An Earth satellite orbits at a radius of 19,600km. If the ratio r^3/T^2 for the Moon is 1.01×10^18⁢km^3/y^2 , what is the orbital period of the satellite?

asked by @kaitlynm26 • about 1 year ago • Physics • 5 pts
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1 answer

According to Kepler's Third Law of planetary motion, the ratio given here for the moon orbiting the Earth should be the same as a satellite. Therefore:

(19600)^3/T^2=1.01x10^18km^3/y^2

Solving for T we get .0027, which is in years. This converts to about 1 Earth day :)

answered by @juanv2 • 12 months ago