## Trending News

# What is the moon's orbital period in earth days?

A 1.4 x 10^21 kg moon orbits a distant planet in a circular orbit of radius 1.5 x 10^8 m. It experiences a 1.1 x 10^19 N gravitational pull from the planet. What is the moon's orbital period in earth days?

### 1 Answer

- billrussell42Lv 72 months ago
F = G m₁m₂/r² = (6.674e-11)(1.4e21)M/(1.5e8)² = 1.1e19N

where M is planet mass

M = (1.1e19)(1.5e8)² / (6.674e-11)(1.4e21)

T = 2π√[R³/GM]

T = 2π√[(1.5e8)³/(6.674e-11)M]

T = 2π√[(1.5e8)³ / (6.674e-11)((1.1e19)(1.5e8)² / (6.674e-11)(1.4e21))]

T = 2π√[(1.5e8)³ (1.4e21) / (1.1e19)(1.5e8)²]

T = 2π√[(1.5e8)³(1.4e21) / (1.1e19)]

you can do the rest. Note that this is in seconds.

Gravitational attraction in newtonsF = G m₁m₂/r²G = 6.674e-11 m³/kgs²m₁ and m₂ are the masses of the two objects in kgr is the distance in meters between their centers

Satellite motion, circularV = √(GM/R)V = √(gR)T = 2π√[R³/GM]g = GM/R² T is period of satellite in sec V = velocity in m/s g = acceleration of gravity in m/s² (9.8 m/s² at ground level) G = 6.673e-11 Nm²/kg² M is mass of central body in kg R is radius of orbit in m