An undiscovered planet, many light-years from Earth, has one moon, which has a nearly circular periodic orbit. If the distance from the center of the moon to the surface of the planet is 2.090×105 km

Question

An undiscovered planet, many light-years from Earth, has one moon, which has a nearly circular periodic orbit. If the distance from the center of the moon to the surface of the planet is 2.090×105 km and the planet has a radius of 3625 km and a mass of 4.15×1022 kg, how long (in days) does it take the moon to make one revolution around the planet? The gravitational constant is 6.67×10−11N·m2/kg.

Answer 1

The force acting on the planet is G*M*m/(R²)

This has to be equal to the centripetal force to keep it in orbit.

Hence, G*M*m/(R^2) = m* V²/ R (call this Eq. 1)

In this case, we can treat V as the speed (and not velocity).

Also,

V = distance / time , where distance = perimeter of the orbit = 2* Pi * R.

Hence, time for revolution = (2* Pi * R) / V (call this eq. 2)

We will use equations 1 and 2 to get the answer now.

From Eq. 1, we have

V^2/R = G * M / R² , which means V = Sqrt(G* M / R)

Hence time =(2 * Pi * R) / Sqrt(G* M / R) (call this eq 3)

Note that R = 3625 +2.090×10⁵

Be careful to convert km to meters above!!

On solving eq 3, we get time = 1.17*10⁷ seconds.

Divide this by (24*60*60) to convert seconds to days.

So we finally get, time = 135.52 days