For a satellite of mass $m$ orbiting Earth at radius $r$ (from the centre), the orbital velocity is:
A$v = \sqrt{\dfrac{GM}{r}}$
B$v = \sqrt{2 GM r}$
C$v = \dfrac{GM}{r^2}$
D$v = G M$
Answer & Solution
Correct answer: A. $v = \sqrt{\dfrac{GM}{r}}$
Set gravitational force equal to centripetal force on the satellite:
$\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$
The $m$ cancels. Solve for $v$: $v^2 = \dfrac{GM}{r}$, so $v = \sqrt{\dfrac{GM}{r}}$.
Orbital velocity at Earth's surface is about $7.9$ km/s (about 28,000 km/h).
Option B looks similar but has a 2 inside the square root — that's the **escape velocity** $v_e = \sqrt{2GM/r}$, which is $\sqrt{2}$ times the orbital speed at the same radius. A textbook trap.
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