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Physics · GRAVITATION
हिन्दीFor a satellite in a circular orbit very close to the surface of the Earth (\( h \ll R_E \)), the ratio of its orbital speed \( V \) to the escape speed \( v_e \) from the surface of the Earth is:
Correct answer · Option 3
\( \frac{1}{\sqrt{2}} \)
Explanation
Orbital speed near Earth is \( V = \sqrt{g R_E} \). Escape speed is \( v_e = \sqrt{2 g R_E} \). Hence, \( \frac{V}{v_e} = \frac{\sqrt{g R_E}}{\sqrt{2 g R_E}} = \frac{1}{\sqrt{2}} \).
