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Gravitation question

2013 · Shift 0 · Q71
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Gravitation question

2013 · Shift 0 · Q71

JEE MainPhysicsGravitationMCQ+4 / −1
What is the minimum energy required to launch a satellite of mass mmm from the surface of a planet of mass MMM and radius RRR in a circular orbit at an altitude of 2R2R2R?
  1. A
    5GmM6R{{5GmM} \over {6R}}6R5GmM​
  2. B
    2GmM3R{{2GmM} \over {3R}}3R2GmM​
  3. C
    GmM2R{{GmM} \over {2R}}2RGmM​
  4. D
    GmM3R{{GmM} \over {3R}}3RGmM​
View written solutionFree

Correct answer: A

  1. Interpret the orbit altitude

The satellite is to be placed in a circular orbit at an altitude of 2R2R2R above the planet's surface.

So, the orbital radius from the center of the planet is r=R+2R=3R.r = R + 2R = 3R.r=R+2R=3R.

  1. Initial energy at the planet's surface

The satellite is launched from the surface. Assuming it starts from rest, its initial total mechanical energy is only gravitational potential energy: Ei=−GMmR.E_i = -\frac{GMm}{R}.Ei​=−RGMm​.

  1. Final energy in circular orbit

For a satellite in a circular orbit of radius rrr, total energy is Ef=−GMm2r.E_f = -\frac{GMm}{2r}.Ef​=−2rGMm​.

Here r=3Rr=3Rr=3R, so Ef=−GMm2(3R)=−GMm6R.E_f = -\frac{GMm}{2(3R)} = -\frac{GMm}{6R}.Ef​=−2(3R)GMm​=−6RGMm​.

  1. Minimum energy required

The minimum external energy needed is the increase in total mechanical energy: ΔE=Ef−Ei.\Delta E = E_f - E_i.ΔE=Ef​−Ei​.

Substitute the values: ΔE=−GMm6R−(−GMmR).\Delta E = -\frac{GMm}{6R} - \left(-\frac{GMm}{R}\right).ΔE=−6RGMm​−(−RGMm​).

ΔE=−GMm6R+GMmR.\Delta E = -\frac{GMm}{6R} + \frac{GMm}{R}.ΔE=−6RGMm​+RGMm​.

Take LCM: ΔE=GMmR(1−16)=GMmR⋅56.\Delta E = \frac{GMm}{R}\left(1 - \frac{1}{6}\right) = \frac{GMm}{R}\cdot\frac{5}{6}. ΔE=RGMm​(1−61​)=RGMm​⋅65​.

Thus, ΔE=5GMm6R.\Delta E = \frac{5GMm}{6R}. ΔE=6R5GMm​.

  1. Match with options

This corresponds to:

  • A: 5GMm6R\dfrac{5GMm}{6R}6R5GMm​

So the correct option is A.

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