# What is the exoplanet's mass?

A starship is in an elliptical orbit in the equatorial plane of an exoplanet. The navigator makes the following observations:

The equatorial angular diameter at periapsis is 16.3060482°

The equatorial angular diameter at apoapsis is 12.0634564°

The polar angular diameter at periapsis is 16.2567986°

The polar angular diameter at apoapsis is 12.0271323°

The period of orbit is 108,000 seconds.

The altitude above surface of the exoplanet at periapsis is 52,568,218 meters.

Find the exoplanet's mass.

### 1 Answer

- 1 year ago
This question mostly involves Euclidean geometry and the laws of Kepler and Newton. I'll give the answers. You can still amuse yourselves figuring out how I calculated it.

The eccentricity of the spaceship's orbit around the exoplanet:

e = 0.1488

The planet's equatorial radius

Rₑ = 8687063.8 meters

The spaceship orbit's periapsis radius

r₀ = 6.12553e+07 meters

The spaceship orbit's semimajor axis

a = 7.19634e+07 meters

The spaceship orbit's apoapsis radius

r₁ = 8.26716e+07 meters

The planet's average radius

R = 8661002.6 meters = 1.3594 R⊕

The planet's geometric volume

V = 2.72140e+21 m³

The planet's polar radius

Rᵨ = 8609114.4 meters

The planet's oblateness

f = 0.0089730

The planet's average density

ρ = 6944.86 kg m⁻³

The planet's mass

M = 1.88998e+25 kg = 3.1645 M⊕

The planet's gravitational parameter,

GM = 1.26139e+15 m³ sec⁻²

The spaceship's orbital speed at periapsis, relative to the planet's center

v₀ = 4863.8 m/s

The spaceship's orbital speed at apoapsis, relative to the planet's center

v₁ = 3603.8 m/s

The planet's surface gravity at the equator, assuming that it isn't rotating

g = 16.816 m sec⁻²

The planet's escape speed from the surface at the equator

vₑ = 17041.3 m/s