Planet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2 . A satellite orbits Planet R47A in a circular orbit of radius 5000 km and period 4.0 hours. What is the radius of Planet R47A?

 

Answer:

The radius of the planet is 2626.4 km

The detailed solution is explained below:

Step-by-step explanation

The force due to the gravitational attraction between the planet and the satellite will be the force causing the rotation. The calculation is shown below:

a_g= 3.45 m/s²

r=5000 km

T=4.0 hours

GMm/?r2?=mω2/0?r

GM=mω2/0?r3

GM=ag?R^2

ag?=ω2/0?r^3/R^2

?ω0?=2π?/4×60×60

3.45=(2π?/4×60×60?)^2(5000×10^3)^3/R^2

?R=262.64×104m

R=2626.4km?