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Physics · Gravitation

NEET (UG) 2024 — Question 40

The minimum energy required to launch a satellite of mass mm from the surface of earth of mass MM and radius RR in a circular orbit at an altitude of 2R2 R from the surface of the earth is:

  1. Option A:

    5GmM6R\frac{5 G m M}{6 R}

    Correct
  2. Option B:

    2GmM3R\frac{2 G m M}{3 R}

  3. Option C:

    GmM2R\frac{G m M}{2 R}

  4. Option D:

    GmM3R\frac{G m M}{3 R}

Answer: A

Step-by-step solution

Apply energy conservation,

Ui+Ki=Ut+KtU_{i}+K_{i}=U_{t}+K_{t}

⇒−GMmR+Ki=−GMm3R+12mv2\Rightarrow \quad-\frac{G M m}{R}+K_{i}=-\frac{G M m}{3 R}+\frac{1}{2} m v^{2}

⇒−GMmR+Ki=−GMm3R+12×m×GM3R \Rightarrow \quad-\frac{G M m}{R}+K_{i}=-\frac{G M m}{3 R}+\frac{1}{2} \times m \times \frac{G M}{3 R}

⇒Ki=−16GMmR+GMmR \Rightarrow \quad K_{i}=-\frac{1}{6} \frac{G M m}{R}+\frac{G M m}{R}

Ki=56GMmR \quad K_{i}=\frac{5}{6} \frac{G M m}{R}

Answer key and solution verified before publishing.

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Exam
NEET (UG) 2024
Paper
2024 paper
Subject
Physics
Chapter
Gravitation
Topic
Motion of Satellites and Escape Speed

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