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Orbital velocity
Concepts tested here
- Orbital speed at height h
All Questions
2015 AIPMT-II 1 question
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A remote-sensing satellite of earth revolves in a circular orbit at a height of $0.25\times10^6$ m above the surface of earth. If earth's radius is $6.38\times10^6$ m and g = 9.8 $ms^{-2}$, then the orbital speed of the satellite is:$v_0 = \sqrt{\frac{GM_e}{R_e + h}} = \sqrt{\frac{gR_e^2}{R_e + h}}$
$v_0 = \sqrt{\frac{9.8\times(6.38\times10^6)^2}{6.63\times10^6}} \approx \sqrt{60\times10^6}$ m/s
$v_0 \approx 7.76\times10^3$ m/s = 7.76 km/s