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A satellite of mass m moves in a circular orbit around the Earth at an altitude x above the Earth's surface. If g denotes the acceleration due to gravity at the Earth's surface, then the satellite's orbital speed is

1. gR² / (R + x)
2. gR / (R - x)
3. gx
4. (gR² / (R + x))^(1/2)

Correct answer: (gR² / (R + x))^(1/2)

Solution

Orbital speed satisfies v^2 = GM/(R+x), and since g = GM/R^2 we get GM = gR^2. Therefore v = sqrt(gR^2/(R+x)).

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