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NEET · Physics · STD 11 - 7. 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. A \(\frac{{2m{g_0}{R^2}}}{{R + h}}\)
  2. B \(-\)\(\;\frac{{2m{g_0}{R^2}}}{{R + h}}\)
  3. C \(\;\frac{{{mg_0}{R^2}}}{{2\left( {R + h} \right)}}\)
  4. D \(-\)\(\;\frac{{{mg_0}{R^2}}}{{2\left( {R + h} \right)}}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(-\)\(\;\frac{{{mg_0}{R^2}}}{{2\left( {R + h} \right)}}\)

Step-by-step Solution

Detailed explanation

           Total energy of satellite at height \(h\) from the earth surface, \(E=PE+KE\)