If earth has a mass nine times and radius twice to that of a planet P. Then $$\frac{v_{e}}{3} \sqrt{x} \mathrm{~ms}^{-1}$$ will be the minimum velocity required by a rocket to pull out of gravitationa

$$ \begin{aligned} & M_E=9 M_P \\\\ & R_E=2 R_P \end{aligned} $$ Escape velocity $=\sqrt{\frac{2GM}{R}}$ For earth $v_e=\sqrt{\frac{2 G M_E}{R_E}}$ $$ \begin{aligned} & \text { For } P, v_e=\sqrt{\frac{\frac{2 G M_E}{9}}{\frac{R_E}{2}}}=\sqrt{\frac{2 G M_E}{R_E} \times \frac{2}{9}} \\\\ & =\frac{v_e \sqrt{2}}{3} \end{aligned} $$

Gravitation — Page 15 | JEE Physics

Q141

A satellite is moving with a constant speed v in circular orbit around the earth. An object of mass ‘m’ is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of ejection, the kinetic energy of the object is -

A mv2

B

{1 \over 2}

mv2

C

{3 \over 2}

mv2

D 2 mv2

Correct Answer

Option A

Solution

At height r from center of earth. orbital velocity =

\sqrt {{{GM} \over r}}

$\therefore$ By energy conservation KE of 'm' +

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

= 0 + 0 (At infinity, PE = KE = 0) $\Rightarrow$ KE of 'm' =

{{{GMm} \over r}}

=

{\left( {\sqrt {{{GM} \over r}} } \right)^2}

m = mv2

Q142

Match List I with List II : .tg .tg LIST I LIST II A. Kinetic energy of planet I.

-\mathrm{GMm} / \mathrm{a}

B. Gravitation Potential energy of sun-planet system II.

\mathrm{GMm} / 2 \mathrm{a}

C. Total mechanical energy of planet III.

\dfrac{\mathrm{Gm}}{\mathrm{r}}

D. Escape energy at the surface of planet for unit mass object IV.

-\mathrm{GMm} / 2 \mathrm{a}

(Where

\mathrm{a}=

radius of planet orbit,

\mathrm{r}=

radius of planet,

\mathrm{M}=

mass of Sun,

\mathrm{m}=

mass of planet) Choose the correct answer from the options given below :

A (A)-(II), (B)-(I), (C)-(IV), (D)-(III)

B (A)-(I), (B)-(II), (C)-(III), (D)-(IV)

C (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

D (A)-(I), (B)-(IV), (C)-(II), (D)-(III)

Correct Answer

Option A

Solution

\text{ K.E }=\frac{G M m}{2 a} \quad \text{(II)}

U_G=\frac{-G M m}{a} \quad \text{(I)}

M . E=\frac{-G M m}{2 a} \quad \text{(IV)}

\text{ and Escape Energy }=\frac{G m}{r} \quad \text{(III)}

Q143

\text{ Match the LIST-I with LIST-II }

.tg .tg List - I List - II A.

\text{ Gravitational constant }

I.

\left[\mathrm{LT}^{-2}\right]

B.

\text{ Gravitational potential energy }

II.

\left[\mathrm{L}^2 \mathrm{~T}^{-2}\right]

C.

\text{ Gravitational potential }

III.

\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]

D.

\text{ Acceleration due to gravity }

IV.

\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]

Choose the correct answer from the options given below:

A A-IV, B-III, C-II, D-I

B A-II, B-IV, C-III, D-I

C A-I, B-III, C-IV, D-II

D A-III, B-II, C-I, D-IV

Correct Answer

Option A

Solution

Here’s how each quantity lines up with its dimensional formula: Gravitational constant,

G

• From Newton’s law

F = G\frac{m_1m_2}{r^2}

•

[G] = \frac{[F]\,[r]^2}{[m]^2} = \frac{(MLT^{-2})\;L^2}{M^2} = M^{-1}L^3T^{-2}

⇒ IV Gravitational potential energy,

U

• As work or

U = mgh

•

[U] = [F]\,[h] = (MLT^{-2})\;L = ML^2T^{-2}

⇒ III Gravitational potential, $\phi$ • Energy per unit mass:

\phi = \frac{U}{m}

•

[\phi] = \frac{ML^2T^{-2}}{M} = L^2T^{-2}

⇒ II Acceleration due to gravity,

g

• Just an acceleration •

[g] = LT^{-2}

⇒ I Matching them gives A–IV, B–III, C–II, D–I That corresponds to Option A.

Q144

The energy required to take a satellite to a height 'h' above Earth surface (radius of Earth = 6.4

\times

103 km) is E1 and kinetic energy required for the satellite to be in a circular orbit at this height is E2. The value of h for which E1 and E2 are equal, is

A 1.6

\times

103 km

B 3.2

\times

103 km

C 6.4

\times

103 km

D 1.28

\times

104 km

Correct Answer

Option B

Solution

Energy required to move a satellite from earth surface to height h is, E1 = Uh $-$ Usurface = $-$

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

= GMm

\left( {{1 \over R} - {1 \over {R + h}}} \right)

=

{{GMm} \over {R(R + h)}} \times h

We know, for sattelite at height h. Centrifigual force = Gravitational force $\Rightarrow$

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

$\Rightarrow$

mv

2 =

{{GMm} \over {R + h}}

$\therefore$

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

=

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

$\therefore$ Kinetic energy (E2) =

{{GMm} \over {2(R + h)}}

Given that, E1 = E2 $\therefore$

{{GMm} \over {R(R + h)}} \times h = {{GMm} \over {2\left( {R + h} \right)}}

$\Rightarrow$

{h \over R} = {1 \over 2}

$\Rightarrow$ h =

{R \over 2}

$\therefore$ h =

{{6.4 \times {{10}^3}} \over 2}

= 3.2 $\times$ 103 km

Q145

A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the nth power of R. If the period of rotation of the particle is T, then :

A T

\propto

Rn/2

B T

\propto

R3/2 for any n

C T

\propto

Rn/2 +1

D T

\propto

R(n+1)/2

Correct Answer

Option D

Solution

We know, Central force in circular motion, F =

m{\omega ^2}R

According to the question,

F \propto {1 \over {{R^n}}}

$\therefore$

m{\omega ^2}R

\propto {1 \over {{R^n}}}

\Rightarrow m{\omega ^2}R = {k \over {{R^n}}}

\Rightarrow {\omega ^2} = {k \over {m{R^{n + 1}}}}

$\therefore$

\omega \propto {1 \over {{R^{{{n + 1} \over 2}}}}}

.......(1) And we know,

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

$\therefore$

T \propto {1 \over \omega }

...... (2) From (1) and (2) we can conclude that,

T \propto {R^{{{n + 1} \over 2}}}

Q146

Given below are two statements: Statement I: If

\mathrm{E}

be the total energy of a satellite moving around the earth, then its potential energy will be

\dfrac{E}{2}

. Statement II: The kinetic energy of a satellite revolving in an orbit is equal to the half the magnitude of total energy

\mathrm{E}

. In the light of the above statements, choose the most appropriate answer from the options given below

A Both Statement I and Statement II are incorrect

B Statement I is incorrect but Statement II is correct

C Statement I is correct but Statement II is incorrect

D Both Statement I and Statement II are correct

Correct Answer

Option A

Solution

A satellite in orbit around a planet is subject to two main forces: gravitational force, which is trying to pull it towards the planet, and its own kinetic energy or inertia, which is trying to keep it moving in a straight line.

The balance of these two forces results in the satellite moving in a circular or elliptical orbit.

The gravitational potential energy ( $U$ ) of the satellite is given by the formula: $U = -\dfrac{GMm}{R}$ where $G$ is the gravitational constant, $M$ is the mass of the Earth, $m$ is the mass of the satellite, and $R$ is the radius of the orbit.

The negative sign indicates that work would have to be done to remove the satellite from the Earth's gravitational influence.

The kinetic energy ( $K$ ) of the satellite is given by the formula: $K = \dfrac{GMm}{2R}$ This is obtained from the fact that for a satellite in stable orbit, the gravitational force must be equal to the centripetal force required to keep the satellite moving in a circle.

From this, we can derive an expression for the velocity of the satellite, and hence its kinetic energy.

The total mechanical energy ( $E$ ) of the satellite, which is the sum of its kinetic and potential energy, is therefore: $E = K + U = \dfrac{GMm}{2R} - \dfrac{GMm}{R} = -\dfrac{GMm}{2R}$ So the potential energy $U$ is $-2E$ , and the kinetic energy $K$ is $-E$ .

Thus, the statement "If $E$ be the total energy of a satellite moving around the earth, then its potential energy will be $2E$ " is correct, and the statement "The kinetic energy of a satellite revolving in an orbit is equal to the half the magnitude of total energy $E$ " is incorrect.

Q147

If earth has a mass nine times and radius twice to that of a planet P. Then

\dfrac{v_{e}}{3} \sqrt{x} \mathrm{~ms}^{-1}

will be the minimum velocity required by a rocket to pull out of gravitational force of

\mathrm{P}

, where

v_{e}

is escape velocity on earth. The value of

x

is

A 1

B 3

C 2

D 18

Correct Answer

Option C

Solution

\begin{aligned} & M_E=9 M_P \\\\ & R_E=2 R_P \end{aligned}

Escape velocity $=\sqrt{\dfrac{2GM}{R}}$ For earth $v_e=\sqrt{\dfrac{2 G M_E}{R_E}}$

\begin{aligned} & \text{ For } P, v_e=\sqrt{\frac{\frac{2 G M_E}{9}}{\frac{R_E}{2}}}=\sqrt{\frac{2 G M_E}{R_E} \times \frac{2}{9}} \\\\ & =\frac{v_e \sqrt{2}}{3} \end{aligned}

Q148

The escape velocity of a body on a planet 'A' is 12 kms

-

1. The escape velocity of the body on another planet 'B', whose density is four times and radius is half of the planet 'A', is :

A 12 kms

-

1

B 24 kms

-

1

C 36 kms

-

1

D 6 kms

-

1

Correct Answer

Option A

Solution

{v_{esc}} - \sqrt {{{2GM} \over R}} = \sqrt {{{2G} \over R} \times \rho \times {4 \over 3}\pi {R^3}}

\Rightarrow {v_{esc}} \propto R\sqrt \rho

\Rightarrow {{{{({v_{esc}})}_B}} \over {{{({v_{esc}})}_A}}} = 1

\Rightarrow {({v_{esc}})_B} = 12

km/s

Q149

Earth has mass 8 times and radius 2 times that of a planet. If the escape velocity from the earth is 11.2 km/s, the escape velocity in km/s from the planet will be:

A 8.4

B 11.2

C 5.6

D 2.8

Correct Answer

Option C

Solution

The escape velocity ( $v_{\text{esc}}$ ) for a planet with mass $M$ and radius $R$ is given by the formula:

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

where $G$ is the gravitational constant.

For Earth, let's denote its mass as $M_E$ and its radius as $R_E$ .

The escape velocity is given as 11.2 km/s.

Therefore:

v_{\text{esc, Earth}} = \sqrt{\frac{2G M_E}{R_E}} = 11.2 \text{ km/s}

For the planet in question, its mass $M_P$ is $M_E/8$ and its radius $R_P$ is $R_E/2$ .

The escape velocity from the planet is:

v_{\text{esc, Planet}} = \sqrt{\frac{2G M_P}{R_P}} = \sqrt{\frac{2G (M_E/8)}{R_E/2}}

Simplifying the expression:

v_{\text{esc, Planet}} = \sqrt{\frac{2G (M_E/8)}{R_E/2}} = \sqrt{\frac{2G M_E \cdot 2}{8R_E}} = \sqrt{\frac{G M_E}{2R_E}}

Now, relate it to the escape velocity from Earth:

v_{\text{esc, Planet}} = \frac{1}{\sqrt{4}} \cdot v_{\text{esc, Earth}}

Calculating further:

v_{\text{esc, Planet}} = \frac{1}{2} \cdot 11.2 \text{ km/s} = 5.6 \text{ km/s}

Thus, the escape velocity from the planet is 5.6 km/s. Option C: 5.6 km/s is the correct answer.

Q150

Given below are two statements : Statement I : The law of gravitation holds good for any pair of bodies in the universe. Statement II : The weight of any person becomes zero when the person is at the centre of the earth. In the light of the above statements, choose the correct answer from the options given below.

A Both Statement I and Statement II are true

B Both Statement I and Statement II are false

C Statement I is true but Statement II is false

D Statement I is false but Statement II is true

Correct Answer

Option A

Solution

Statement I is true.

Newton's law of universal gravitation states that every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between their centers.

This law applies to any pair of bodies in the universe.

Statement II is also true.

The weight of an object is the force of gravity acting on it.

If a person were at the center of the Earth, the gravitational pull from all the surrounding mass of the Earth would cancel out, resulting in zero net gravitational force and therefore zero weight.

Therefore, Option A: Both Statement I and Statement II are true, is the correct answer.