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Difficulty: HardGravitational Field and Orbits

A satellite of mass mm moves in a circular orbit around a uniform spherical planet of radius RR. If the height of the satellite above the planet's surface is h=2Rh = 2R and the acceleration due to gravity at the planet's surface is gg, which of the following expressions represents the orbital speed of the satellite?

  1. gR3\sqrt{\frac{gR}{3}}Answer
  2. B
    gR2\sqrt{\frac{gR}{2}}
  3. C
    2gR3\sqrt{\frac{2gR}{3}}
  4. D
    3gR\sqrt{3gR}

Answer

The orbital speed of the satellite is gR3\sqrt{\frac{gR}{3}}.
The total orbital radius from the center of the planet is r=R+h=R+2R=3Rr = R + h = R + 2R = 3R. Since surface gravity is g=GMR2g = \frac{GM}{R^2}, we have GM=gR2GM = gR^2. Substituting these into the orbital velocity expression v=GMrv = \sqrt{\frac{GM}{r}} yields v=gR23R=gR3v = \sqrt{\frac{gR^2}{3R}} = \sqrt{\frac{gR}{3}}, which makes the expression gR3\sqrt{\frac{gR}{3}} correct.

Step-by-Step Solution

1
Determine the total orbital radius from the center of the planet.
r=R+h=R+2R=3Rr = R + h = R + 2R = 3R
Gravitational attraction and circular orbital radii are always measured from the center of mass of the primary body, not its surface.
2
Relate the gravitational constant GG and planet mass MM to surface gravity gg.
g=GMR2  ⟹  GM=gR2g = \frac{GM}{R^2} \implies GM = gR^2
At the surface of a spherical planet of radius RR, the gravitational field strength is gg.
3
Substitute r=3Rr = 3R and GM=gR2GM = gR^2 into the circular orbital velocity equation v=GMrv = \sqrt{\frac{GM}{r}}.
v=gR23R=gR3v = \sqrt{\frac{gR^2}{3R}} = \sqrt{\frac{gR}{3}}
Equating centripetal force to gravitational force mv2r=GMmr2\frac{m v^2}{r} = \frac{G M m}{r^2} yields v=GMrv = \sqrt{\frac{GM}{r}}.

Key Concept

Orbital velocity of a satellite in terms of surface gravitational acceleration and orbital radius
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Topics

Gravitational Field and Orbits

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