Find the speed of an orbiter in a circular orbit that is just above the surface...
Find the speed of an orbiter in a circular orbit that is just above the surface of a planet, given that the orbiter's acceleration is equal to the planets gravitational acceleration of 1.85 m/s^2. The radius of the planet is 1.54*10^6 m.
Homework Answers
Solution)
We know, acceleration,.a= v^2/r
So, v= root aR
V= root (1.85* 1.54*10^6) = 2.84 * 10^6 m/s (Ans)
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