Gravitation - Result Question 55
58. The mean radius of earth is $R$, its angular speed on its own axis is $\omega$ and the acceleration due to gravity at earth’s surface is $g$. What will be the radius of the orbit of a geostationary satellite?
(a) $(R^{2} g / \omega^{2})^{1 / 3}$
(b) $(R g / \omega^{2})^{1 / 3}$
(c) $(R^{2} \omega^{2} / g)^{1 / 3}$
(d) $(R^{2} g / \omega)^{1 / 3}$
[1992]
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Answer:
Correct Answer: 58. (a)
Solution:
(a) $T=\frac{2 \pi r}{v_0}=\frac{2 \pi r}{(g R^{2} / r)^{1 / 2}}=\frac{2 \pi r^{3 / 2}}{\sqrt{g R^{2}}}=\frac{2 \pi}{\omega}$
Hence, $r^{3 / 2}=\frac{\sqrt{g R^{2}}}{\omega}$ or $r^{3}=\frac{g R^{2}}{\omega^{2}}$ or, $r=(g R^{2} / \omega^{2})^{1 / 3}$
