Question

A satellite is revolving round the earth with orbital speed ' \(V_0\) '. If it stops suddenly, the speed with which it will strike the surface of the earth would be ( \(V_e\) = escape velocity of a particle on earth's surface)

• A. \(\frac{V_e^2}{V_0}\)
• B. \(2 V_0\)
• C. \(\sqrt{V_e^2 - V_0^2}\)
• D. \(\sqrt{V_e^2 - 2 V_0^2}\)

Hint:

Relation: $V_e = \sqrt{2}\, V_0$

Answer 1

The Correct Option is C

Concept: Energy conservation: \[ \text{Total energy} = \text{constant} \] Also: \[ V_0 = \sqrt{\frac{GM}{R}}, \quad V_e = \sqrt{\frac{2GM}{R}} \]

Step 1: Initial energy (satellite in orbit). \[ E_i = -\frac{GMm}{2R} \]

Step 2: Final energy (at Earth's surface). \[ E_f = \frac{1}{2}mv^2 - \frac{GMm}{R} \]

Step 3: Apply conservation. \[ -\frac{GMm}{2R} = \frac{1}{2}mv^2 - \frac{GMm}{R} \]

Step 4: Solve. \[ \frac{1}{2}mv^2 = \frac{GMm}{2R} \Rightarrow v^2 = \frac{GM}{R} \] Now: \[ V_e^2 = \frac{2GM}{R}, \quad V_0^2 = \frac{GM}{R} \] \[ v = \sqrt{V_e^2 - V_0^2} \]

Step 5: Conclusion.
Velocity on striking earth = $\sqrt{V_e^2 - V_0^2}$ Final Answer: Option (C)