1. -mg₀R²/2(R+h) 2. 2mg₀R²/R+h 3. -2mg₀R²/R+h 4. mg₀R²/2(R+h)

Answer: 1. -mg₀R²/2(R+h) For a satellite in orbit, total energy is half of its (negative) potential energy: E=-GMm/2(R+h) . Use g₀=GM/R² , so GM=g₀R² . E=-mg₀R²/2(R+h) Option (1). Options (2) and (4) are positive, which is impossible for a bound orbit; option (3) is twice the correct magnitude.

Topic: Motion of a satellite: orbital velocity, time period and energy· Gravitation

A satellite of mass m is orbiting the earth (of radius R) at a height h from its surface. The total energy of the satellite in terms of g_0, the value of acceleration due to gravity at the earth's surface, is

1. $-\frac{mg_0R^2}{2(R+h)}$

2. $\frac{2mg_0R^2}{R+h}$

3. $-\frac{2mg_0R^2}{R+h}$

4. $\frac{mg_0R^2}{2(R+h)}$

Doubtly Asked 30 Oct 2026 NEET UG 2016 Phase 2 Practice test neet Save WhatsApp Report

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Answer: 1. $-\frac{mg_0R^2}{2(R+h)}$

For a satellite in orbit, total energy is half of its (negative) potential energy: $E=-\dfrac{GMm}{2(R+h)}$.

Use $g_0=\dfrac{GM}{R^2}$, so $GM=g_0R^2$.

$$E=-\frac{mg_0R^2}{2(R+h)}$$

Option (1). Options (2) and (4) are positive, which is impossible for a bound orbit; option (3) is twice the correct magnitude.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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A satellite of mass m is orbiting the earth (of radius R) at a height h from its surface. The total energy of the satellite in terms of g_0, the value of acceleration due to gravity at the earth's surface, is

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