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Physics · PYQ NEET UG 2016(2)

हिन्दी

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

Correct answer · Option 3

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

Explanation

Total energy of satellite \( E = -\frac{GMm}{2r} \). Substituting \( GM = g_0 R^2 \) and \( r = R+h \), we get \( E = -\frac{mg_0R^2}{2(R+h)} \).
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