A satellite is put into a circular orbit very close to the surface of the Earth, just outside the atmosphere. What is its orbital speed, and how is this speed related to the acceleration due to gravity and the radius of the Earth?
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Set the gravitational force equal to the centripetal force required for circular motion, then use \( g = GM/R^2 \) to remove G and M from the expression.
A different way to reach the same result is to think in terms of the time period of a satellite orbiting close to the Earth.
For a satellite going around in a circle of radius R with speed v, the time it takes to complete one revolution is \( T = \dfrac{2\pi R}{v} \).
Near the Earth's surface, this time period works out to about 84.6 minutes, sometimes called the period of a ground grazing satellite.
Rearranging the time period formula gives \( v = \dfrac{2\pi R}{T} \). Using \( R = 6.4 \times 10^6 \ m \) and \( T \approx 5076 \) seconds, the speed comes out to roughly 7.9 kilometre per second, matching the value obtained from \( v = \sqrt{gR} \).
Physically, this speed is the exact balance point where the pull of gravity is just enough to curve the satellite's straight line path into a circle that follows the curvature of the Earth, so the satellite keeps falling toward the Earth but always misses it.