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

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  • 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

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