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JEE Main 2024PhysicsGravitationEscape Speed And Motion Of SatellitesmediumMCQ

JEE Main 2024 — Gravitation Question with Solution

From: JEE Main 2024 (Online) 9th April Evening Shift

Question

A satellite of $1 0^{3} kg$ mass is revolving in circular orbit of radius $2 R$ . If $\frac{1 0 ^{4} R}{6} J$ energy is supplied to the satellite, it would revolve in a new circular orbit of radius

(use $g = 10 m / s^{2}, R =$ radius of earth)

Choose an option

Show full solutionCorrect option: B

Correct answer

B6 R

Step-by-step explanation

To determine the new radius of the orbit after the energy is supplied to the satellite, we need to compare the initial and final energies of the satellite in its orbit around the Earth. We will use the formula for total energy (the sum of kinetic and potential energy) of a satellite in a circular orbit:

The total energy, $E$ , of a satellite of mass $m$ orbiting at a distance $r$ from the center of the Earth is given by:

$E = - \frac{GM m}{2 r}$

where:

$G$ is the gravitational constant
$M$ is the mass of the Earth
$r$ is the radius of the orbit

Given, the initial radius of orbit is $2 R$ . Therefore, the initial total energy $E_{i}$ is:

$E_{i} = - \frac{GM m}{2 \times 2 R} = - \frac{GM m}{4 R}$

Now, the energy supplied to the satellite is given as $\frac{1 0 ^{4} R}{6} J$ . The new total energy $E_{f}$ will be the sum of the initial energy and the supplied energy:

$E_{f} = E_{i} + Energy Supplied$

Substituting the values:

$E_{f} = - \frac{GM m}{4 R} + \frac{1 0 ^{4} R}{6}$

The final energy formula for a new orbit radius $r_{f}$ is similar to the initial energy formula:

$E_{f} = - \frac{GM m}{2 r _{f}}$

By equating the two expressions for $E_{f}$ , we get:

$- \frac{GM m}{2 r _{f}} = - \frac{GM m}{4 R} + \frac{1 0 ^{4} R}{6}$

Rearrange the equation to solve for $r_{f}$ :

$\frac{GM m}{2 r _{f}} = \frac{GM m}{4 R} - \frac{1 0 ^{4} R}{6}$

Since $GM m = g R^{2} m$ (using gravitational acceleration $g$ and radius of Earth $R$ ), we can substitute this to simplify the expression:

$\frac{g R ^{2} m}{2 r _{f}} = \frac{g R ^{2} m}{4 R} - \frac{1 0 ^{4} R}{6}$

By cancelling the common terms and rearranging:

$\frac{1}{2 r _{f}} = \frac{1}{4 R} - \frac{1 0 ^{4} R}{6 g R ^{2} m}$

Recall that $g = 10 m/ s^{2}$ and the mass of the satellite $m = 1 0^{3} kg$ :

$\frac{1}{2 r _{f}} = \frac{1}{4 R} - \frac{1 0 ^{4} \times R}{6 \times 10 \times R ^{2} \times 1 0 ^{3}}$

Simplifying the second term further:

$\frac{1 0 ^{4} \times R}{60 \times R ^{2} \times 1 0 ^{3}} = \frac{10 \times R}{60 \times R ^{2}} = \frac{1}{6 R}$

So, the final equation becomes:

$\frac{1}{2 r _{f}} = \frac{1}{4 R} - \frac{1}{6 R}$

Finding a common denominator for the right-hand side:

$\frac{1}{2 r _{f}} = \frac{3 - 2}{12 R} = \frac{1}{12 R}$

Therefore:

$2 r_{f} = 12 R$

Thus:

$r_{f} = 6 R$

Hence, the new radius of the circular orbit is 6R.

The correct answer is:

Option B: 6R

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About this question

This is a previous-year question from JEE Main 2024, covering the Gravitation chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.