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Q.

The energy required to take a satellite to a height 'h' above Earth surface (radius of Earth $= 6.4 \times 10^{3} km$ ) is E₁ and kinetic energy required for the satellite to be in a circuit orbit at this is E₂ . The value of h for which E₁and E₂ are equal, is

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a

$1.28 \times 10^{4} km$

b

$6.4 \times 10^{3} km$

c

$3.2 \times 10^{3} km$

d

$1.6 \times 10^{3} km$

answer is C.

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Detailed Solution

$\begin{array}{l} E_{1} = - \frac{GMm}{R + h} - (- \frac{GMm}{R}) = GMm [\frac{1}{R} - \frac{1}{R + h}] \\ E_{2} = \frac{GMm}{2 (R + h)}, E_{1} = E_{2}, \frac{1}{R} - \frac{1}{R + h} = \frac{1}{2 (R + h)} \\ \frac{1}{R} = \frac{3}{2 (R + h)}, h = \frac{R}{2} = 3 .2 \times 10^{3} km \end{array}$

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