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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
- A \(\frac{{2m{g_0}{R^2}}}{{R + h}}\)
- B \(-\)\(\;\frac{{2m{g_0}{R^2}}}{{R + h}}\)
- C \(\;\frac{{{mg_0}{R^2}}}{{2\left( {R + h} \right)}}\)
- D \(-\)\(\;\frac{{{mg_0}{R^2}}}{{2\left( {R + h} \right)}}\)
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, \(E=PE+KE\)
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