Gravitation Questions — NEET Physics

Q1

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

A

2 m g R

B

m g \dfrac{R}{4}

C

m g R

D

m g \dfrac{R}{2}

Correct Answer

Option D

Solution

W.D. $=U_2-U_1$

\begin{aligned} & =-\frac{G M m}{R+R}-\left(\frac{-G M m}{R}\right) \\ & =-\frac{G M m}{2 R}+\frac{G M m}{R} \\ & =\frac{G M m}{2 R}=\frac{m g R}{2} \end{aligned}

Q2

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?

A 172 earth days

B 124 earth days

C 88 earth days

D 225 earth days

Correct Answer

Option C

Solution

To determine the length of one year on Mercury, we use Kepler's Third Law, which states that the square of the orbital period $T$ is proportional to the cube of the mean radius $R$ of the orbit: $T^2 \propto R^3$ Given: The radius of Mars' orbit around the Sun, $R' = 4R$ , where $R$ is the radius of Mercury's orbit.

The Martian year $T' = 687$ Earth days.

First, apply Kepler's Third Law to find the ratio of the orbital periods: $\left(\dfrac{T'}{T}\right)^2 = \left(\dfrac{R'}{R}\right)^3 = \left(\dfrac{4R}{R}\right)^3 = 4^3 = 64$ From this, we can solve for the ratio of the periods: $\dfrac{T'}{T} = 8$ This means that Mars takes 8 times longer to orbit the Sun compared to Mercury.

Therefore, the length of one year on Mercury is: $T = \dfrac{T'}{8} = \dfrac{687}{8} \approx 85.88 \text{ days}$ Hence, one year on Mercury is approximately 86 Earth days.

Q3

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:

A 32 N

B 36 N

C 16 N

D 27 N

Correct Answer

Option D

Solution

To find the gravitational force on a body at a height equal to one-third the Earth's radius from its surface, we start with the weight of the body on the Earth’s surface, which is 48 N.

The gravitational force at the surface, $W$ , is given by: $W = mg$ Where $g = \dfrac{GM}{R^2}$ .

At a height $h$ above the Earth's surface, the gravitational force $g_h$ is: $g_h = \dfrac{GM}{(R+h)^2}$ To find the weight $W_h$ at this height, the ratio of gravitational forces at height $h$ and at the surface is: $\dfrac{W_h}{W} = \dfrac{g_h}{g} = \dfrac{R^2}{(R+h)^2}$ Given $h = \dfrac{R}{3}$ , substitute $h$ into the equation: $\dfrac{W_h}{W} = \dfrac{R^2}{\left(R+\dfrac{R}{3}\right)^2} = \dfrac{R^2}{\left(\dfrac{4R}{3}\right)^2} = \dfrac{9}{16}$ Thus, the new weight $W_h$ is: $W_h = \dfrac{9}{16} \times W = \dfrac{9}{16} \times 48 \, \text{N}$ $W_h = 27 \, \text{N}$ Therefore, at this height, the gravitational force experienced by the body is 27 N.

Q4

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:

A

\dfrac{v}{3}

B

\dfrac{2 v}{3}

C

\dfrac{3 v}{4}

D

\dfrac{9 v}{4}

Correct Answer

Option C

Solution

Escape velocity of object from planet is given by

\begin{aligned} & \left(v_e\right)_p=\sqrt{\frac{2 G M_p}{R_p}} \\ & \therefore \quad\left(v_e\right)_p \propto \sqrt{\frac{M_p}{R_p}} \end{aligned}

Now,

M_p=9 M_e

and

R_\rho=16 R_e

(given)

\begin{aligned} & \frac{\left(v_e\right)_p}{\left(v_e\right)_e}=\sqrt{\frac{M_p}{R_p} \times \frac{R_e}{M_e}}=\sqrt{\frac{9 M_e}{16 R_e} \times \frac{R_e}{M_e}}=\sqrt{\frac{9}{16}}=\frac{3}{4} \\ & \Rightarrow \quad\left(v_e\right)_p=\frac{3}{4} v \end{aligned}

Q5

An object of mass

100 \mathrm{~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)

A 49 N

B 89 N

C 5 N

D 10 N

Correct Answer

Option A

Solution

M g^{\prime}=M g \frac{R^2}{(R+h)^2}

At

A

\quad M g^{\prime}=M g \frac{R^2}{(R+2 R)^2}=\frac{M g}{9}

At

B \quad M g^{\prime}=M g \frac{R}{\left(R+\frac{3 R}{2}\right)^2}=\frac{M g \cdot 4}{25}

Change in weight

=M g \frac{4}{25}-\frac{M g}{9}=49 \mathrm{~N}

Q6

The mass of a planet is

\dfrac{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:

A

19.6 \mathrm{~m} \mathrm{~s}^{-2}

B

9.8 \mathrm{~m} \mathrm{~s}^{-2}

C

4.9 \mathrm{~m} \mathrm{~s}^{-2}

D

3.92 \mathrm{~m} \mathrm{~s}^{-2}

Correct Answer

Option D

Solution

The acceleration due to gravity (g) on a planet is given by the formula:

g = G \frac{M}{R^2}

where: $G$ is the universal gravitational constant, $M$ is the mass of the planet, and $R$ is the radius of the planet.

In this question, we know that: The mass of the planet $M_p$ is $\dfrac{1}{10}$ of the mass of the earth $M_e$ , so $M_p = \dfrac{M_e}{10}$ .

The diameter of the planet is half that of earth.

Since the diameter is twice the radius, a diameter half that of earth implies the radius $R_p$ is also half the radius of the earth $R_e$ .

Therefore, $R_p = \dfrac{R_e}{2}$ .

Using the formula for acceleration due to gravity and substituting the above information:

g_p = G \frac{M_p}{R_p^2} = G \frac{\frac{M_e}{10}}{\left(\frac{R_e}{2}\right)^2}

Simplify the expression:

g_p = G \frac{M_e}{10} \cdot \frac{4}{R_e^2} = \frac{4}{10} G \frac{M_e}{R_e^2}

Knowing that $G \dfrac{M_e}{R_e^2}$ is the acceleration due to gravity on Earth $g_e \approx 9.8 \, \mathrm{m/s}^2$ , we substitute this value into the equation:

g_p = \frac{4}{10} \cdot 9.8 \, \mathrm{m/s}^2 = 3.92 \, \mathrm{m/s}^2

Thus, the acceleration due to gravity on the planet is $3.92 \, \mathrm{m/s}^2$ , which corresponds to: Option D:

3.92 \mathrm{~m/s}^2

Q7

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:

A

\dfrac{5 G m M}{6 R}

B

\dfrac{2 G m M}{3 R}

C

\dfrac{G m M}{2 R}

D

\dfrac{G m M}{3 R}

Correct Answer

Option A

Solution

Apply energy conservation,

\begin{aligned} & U_i+K_i=U_f+K_f \\ & \Rightarrow \quad-\frac{G M m}{R}+K_i=-\frac{G M m}{3 R}+\frac{1}{2} m v^2 \\ & \Rightarrow \quad-\frac{G M m}{R}+K_i=-\frac{G M m}{3 R}+\frac{1}{2} \times m \times \frac{G M}{3 R} \\ & \Rightarrow \quad K_i=-\frac{1}{6} \frac{G M m}{R}+\frac{G M m}{R} \\ & K_i=\frac{5}{6} \frac{G M m}{R} \end{aligned}

Q8

The escape velocity of a body on the earth surface is

11.2 \mathrm{~km} / \mathrm{s}

. If the same body is projected upward with velocity

22.4 \mathrm{~km} / \mathrm{s}

, the velocity of this body at infinite distance from the centre of the earth will be:

A

11.2 \sqrt{2} \mathrm{~km} / \mathrm{s}

B Zero

C

11.2 \mathrm{~km} / \mathrm{s}

D

11.2 \sqrt{3} \mathrm{~km} / \mathrm{s}

Correct Answer

Option D

Solution

V_{\infty}=\sqrt{V^2-V_e^2}

Given than

\mathrm{V}=2 \mathrm{~V}_e

So,

\begin{aligned} & \mathrm{V}_{\infty}=\sqrt{\left(2 \mathrm{~V}_e\right)^2-\mathrm{V}_e^2} \\ & \mathrm{~V}_{\infty}=\sqrt{3} \mathrm{~V}_e=11.2 \sqrt{3} \mathrm{~km} / \mathrm{s}\end{aligned}

Q9

If

\mathrm{R}

is the radius of the earth and

\mathrm{g}

is the acceleration due to gravity on the earth surface. Then the mean density of the earth will be :

A

\dfrac{\pi \mathrm{RG}}{12 \mathrm{~g}}

B

\dfrac{3 \pi R}{4 g G}

C

\dfrac{3 g}{4 \pi R G}

D

\dfrac{4 \pi \mathrm{G}}{3 g R}

Correct Answer

Option C

Solution

\begin{aligned} & g=\frac{4}{3} \pi \mathrm{GR} \rho \\ & \rho=\frac{3 g}{4 \pi \mathrm{GR}} \end{aligned}

Q10

Two bodies of mass

m

and

9 m

are placed at a distance

R

. The gravitational potential on the line joining the bodies where the gravitational field equals zero, will be (

G=

gravitational constant) :

A

-\dfrac{12 G m}{R}

B

-\dfrac{16 G m}{R}

C

-\dfrac{20 G m}{R}

D

-\dfrac{8 G m}{R}

Correct Answer

Option B

Solution

Position of Neutral point (Zero Gravitational Field)

r_{1}=\frac{\sqrt{m_{1}} R}{\sqrt{m_{1}}+\sqrt{m_{2}}}=\frac{\sqrt{m} R}{\sqrt{m}+\sqrt{9 m}}=\frac{R}{4}

\mathrm{r}_{2}=\mathrm{R}-\mathrm{R} / 4=3 \mathrm{R} / 4

Now Gravitational potential at point

\mathrm{P}

\begin{aligned} & V_{P}=-\frac{G M}{R / 4}-\frac{9(\mathrm{GM})}{3 \mathrm{R} / 4} \\ & =\frac{-16 \mathrm{GM}}{\mathrm{R}} \end{aligned}