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Per NCERT §7.9 eqn 7.35, the orbital speed of a satellite at height h above Earth's surface is which?
Av = √(2 G M / R)
Bv = G M / (R + h)
Cv = √(G M R)
Dv = √[G M / (R + h)]
Answer & Solution
Correct answer: D. v = √[G M / (R + h)]
NCERT eqn 7.35: orbital speed v = √(GM/(R+h)). For near-Earth (h=0), v_0 = √(gR) ≈ 7.9 km/s.
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