Gravitation | JEE Main previous year questions with Solutions

Q 1 :

Two satellite A and B go round a planet in circular orbits having radii 4R and R respectively. If the speed of A is 3v, the speed of B will be [2024]

$6 v$
$12 v$
$\frac{4}{3} v$
$3 v$

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(1)

Q 2 :

A satellite revolving around a planet in stationary orbit has time period 6 hours. The mass of planet is one-fourth the mass of earth. The radius orbit of planet is

(Given = Radius of geo-stationary orbit for earth is $4.2 \times 10^{4}$ km) [2024]

$1.05 \times 10^{4}$ km
$1.4 \times 10^{4}$ km
$1.68 \times 10^{5}$ km
$8.4 \times 10^{4}$ km

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(1)

By Kepler's law, $T^{2} = \frac{4 π^{2} R^{3}}{G M} \Rightarrow T^{2} \propto \frac{R^{3}}{M}$

${(\frac{T_{1}}{T_{2}})}^{2} = {(\frac{R_{1}}{R_{2}})}^{3} (\frac{M_{2}}{M_{1}})$

${(\frac{24}{6})}^{2} = {(\frac{4.2 \times 10^{4}}{R_{2}})}^{3} \times (\frac{\frac{M_{E}}{4}}{M_{E}})$

$4^{2} \times 4 = {(\frac{4.2 \times 10^{4}}{R_{2}})}^{3} \Rightarrow R_{2} = 1.05 \times 10^{4}$ km

Q 3 :

Correct formula for height of a satellite from Earth's surface is _____. [2024]

${(\frac{T^{2} R^{2} g}{4 π^{2}})}^{1 / 3} - R$
${(\frac{T^{2} R^{2} g}{4 π^{2}})}^{1 / 2} - R$
${(\frac{T^{2} R^{2}}{4 π^{2} g})}^{1 / 3} - R$
${(\frac{T^{2} R^{2} g}{4 π^{2}})}^{- 1 / 3} + R$

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(1)

$\frac{m v^{2}}{(R + h)} = \frac{G M m}{{(R + h)}^{2}} \Rightarrow v = \sqrt{\frac{G M}{R + h}}$

$Now T = \frac{2 π (R + h)}{\sqrt{\frac{G M}{R + h}}}$

$\Rightarrow T = \frac{2 π {(R + h)}^{3 / 2}}{\sqrt{G M}} = \frac{2 π {(R + h)}^{3 / 2}}{R \sqrt{g}}$

$\Rightarrow T^{2} R^{2} g = 4 π^{2} (R + h^{3})$

$\Rightarrow {(\frac{T^{2} R^{2} g}{4 π^{2}})}^{1 / 3} - R = h$

Q 4 :

A satellite of mass $\frac{M}{2}$ is revolving around earth in a circular orbit at a height of $\frac{R}{3}$ from earth surface. The angular momentum of the satellite is $M \sqrt{\frac{G M R}{x}}$ . The value of x is ________, where M and R are the mass and radius of earth, respectively. (G is the gravitational constant) [2025]

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(3)

(i) If earth is assumed to be stationary

Orbital velocity $v_{0} = \sqrt{\frac{G M}{4 R / 3}} = \sqrt{\frac{3 G M}{4 R}}$

Angular momentum of satellite

$L = \frac{M}{2} v_{0} \frac{4 R}{3}$

$L = \frac{M}{2} \cdot \sqrt{\frac{3 G M}{4 R}} \cdot \frac{4 R}{3} = M \sqrt{\frac{G M R}{3}}$

$\Rightarrow x = 3$

Q 5 :

A satellite of mass 1000 kg is launched to revolve around the earth in an orbit at a height of 270 km from the earth's surface. Kinetic energy of the satellite in this orbit is ___ $\times 10^{10} J$ . (Mass of earth $= 6 \times 10^{24} kg$ , Radius of earth $= 6.4 \times 10^{6} m$ , Gravitational constant $= 6.67 \times 10^{- 11} {Nm}^{2} {kg}^{- 2}$ ) [2025]

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(3)

$K = - \frac{U}{2} = \frac{G M m}{2 r}$

$\Rightarrow K = \frac{6.67 \times 10^{- 11} \times 6 \times 10^{24} \times 10^{3}}{2 \times 6670 \times 10^{3}} = 3 \times 10^{10} J$

Q 6 :

The time period of a satellite of earth is 24 hours. If the separation between the earth and the satellite is decreased to one fourth of the previous value, then its new time period will become [2023]

4 hours
6 hours
12 hours
3 hours

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(4)

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

$\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{R_{1}^{3}}{R_{2}^{3}}$ $\Rightarrow {(\frac{T_{1}}{T_{2}})}^{2} = {(\frac{R}{R / 4})}^{3}$

$∴$ $\frac{T_{1}^{2}}{T_{2}^{2}} = 64$

$\Rightarrow T_{2}^{2} = \frac{T_{1}^{2}}{64}$

$\Rightarrow T_{2} = \frac{24}{8} = 3 hr$

Q 7 :

Given below are two statements: [2023]

Statement I: If E be the total energy of a satellite moving around the earth, then its potential energy will be $\frac{E}{2} .$

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

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

Both Statement I and Statement II are incorrect
Both Statement I and Statement II are correct
Statement I is incorrect but Statement II is correct
Statement I is correct but Statement II is incorrect

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(1)

$Energy of satellite in orbit E = \frac{- G M m}{2 R}$

$PE of satellite in orbit U = \frac{- G M m}{R}$

$\Rightarrow U = 2 E$

$KE of satellite in orbit K = E - U$

$K = \frac{G M m}{2 R} = (- E)$

Q 8 :

The orbital angular momentum of a satellite is $L$ . When it is revolving in a circular orbit at height $h$ from earth surface, If the distance of satellite from the earth centre is increased by eight times to its initial value, then the new angular momentum will be [2023]

$8 L$
$9 L$
$4 L$
$3 L$

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(4)

$L = m v r$

$\Rightarrow v = \sqrt{\frac{G M_{e}}{r}}$

$L = m \sqrt{\frac{G M_{e}}{r} \cdot r}$ $\Rightarrow L \propto r^{1 / 2}$

$Now distance from centre is increased by 8 times.$

$So new distance from center = r + 8 r = 9 r$

$Now angular momentum L' \propto {(9 r)}^{1 / 2}$

$\frac{L}{L'} = \frac{r^{1 / 2}}{{(9 r)}^{1 / 2}} = \frac{1}{3}$ $\Rightarrow L' = 3 L$

Q 9 :

Two satellites of masses $m$ and $3 m$ revolve around the earth in circular orbits of radii $r$ and $3 r$ respectively. The ratio of orbital speeds of the satellites respectively is [2023]

1 : 1
3 : 1
$\sqrt{3}$ : 1
9 : 1

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(3)

$v = \sqrt{\frac{G M}{r}}$

$v \propto \frac{1}{\sqrt{r}}$

$\Rightarrow \frac{v_{1}}{v_{2}} = \sqrt{\frac{r_{2}}{r_{1}}} = \sqrt{\frac{3 r}{r}} = \sqrt{3} : 1$

Q 10 :

The time period of a satellite, revolving above earth’s surface at a height equal to $R$ will be (Given $g = π^{2} {m/s}^{2}$ , R = radius of earth) [2023]

$\sqrt{4 R}$
$\sqrt{8 R}$
$\sqrt{32 R}$
$\sqrt{2 R}$

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(3)

$\frac{m v^{2}}{2 R} = \frac{G M m}{{(2 R)}^{2}}$

$\Rightarrow v = \sqrt{\frac{G M}{2 R}} = \sqrt{\frac{R g}{2}}$

$T = \frac{2 π (2 R)}{v} = \frac{4 π R \sqrt{2}}{\sqrt{R g}} = \sqrt{32 R}$

Topic Question Set

Q 1 :

Two satellite A and B go round a planet in circular orbits having radii 4R and R respectively. If the speed of A is 3v, the speed of B will be [2024]

Q 2 :

A satellite revolving around a planet in stationary orbit has time period 6 hours. The mass of planet is one-fourth the mass of earth. The radius orbit of planet is

(Given = Radius of geo-stationary orbit for earth is $4.2 \times 10^{4}$ km) [2024]

Q 3 :

Correct formula for height of a satellite from Earth's surface is _____. [2024]

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Q 6 :

The time period of a satellite of earth is 24 hours. If the separation between the earth and the satellite is decreased to one fourth of the previous value, then its new time period will become [2023]

Q 8 :

The orbital angular momentum of a satellite is $L$ . When it is revolving in a circular orbit at height $h$ from earth surface, If the distance of satellite from the earth centre is increased by eight times to its initial value, then the new angular momentum will be [2023]

Q 9 :

Two satellites of masses $m$ and $3 m$ revolve around the earth in circular orbits of radii $r$ and $3 r$ respectively. The ratio of orbital speeds of the satellites respectively is [2023]

Q 10 :

The time period of a satellite, revolving above earth’s surface at a height equal to $R$ will be (Given $g = π^{2} {m/s}^{2}$ , R = radius of earth) [2023]

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Topic Question Set

Q 3 : Correct formula for height of a satellite from Earth's surface is _____. [2024] .question-row { display: flex; align-items: flex-start; gap: 8px; } .q-no { font-size: 17px; white-space: nowrap; margin-left: -15px; } .q-text { font-size: 18px; flex: 1; } .q-no, .q-text p { line-height: 30px; }

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Q 3 :

Correct formula for height of a satellite from Earth's surface is _____. [2024]