Class 11th Gravitation Physics Physics Gravitation

The minimum energy required to launch a satellite of mass $m$ from the surface of earth of mass $M$ and radius $R$ in a circular orbit at an altitude of $2 R$ from the surface of the earth is:

Option 1 - <p><span contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>5</mn> <mi>G</mi> <mi>m</mi> <mi>M</mi> </mrow> <mrow> <mn>6</mn> <mi>R</mi> </mrow> </mfrac> </mrow> </math> </span></p>

Option 2 - <p><span contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>2</mn> <mi>G</mi> <mi>m</mi> <mi>M</mi> </mrow> <mrow> <mn>3</mn> <mi>R</mi> </mrow> </mfrac> </mrow> </math> </span></p>

Option 3 - <p><span contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mi>G</mi> <mi>m</mi> <mi>M</mi> </mrow> <mrow> <mn>2</mn> <mi>R</mi> </mrow> </mfrac> </mrow> </math> </span></p>

Option 4 - <p><span contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mi>G</mi> <mi>m</mi> <mi>M</mi> </mrow> <mrow> <mn>3</mn> <mi>R</mi> </mrow> </mfrac> </mrow> </math> </span></p>

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Answered by

Alok Kumar Singh | Contributor-Level 10

5 months ago

Correct Option - 1

Detailed Solution:

Apply energy conservation,

$U_{i} + K_{i} = U_{f} + K_{f}$

$\Rightarrow - \frac{G M m}{R} + K_{i} = - \frac{G M m}{3 R} + \frac{1}{2} m v^{2}$

$\Rightarrow - \frac{G M m}{R} + K_{i} = - \frac{G M m}{3 R} + \frac{1}{2} * m * \frac{G M}{3 R}$

$\Rightarrow K_{i} = - \frac{1}{6} \frac{G M m}{R} + \frac{G M m}{R}$

$K_{i} = \frac{5}{6} \frac{G M m}{R}$

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