Gravitation — Page 11 | JEE Physics

Q101

A body is released from a height equal to the radius

(\mathrm{R})

of the earth. The velocity of the body when it strikes the surface of the earth will be (Given

g=

acceleration due to gravity on the earth.)

A

\sqrt{\dfrac{g R}{2}}

B

\sqrt{4 g R}

C

\sqrt{2 g R}

D

\sqrt{g R}

Correct Answer

Option D

Solution

By conservation of mechanical energy

\begin{aligned} & \mathrm{U}_{\mathrm{i}}+\mathrm{K}_{\mathrm{i}}=\mathrm{U}_{\mathrm{f}}+\mathrm{K}_{\mathrm{i}} \\\\ & -\frac{\mathrm{GMm}}{2 \mathrm{R}}+0=-\frac{\mathrm{GMm}}{\mathrm{R}}+\frac{1}{2} \mathrm{mv}^2 \\\\ & \frac{\mathrm{GMm}}{2 \mathrm{R}}=\frac{1}{2} \mathrm{mv}^2 \\\\ & \mathrm{v}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}=\sqrt{\mathrm{gR}} \end{aligned}

Q102

Given below are two statements: Statement I : For a planet, if the ratio of mass of the planet to its radius increases, the escape velocity from the planet also increases. Statement II : Escape velocity is independent of the radius of the planet. In the light of above statements, choose the most appropriate answer form the options given below

A Both Statement I and Statement II are correct

B Statement I is correct but statement II is incorrect

C Both Statement I and Statement II are incorrect

D Statement I is incorrect but statement II is correct

Correct Answer

Option B

Solution

Statement I suggests that the escape velocity of a planet increases with an increase in the ratio of its mass to its radius.

This statement is consistent with the formula for escape velocity, which shows that the escape velocity of a planet is directly proportional to the square root of its mass and inversely proportional to the square root of its radius.

On the other hand, Statement II suggests that the escape velocity is independent of the radius of the planet.

This statement is not consistent with the formula for escape velocity, which clearly shows that the escape velocity is inversely proportional to the square root of the radius of the planet.

Hence, the most appropriate answer is: Statement I is true, but Statement II is false.

Q103

The radii of two planets 'A' and 'B' are 'R' and '4R' and their densities are

\rho

and

\rho / 3

respectively. The ratio of acceleration due to gravity at their surfaces

\left(g_{A}: g_{B}\right)

will be:

A 3 : 16

B 4 : 3

C 1 : 16

D 3 : 4

Correct Answer

Option D

Solution

The acceleration due to gravity at the surface of a planet can be expressed as:

g \propto \rho R

Now let's find the ratio of acceleration due to gravity at the surfaces of planets A and B:

\frac{g_A}{g_B} = \frac{\rho_A R_A}{\rho_B R_B}

Given that the densities are $\rho$ and

\frac{\rho}{3}

and the radii are

R

and

4R

for planets A and B, respectively, we have:

\frac{g_A}{g_B} = \frac{\rho \cdot R}{\left(\frac{\rho}{3}\right) \cdot (4R)} = \frac{3}{4}

So, the ratio of acceleration due to gravity at their surfaces is 3 : 4

Q104

Two planets A and B of radii

\mathrm{R}

and 1.5 R have densities

\rho

and

\rho / 2

respectively. The ratio of acceleration due to gravity at the surface of

\mathrm{B}

to

\mathrm{A}

is:

A 2 : 1

B 2 : 3

C 4 : 3

D 3 : 4

Correct Answer

Option D

Solution

The acceleration due to gravity at the surface of a planet is given by the formula:

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

where

G

is the gravitational constant,

M

is the mass of the planet, and

R

is the radius of the planet. In this problem, we are given two planets A and B, with radii

R_A

and

R_B=1.5R_A

, and densities

\rho_A

and

\rho_B=\frac{\rho_A}{2}

, respectively. We can use the formula above to calculate the acceleration due to gravity at the surface of each planet:

g_A=\frac{GM_A}{R_A^2}=\frac{4}{3}\pi G\rho_AR_A

and

g_B=\frac{GM_B}{R_B^2}=\frac{4}{3}\pi G\rho_B R_B=\frac{4}{3}\pi G\frac{\rho_A}{2}1.5R_A=\frac{3}{2}g_A

To find the ratio of the accelerations due to gravity at the surfaces of planets A and B, we can divide the expression for

g_B

by the expression for

g_A

:

\frac{g_B}{g_A}=\frac{\frac{3}{2}g_A}{g_A}=\frac{3}{2}

So, we can see that the acceleration due to gravity at the surface of planet B is

\frac{3}{2}

times larger than that at the surface of planet A.

However, the question asks for the ratio of the gravitational accelerations at the surfaces of the planets, which is equal to the ratio of the densities times the ratio of the radii:

\frac{g_B}{g_A}=\frac{\rho_B}{\rho_A}\cdot\frac{R_B}{R_A}=\frac{1/2}{1}\cdot\frac{1.5R_A}{R_A}=\frac{3}{4}

Therefore, the ratio of the gravitational accelerations at the surfaces of the planets is

3/4

, which is our final answer.

Q105

A planet having mass

9 \mathrm{Me}

and radius

4 \mathrm{R}_{\mathrm{e}}

, where

\mathrm{Me}

and

\mathrm{Re}

are mass and radius of earth respectively, has escape velocity in

\mathrm{km} / \mathrm{s}

given by: (Given escape velocity on earth

\mathrm{V}_{\mathrm{e}}=11.2 \times 10^{3} \mathrm{~m} / \mathrm{s}

)

A 33.6

B 11.2

C 16.8

D 67.2

Correct Answer

Option C

Solution

The escape velocity on a planet is given by the following formula:

v_{esc} = \sqrt{\frac{2GM}{R}}

where $v_{esc}$ is the escape velocity, $G$ is the gravitational constant, $M$ is the mass of the planet, and $R$ is the radius of the planet.

For Earth, we are given that $v_e = 11.2 \times 10^3$ m/s.

We know that the mass of the planet in question is $9M_e$ and its radius is $4R_e$ .

Let's find the escape velocity of this planet:

v_{esc} = \sqrt{\frac{2G(9M_e)}{4R_e}}

Divide both sides by the Earth's escape velocity formula:

\frac{v_{esc}}{v_e} = \frac{\sqrt{\frac{2G(9M_e)}{4R_e}}}{\sqrt{\frac{2GM_e}{R_e}}}

Simplify:

\frac{v_{esc}}{11.2 \times 10^3 \text{ m/s}} = {\sqrt{\frac{9}{4}}}

\frac{v_{esc}}{11.2 \times 10^3 \text{ m/s}} = \frac{3}{2}

Now, solve for $v_{esc}$ :

v_{esc} = 11.2 \times 10^3 \text{ m/s} \times \frac{3}{2}

v_{esc} = 16.8 \times 10^3 \text{ m/s}

Converting this to km/s, we get:

v_{esc} = 16.8 \text{ km/s}

Q106

The ratio of escape velocity of a planet to the escape velocity of earth will be:- Given: Mass of the planet is 16 times mass of earth and radius of the planet is 4 times the radius of earth.

A

1: 4

B

1: \sqrt{2}

C

4: 1

D

2: 1

Correct Answer

Option D

Solution

The escape velocity of a planet or a celestial body is given by:

v_e = \sqrt{\frac{2GM}{r}}

where

G

is the gravitational constant,

M

is the mass of the planet, and

r

is the radius of the planet. Let the subscripts "p" and "e" denote the planet and earth, respectively. Then, we have:

\frac{v_{e,p}}{v_{e,e}} = \frac{\sqrt{\frac{2G M_p}{r_p}}}{\sqrt{\frac{2G M_e}{r_e}}} = \sqrt{\frac{M_p r_e}{M_e r_p}}

Substituting the given values, we get:

\frac{v_{e,p}}{v_{e,e}} = \sqrt{\frac{16 \cdot 1}{1 \cdot 4}} = \sqrt{4} = 2

Therefore, the ratio of the escape velocity of the planet to that of earth is 2 : 1.

Q107

Two satellites

\mathrm{A}

and

\mathrm{B}

move round the earth in the same orbit. The mass of

\mathrm{A}

is twice the mass of

\mathrm{B}

. The quantity which is same for the two satellites will be

A Potential energy

B Kinetic energy

C Total energy

D Speed

Correct Answer

Option D

Solution

The quantity which is same for the two satellites will be speed.

The orbital speed of a satellite is independent of the mass of the satellite, but it depends on the radius of the orbit.

Potential energy, kinetic energy and total energy depend on the mass of the the satellite.

Therefore, the only quantity that is the same for the two satellites is speed.

Q108

A space ship of mass

2 \times 10^{4} \mathrm{~kg}

is launched into a circular orbit close to the earth surface. The additional velocity to be imparted to the space ship in the orbit to overcome the gravitational pull will be (if

g=10 \mathrm{~m} / \mathrm{s}^{2}

and radius of earth

=6400 \mathrm{~km}

):

A

7.9(\sqrt{2}-1) \mathrm{km} / \mathrm{s}

B

11.2(\sqrt{2}-1) \mathrm{km} / \mathrm{s}

C

7.4(\sqrt{2}-1) \mathrm{km} / \mathrm{s}

D

8(\sqrt{2}-1) \mathrm{km} / \mathrm{s}

Correct Answer

Option D

Solution

To find the additional velocity required to overcome the gravitational pull and launch the spaceship into orbit, we first need to find the orbital velocity.

The formula for orbital velocity (v) is given by:

v = \sqrt{\frac{GM}{r}}

where G is the gravitational constant, M is the mass of Earth, and r is the distance from the center of the Earth to the spaceship (which is the sum of the Earth's radius and the altitude of the spaceship's orbit).

In this problem, the spaceship is orbiting close to Earth's surface, so we can approximate r as the Earth's radius.

Given that g = 10 m/s² and Earth's radius R = 6400 km, we can relate the gravitational constant G and the mass of Earth M through the formula:

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

Now, we can find the orbital velocity:

v = \sqrt{\frac{gR^2}{R}} = \sqrt{gR}

Converting the Earth's radius to meters:

R = 6400 \times 10^3 \mathrm{~m}

Plugging in the values:

v = \sqrt{10 \times 6400 \times 10^3} = 8 \times 10^3 \mathrm{~m/s}

Now, we need to find the additional velocity required to overcome the gravitational pull.

To do this, we can use the formula for escape velocity:

v\text{escape} = \sqrt{2} \times v\text{orbital}

Finding the additional velocity required:

\Delta v = v\text{escape} - v\text{orbital} = (\sqrt{2} - 1) \times v_\text{orbital}

Plugging in the values:

\Delta v = (\sqrt{2} - 1) \times 8 \times 10^3 \mathrm{~m/s}

Converting the velocity to km/s:

\Delta v = (\sqrt{2} - 1) \times 8 \mathrm{~km/s}

Thus, the additional velocity required to overcome the gravitational pull and launch the spaceship into orbit is:

8(\sqrt{2}-1) \mathrm{~km/s}

Q109

If

\mathrm{V}

is the gravitational potential due to sphere of uniform density on it's surface, then it's value at the center of sphere will be:-

A

\dfrac{3 \mathrm{~V}}{2}

B

\dfrac{\mathrm{V}}{2}

C

\dfrac{4}{3} \mathrm{~V}

D

\mathrm{V}

Correct Answer

Option A

Solution

The gravitational potential (V) due to a sphere of uniform density at a distance r from its center is given by:

V(r) = \frac{GM}{2R^3} \left(3R^2 - r^2\right)

At the surface of the sphere (r = R), the gravitational potential is:

V = \frac{GM}{R}

Now, let's find the gravitational potential at the center of the sphere (r = 0):

V(0) = \frac{GM}{2R^3} \left(3R^2 - 0^2\right) = \frac{3GM}{2R}

The gravitational potential at the center of the sphere is

\frac{3}{2}

times the potential at the surface of the sphere. Therefore:

V(0) = \frac{3V}{2}

Q110

The time period of a satellite, revolving above earth's surface at a height equal to

\mathrm{R}

will be (Given

g=\pi^{2} \mathrm{~m} / \mathrm{s}^{2}, \mathrm{R}=

radius of earth)

A

\sqrt{32 R}

B

\sqrt{4 \mathrm{R}}

C

\sqrt{8 R}

D

\sqrt{2 R}

Correct Answer

Option A

Solution

For a satellite orbiting the Earth at a height equal to Earth's radius, its distance from the center of the Earth will be 2R, where R is the radius of the Earth.

Using the formula for the gravitational force:

F = G\frac{Mm}{r^2}

where F is the gravitational force, G is the gravitational constant, M is the mass of the Earth, m is the mass of the satellite, and r is the distance from the center of the Earth.

The centripetal force acting on the satellite is given by:

F_c = \frac{mv^2}{r}

Equating the gravitational force and the centripetal force, we get:

G\frac{Mm}{(2R)^2} = \frac{mv^2}{2R}

Solving for the orbital speed v, we get:

v^2 = \frac{GM}{2R}

The circumference of the satellite's orbit is given by:

C = 2\pi(2R) = 4\pi R

The time period T of the satellite's orbit can be calculated as the ratio of the circumference to the orbital speed:

T = \frac{C}{v} = \frac{4\pi R}{\sqrt{\frac{GM}{2R}}}

Given that the acceleration due to gravity at the Earth's surface is

g = \pi^2 \,\mathrm{m/s^2}

, we can express the gravitational constant G in terms of the Earth's radius R and mass M:

g = \frac{GM}{R^2} \Rightarrow GM = gR^2 = \pi^2 R^2

Substituting the expression for GM into the equation for the time period T, we get:

T = \frac{4\pi R}{\sqrt{\frac{\pi^2 R^2}{2R}}} = \frac{4\pi R}{\sqrt{\frac{\pi^2 R}{2}}}

T = \frac{4\pi R}{\sqrt{\pi^2}\sqrt{\frac{R}{2}}} = \frac{4\pi R}{\pi\sqrt{\frac{R}{2}}} = \frac{4R}{\sqrt{\frac{R}{2}}}

Multiplying the numerator and denominator by

\sqrt{2}

, we get:

T = \frac{4R\sqrt{2}}{\sqrt{R}} = 4\sqrt{2R} = \sqrt{32R}