Question

At a height H from the surface of earth, the total energy of a satellite is equal to the potential energy of a body of equal mass at a height 3R from the surface of the earth(R = radius of the earth). The value of H is [EAMCET 2014]

Moderate

Solution

We know that, the total energy of the satellite,
$E = - \frac{1}{2} \frac{G M_{e} m}{R}$
At a height H from the surface of the earth, the total energy of satellite,
$E = - \frac{1}{2} \frac{G M_{e} m}{(R + H)}$ …(i)
and the potential energy of the satellite,
$U = - \frac{G M_{e} m}{(R + 3 R)} = - \frac{G M_{e} m}{4 R}$ …(ii)
From given conditlon, $E = U \Rightarrow - \frac{1}{2} \cdot \frac{G M_{e} m}{(R + H)} = - \frac{G M_{e} m}{4 R}$
$\Rightarrow \frac{1}{2 R + 2 H} = \frac{1}{4 R} \Rightarrow 2 H = 4 R - 2 R \Rightarrow H = R$

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