A satellite circles planet Roton every 2.8 h in an orbit having a radius of 1.2* 107 m. If the radius of Roton is 5.0* 106 m, what is the magnitude of the free-fall acceleration on the surface of Roton?
A)31 m/s2
B)27 m/s2
C)34 m/s2
D)40 m/s2
E)19 m/s2

Question

Quizplus · Accepted Answer

The gravitational force provides the centripetal force for the satellite. Using the formula for centripetal acceleration and the gravitational force, we can solve for the gravitational acceleration at the surface of Roton. The formula is g = (4π²r³)/(T²R²), where r is the orbit radius, T is the period, and R is the planet's radius. Plugging in the values gives approximately 27 m/s².

A Satellite Circles Planet Roton Every 2