Gravitation | JEE Main previous year questions with Solutions

Q 11 :

Two satellites A and B move round the earth in the same orbit. The mass of A is twice the mass of B. The quantity which is same for the two satellites will be [2023]

Potential energy
Total energy
Kinetic energy
Speed

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(4)

$P.E. = - \frac{G M_{p} M_{A}}{R}$

$K.E. = + \frac{G M_{p} M_{A}}{2 R}$

$T.E. = - \frac{G M_{p} M_{A}}{2 R}$

$Speed = v = \sqrt{\frac{G M_{p}}{R}}$

$Speed of satellite is independent of mass of satellite.$

Q 12 :

Initially a satellite of 100 kg is in a circular orbit of radius $1.5 R_{E}$ . This satellite can be moved to a circular orbit of radius $3 R_{E}$ by supplying $α \times 10^{6} J$ of energy. The value of $α$ is______.

$($ Take Radius of Earth $R_{E} = 6 \times 10^{6} m and 𝑔 = 10 m / s^{2})$ [2026]

100
500
1000
150

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(3)

Energy of a satellite in a circular orbit is given as

$E = \frac{- G M_{E} m}{2 r}; r = radius of circular orbit$

$Required energy to be supplied = Δ E = E_{f} - E_{i}$

$Δ E = (\frac{- G M_{E} m}{2 (3 R_{E})}) - (\frac{- G M_{E} m}{2 (1.5 R_{E})})$

$= \frac{G M_{E} m}{6 R_{E}}$

$Now, g = \frac{G M_{E}}{R_{E}^{2}} \Rightarrow \frac{G M_{E}}{R_{E}} = g R_{E}$

$∴$ $Δ E = \frac{1}{6} g m R_{E}$

$= \frac{1}{6} \times 10 \times 100 \times 6 \times 10^{6}$

$= 1000 \times 10^{6}$

$α = 1000$

Q 13 :

Given below are two statements :

Statement I : A satellite is moving around earth in the orbit very close to the earth surface. The time period of revolution of satellite depends upon the density of earth.

Statement II : The time period of revolution of the satellite is $T = 2 π \sqrt{\frac{R_{e}}{g}}$ (for satellite very close to the earth surface), where $R_{e}$ radius of earth and g acceleration due to gravity.

In the light of the above statements, choose the correct answer from the options given below : [2026]

Statement I is true but Statement II is false
Statement I is false but Statement II is true
Both Statement I and Statement II are true
Both Statement I and Statement II are false

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(3)

$T = 2 π \sqrt{\frac{R^{3}}{G M}}$

$∵$ $M = ρ \cdot \frac{4}{3} π R^{3}$

$∴$ $T = 2 g p \sqrt{\frac{1}{G ρ \frac{4}{3} π}}$

$Statement I is correct.$

$And$ $∴$ $\frac{G M}{R^{2}} = g$

$∴$ $T = 2 π \sqrt{\frac{R}{g}}$

$Statement II is correct$

$Ans. (3)$

Topic Question Set

Q 11 :

Two satellites A and B move round the earth in the same orbit. The mass of A is twice the mass of B. The quantity which is same for the two satellites will be [2023]

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Topic Question Set

Q 11 : Two satellites A and B move round the earth in the same orbit. The mass of A is twice the mass of B. The quantity which is same for the two satellites will be [2023]

Q 12 : Initially a satellite of 100 kg is in a circular orbit of radius 1.5RE . This satellite can be moved to a circular orbit of radius 3RE by supplying α×106 J of energy. The value of α is______. (Take Radius of Earth RE=6×106m and 𝑔=10m/s2) [2026]

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Q 11 :

Two satellites A and B move round the earth in the same orbit. The mass of A is twice the mass of B. The quantity which is same for the two satellites will be [2023]