - {{GMm} \over R}}

Energy of the satellite at 2R distance from the surface of the planet while moving with velocity v Ef =

{1 \over 2}m{v^2}

+

\left( { - {{GMm} \over {R + 2R}}} \right)

In the orbital of planet, the centripetal force is provided by the gravitational force $\therefore$

{{m{v^2}} \over {R + 2R}} = {{GMm} \over {{{\left( {R + 2R} \right)}^2}}}

\Rightarrow {v^2} = {{GM} \over {3R}}

$\therefore$ Ef =

{1 \over 2}m{v^2}

+

\left( { - {{GMm} \over {R + 2R}}} \right)

= {1 \over 2}m{{GM} \over {3R}} - {{GMm} \over {3R}}

=

- {{GMm} \over {6R}}

$\therefore$ Minimum energy required required to launch the satellite = Ef - Ei =

- {{GMm} \over {6R}}

-

\left( { - {{GMm} \over R}} \right)

=

{{5GMm} \over {6R}}