Gravitation Questions — JEE Physics

Q1

The masses and radii of the earth and moon are (M1, R1) and (M2, R2) respectively. Their centres are at a distance 'r' apart. Find the minimum escape velocity for a particle of mass 'm' to be projected from the middle of these two masses :

A

V = {1 \over 2}\sqrt {{{4G({M_1} + {M_2})} \over r}}

B

V = \sqrt {{{4G({M_1} + {M_2})} \over r}}

C

V = {1 \over 2}\sqrt {{{2G({M_1} + {M_2})} \over r}}

D

V = {{\sqrt {2G} ({M_1} + {M_2})} \over r}

Correct Answer

Option B

Solution

{1 \over 2}m{V^2} - {{G{M_1}m} \over {r/2}} - {{G{M_2}m} \over {r/2}} = 0

{1 \over 2}m{V^2} = {{2Gm} \over r}({M_1} + {M_2})

V = \sqrt {{{4G({M_1} + {M_2})} \over r}}

Option (b)

Q2

The kinetic energy needed to project a body of mass

m

from the earth surface (radius

R

) to infinity is

A

mgR/2

B

2mgR

C

mgR

D

mgR/4

Correct Answer

Option C

Solution

The required velocity is called escape velocity (

{v_e}

) to leave the earth surface of a body.

{v_e} =

escape velocity

= \sqrt {2gR}

Kinetic Energy

K.E = {1 \over 2}mv_e^2

$\therefore$

K.E = {1 \over 2}m \times 2gR = mgR

Q3

A satellite is in an elliptical orbit around a planet P. It is observed that the velocity of the satellite when it is farthest from the planet is 6 times less than that when it is closest to the planet. The ratio of distances between the satellite and the planet at closest and farthest points is:

A 1 : 2

B 1 : 3

C 1 : 6

D 3 : 4

Correct Answer

Option C

Solution

By angular momentum conservation Li = Lf rminvmax = rmaxvmin .....(i) Given, vmax = 6vmin From equation (i),

{{{r_{\min }}} \over {{r_{\max }}}}

=

{{{v_{\min }}} \over {{v_{\max }}}} = {1 \over 6}

Q4

The mass density of a planet of radius R varies with the distance r from its centre as

\rho

(r) =

{\rho _0}\left( {1 - {{{r^2}} \over {{R^2}}}} \right)

. Then the gravitational field is maximum at :

A

r = {1 \over {\sqrt 3 }}R

B r = R

C

r = \sqrt {{3 \over 4}} R

D

r = \sqrt {{5 \over 9}} R

Correct Answer

Option D

Solution

dm = $\rho$ dv

\int {dm}

= \int {{\rho _0}} 4\pi {r^2}dr

$\Rightarrow$ M =

\int {{\rho _0}} 4\pi {r^2}dr

Also E =

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

\Rightarrow E{r^2} = 4\pi G\,\int\limits_0^r {{\rho _0}} \left( {1 - {{{r^2}} \over {{R^2}}}} \right){r^2}dr

\Rightarrow E = 4\pi G{\rho _0}\left( {{{{r}} \over 3} - {{{r^3}} \over {5{R^2}}}} \right)

For maximum E,

{{dE} \over {dr}} = 0

{1 \over 3} - {{3{r^2}} \over {5{R^2}}}

= 0 $\therefore$

r = \sqrt {{5 \over 9}} R

Q5

A geostationary satellite is orbiting around an arbitrary planet 'P' at a height of 11R above the surface of 'P', R being the radius of 'P'. The time period of another satellite in hours at a height of 2R from the surface of 'P' is _________. 'P' has the time period of 24 hours.

A 3

B 5

C

6\sqrt 2

D

{6 \over {\sqrt 2 }}

Correct Answer

Option A

Solution

From Kepler's law

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

{\left( {{{24} \over T}} \right)^2} = {\left( {{{12R} \over {3R}}} \right)^3}

T = 3 sec

Q6

The percentage decrease in the weight of a rocket, when taken to a height of

32 \mathrm{~km}

above the surface of earth will, be :

(

Radius of earth

=6400 \mathrm{~km})

A 1%

B 3%

C 4%

D 0.5%

Correct Answer

Option A

Solution

$\because$

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

\Rightarrow {{\Delta g} \over g} = 2{{\Delta r} \over r}

\Rightarrow {{\Delta g} \over g} \times 100 = 2 \times {{32} \over {6400}} \times 100\% = 1\%

$\Rightarrow$ % decrease in weight = 1%

Q7

For a body projected at an angle with the horizontal from the ground, choose the correct statement.

A Gravitational potential energy is maximum at the highest point.

B The vertical component of momentum is maximum at the highest point.

C The horizontal component of velocity is zero at the highest point.

D The Kinetic Energy (K.E.) is zero at the highest point of projectile motion.

Correct Answer

Option A

Solution

At highest point height is maximum and vertical component of velocity is zero.

So momentum is zero.

At highest point horizontal component of velocity will not be zero but vertical component of velocity is equal to zero and because of this K.E. will not be equal to zero.

Gravitational potential energy is maximum at highest point and equal to $\mathrm{mgH}=

mg\left( {{{{u^2}{{\sin }^2}\theta } \over {2g}}} \right)

Q8

The acceleration due to gravity on the surface of earth is

\mathrm{g}

. If the diameter of earth reduces to half of its original value and mass remains constant, then acceleration due to gravity on the surface of earth would be :

A g/4

B 2g

C g/2

D 4g

Correct Answer

Option D

Solution

\begin{aligned} & g=\frac{G M}{R^2} \Rightarrow g \propto \frac{1}{R^2} \\ & \frac{g_2}{g_1}=\frac{R_1^2}{R_2^2} \\ & g_2=4 g_1\left(R_2=\frac{R_1}{2}\right) \end{aligned}

Q9

Two planets

A

and

B

having masses

m_1

and

m_2

move around the sun in circular orbits of

r_1

and

r_2

radii respectively. If angular momentum of

A

is

L

and that of

B

is

3 \mathrm{~L}

, the ratio of time period

\left(\dfrac{T_A}{T_B}\right)

is:

A

\left(\dfrac{r_2}{r_1}\right)^{\dfrac{3}{2}}

B

27\left(\dfrac{m_1}{m_2}\right)^3

C

\left(\dfrac{r_1}{r_2}\right)^3

D

\dfrac{1}{27}\left(\dfrac{m_2}{m_1}\right)^3

Correct Answer

Option D

Solution

\begin{aligned} & \frac{v_1}{v_2}=\sqrt{\frac{r_2}{r_1}} \quad \text{.... (i)}\\ & m_1 v_1 r_1=h \\ & m_2 v_2 r_2=32 \\ & \Rightarrow \frac{v_1}{v_2}=\frac{1}{3} \frac{m_2}{m_1} \frac{r_2}{r_1} \quad \text{.... (ii)} \end{aligned}

From (i) & (ii)

\begin{aligned} & \sqrt{\frac{r_2}{r_1}}=\frac{1}{3} \frac{m_2}{m_1} \frac{r_2}{r_1} \\ & \frac{3 m_1}{m_2}=\sqrt{\frac{r_2}{r_1}} \\ & \frac{T_1}{T_2}=\left(\frac{r_1}{r_2}\right)^{3 / 2}=\left(\frac{m_2}{3 m_1}\right)=\frac{1}{27}\left(\frac{m_2}{m_1}\right)^3 \end{aligned}

Q10

The distance between Sun and Earth is R. The duration of year if the distance between Sun and Earth becomes 3R will be :

A

\sqrt 3

years

B 3 years

C 9 years

D 3

\sqrt 3

years

Correct Answer

Option D

Solution

We know that T2 $\propto$ R3

\Rightarrow {\left( {{{T'} \over T}} \right)^2} = {\left( {{{3R} \over R}} \right)^3}

\Rightarrow {{T'} \over T} = 3\sqrt 3

\Rightarrow T' = 3\sqrt 3

years