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