Question

In another solar system, a planet has a moon that is 4.0 × 105 m in diameter. Measurements reveal that this moon takes 3.0 x 105 s to make each orbit of diameter 1.8 × 108 m. What is the mass of the planet? (G = 6.67 × 10-11 N ∙ m2/kg2)

Answer 1

`M= 4.8*10^(24)kg`

Step-by-step explanation:

To solve this problem we need to apply the Kepler's third law, which say,

`T^2 = (4\pi^2R^3)/(GM)`

Where ,

T= Period

R = Radius

G =Gravitational constant

M = Mass

We have all that values, then replacing,

`(3*10^5)^2 = 4\pi^2((0.9*10^8)^3)/(6.67*10^(-11)M)`

Solving for M,

`M = 4\pi^2((0.9*10^8)^3)/(6.67*10^(-11)((3*10^5)^2 ))`

`M= 4.8*10^(24)kg`

Note that
`0.9*10^8` is used because was provided the value of the diameter, not the radius which is equal to the half.