MHT CET · Physics · Gravitation
The orbital velocity of an artificial satellite in a circular orbit just above the earth's surface is ' \(\mathrm{V}\) '. For the satellite orbiting at an altitude of half the earth's radius, the orbital velocity is
- A \(\frac{3}{2} \mathrm{~V}\)
- B \(\sqrt{\frac{3}{2}} \mathrm{~V}\)
- C \(\sqrt{\frac{2}{3}} \mathrm{~V}\)
- D \(\frac{2}{3} \mathrm{~V}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{\frac{2}{3}} \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
The orbital velocity near the surface of the earth
\(\mathrm{V}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\)
At an altitude \(\frac{R}{2}\), the orbital velocity
\(V^{\prime}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}+\frac{\mathrm{R}}{2}}}=\sqrt{\frac{2 \mathrm{GM}}{3 \mathrm{R}}}=\sqrt{\frac{2}{3}} \mathrm{~V}\)
\(\mathrm{V}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\)
At an altitude \(\frac{R}{2}\), the orbital velocity
\(V^{\prime}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}+\frac{\mathrm{R}}{2}}}=\sqrt{\frac{2 \mathrm{GM}}{3 \mathrm{R}}}=\sqrt{\frac{2}{3}} \mathrm{~V}\)
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