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Physics · GRAVITATION
हिन्दीA satellite moves from a circular orbit of radius \( r_1 = 4 R_E \) to a lower circular orbit of radius \( r_2 = R_E \). The ratio of its final orbital speed to its initial orbital speed is:
Correct answer · Option 2
\( 2 : 1 \)
Explanation
Orbital speed is given by \( V = \sqrt{\frac{G M_E}{r}} \). Thus, \( \frac{V_2}{V_1} = \sqrt{\frac{r_1}{r_2}} = \sqrt{\frac{4 R_E}{R_E}} = 2 \). The ratio is \( 2 : 1 \).
