The minimum and maximum distance of a satellite from the centre of the earth are 2R and 4R respectively. Where R= 6400km is the radius of the earth and $M = 6 \times 10^{24} kg$ is the mass of earth, its maximum speed is _______km/s

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Detailed Solution

By conservation of angular momentum

$\begin{matrix} m V_{1} \times 2 R = m V_{2} \times 4 R \\ V_{1} = 2 V_{2} \dots \dots \dots \dots (i) \end{matrix}$

By conservation of energy

$- \frac{G M m}{2 R} + \frac{1}{2} m V_{1}^{2} = - \frac{G M m}{4 R} + \frac{1}{2} m V_{2}^{2} \dots \dots (i i)$

From (i) and (ii) $V_{1} = \sqrt{\frac{2 G M}{3 R}}$

$= \sqrt{\frac{2 \times 6.67 \times 10^{- 11} \times 6 \times 10^{24}}{3 \times 6.4 \times 10^{4}}} = 6.45 k m / s \approx 6 k m / s$

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