Gravitation — Page 2 | JEE Physics

Q11

The approximate height from the surface of earth at which the weight of the body becomes

{1 \over 3}

of its weight on the surface of earth is : [Radius of earth R = 6400 km and

\sqrt 3

= 1.732]

A 3840 km

B 4685 km

C 2133 km

D 4267 km

Correct Answer

Option B

Solution

According to the given information

{{GM} \over {{{(R + h)}^2}}} = {1 \over 3} \times {{GM} \over {{R^2}}}

\Rightarrow R + h = \sqrt 3 R

\Rightarrow h = (\sqrt 3 - 1)R \simeq 4685

km

Q12

An object of mass

1 \mathrm{~kg}

is taken to a height from the surface of earth which is equal to three times the radius of earth. The gain in potential energy of the object will be [If,

\mathrm{g}=10 \mathrm{~ms}^{-2}

and radius of earth

=6400 \mathrm{~km}

]

A 48 MJ

B 24 MJ

C 36 MJ

D 12 MJ

Correct Answer

Option A

Solution

$\Delta U=U_{f}-U_{i}$ $\begin{aligned} &=-\dfrac{G M m}{4 R}+\dfrac{G M m}{R} \\\\ &=\dfrac{3 G M m}{4 R}=\dfrac{3}{4} m g R \\\\ &=48 \mathrm{MJ} \end{aligned}$

Q13

Suppose that the angular velocity of rotation of earth is increased. Then, as a consequence :

A Weight of the object, everywhere on the earth, will increase.

B Weight of the object, everywhere on the earth, will decrease.

C There will be no change in weight anywhere on the earth.

D Except at poles, weight of the object on the earth will decrease.

Correct Answer

Option D

Solution

With rotation of earth the effect on acceleration due to gravity vary as g' = g $-$ $\omega$ 2 R cos2 $\theta$ Here $\theta$ is lattitude.

At poles $\theta$ = 90o so those will be no change in gravity.

At all other point $\omega$ will increase so g' will decrease.

Q14

A solid sphere of mass 'M' and radius 'a' is surrounded by a uniform concentric spherical shell of thickness 2a and mass 2M. The gravitational field at distance '3a' from the centre will be :

A

{{GM} \over {3{a^2}}}

B

{{2GM} \over {9{a^2}}}

C

{{GM} \over {9{a^2}}}

D

{{2GM} \over {3{a^2}}}

Correct Answer

Option A

Solution

We use Gauss’s Law for gravitation g.4 $\pi$ r2 = (Mass enclosed) 4 $\pi$ G

g = {{3M4\pi G} \over {4\pi {{(3a)}^2}}} = {{GM} \over {3{a^2}}}

Q15

If

g

is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass

m

raised from the surface of the earth to a height equal to the radius

R

of the earth is

A

{1 \over 4}mgR

B

2mgR

C

{1 \over 2}mgR

D

mgR

Correct Answer

Option C

Solution

Gravitational potential energy on the earth surface of a body U =

-{{GmM} \over R}

And at the height h from the earth surface the potential energy

{U_h} = - {{GmM} \over {R + h}}

=

- {{GmM} \over {2R}}

[ as h = R ] So the gain in the potential energy

\Delta U = {U_h} - U

$\therefore$

\Delta U = {{ - GmM} \over {2R}} + {{GmM} \over R};

$\Rightarrow$

\Delta U = {{GmM} \over {2R}}

Now

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

\,\,\,

$\therefore$

{\mkern 1mu} {{GM} \over R} = gR

$\therefore$

\Delta U = {1 \over 2}mgR

Q16

If suddenly the gravitational force of attraction between Earth and a satellite revolving around it becomes zero, then the satellite will

A continue to move in its orbit with same velocity

B move tangentially to the original orbit with the same velocity

C become stationary in its orbit

D move towards the earth

Correct Answer

Option B

Solution

When gravitational force of attraction between Earth and a satellite revolving around it becomes zero, then the centripetal force becomes zero.

So the satellite will move tangentially to the original orbit with the same velocity as it has at the instant when gravitational force becomes zero.

Q17

Energy required to move a body of mass

m

from an orbit of radius

2R

to

3R

is

A

{{GMm} \over {12{R^2}}}

B

{{GMm} \over {3{R^2}}}

C

{{GMm} \over {8R}}

D

{{GMm} \over {6R}}

Correct Answer

Option D

Solution

Gravitational potential energy E =

- {{GMm} \over r}

where M = mass of earth m = mass of body r = radius of earth Energy required to move a body of mass

m

from an orbit of radius

2R

to

3R

$=$ (Potential energy of the Earth-mass system when mass is at distance

3R

) $-$ (Potential energy of the Earth-mass system when mass is at distance

2R

)

= {{ - GMm} \over {3R}} - \left( {{{ - GMm} \over {2R}}} \right)

= {{ - GMm} \over {3R}} + {{GMm} \over {2R}}

= {{ - 2GMm + 3GMm} \over {6R}} = {{GMm} \over {6R}}

Q18

The time period of satellite of earth is

5

hours. If the separation between the earth and the satellite is increased to

4

times the previous value, the new time period will become

A

10

hours

B

80

hours

C

40

hours

D

20

hours

Correct Answer

Option C

Solution

According to kepler's law,

{T^2} \propto {R^3}

$\therefore$

{{T_1^2} \over {T_2^2}} = {{R_1^3} \over {R_2^3}}

$\Rightarrow$

{T_2} = {T_1}{\left( {{{{R_2}} \over {{R_1}}}} \right)^{{{3}/{2}}}} = 5 \times {\left[ {{{4R} \over R}} \right]^{{{3}/{2}}}}

= 5 \times {2^3} = 40\,\,

hour

Q19

The escape velocity for a body projected vertically upwards from the surface of earth is

11

km/s.

If the body is projected at an angle of

{45^ \circ }

with the vertical, the escape velocity will be

A

11\sqrt 2 \,\,km/s

B

22

km/s

C

11

km/s

D

{{11} \over {\sqrt 2 }}km/s

Correct Answer

Option C

Solution

We know, Escape velocity,

{v_e} = \sqrt {2gR}

So the escape velocity is independent of the angle at which the body is projected, hence it will remain same as 11 km/s.

Q20

The time period of an earth satellite in circular orbit is independent of

A both the mass and radius of the orbit

B radius of its orbit

C the mass of the satellite

D neither the mass of the satellite nor the radius of its orbit

Correct Answer

Option C

Solution

For satellite, gravitational force = centripetal force $\therefore$

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

x=

height of satellite from earth surface

m=

mass of satellite

\Rightarrow {v^2} = {{GM} \over {\left( {R + x} \right)}}

or

v = \sqrt {{{GM} \over {R + x}}}

We know,

T = {{2\pi } \over \omega }

T = {{2\pi \left( {R + x} \right)} \over v}

[ as

\omega = {v \over r}

]

= {{2\pi \left( {R + x} \right)} \over {\sqrt {{{GM} \over {R + x}}} }}

which is independent of mass of satellite