Can you find the mass of the exoplanet?

Given...

That a spaceship orbits in the equatorial plane of an exoplanet , and an observer on the spaceship measures the following:

The exoplanet's angular equatorial diameter at periapsis

Dₑ₀ = 16.3060482°

The exoplanet's angular equatorial diameter at apoapsis

Dₑ₁ = 12.0634564°

The exoplanet's angular polar diameter at periapsis

Dᵨ₀ = 16.2567986°

The exoplanet's angular polar diameter at apoapsis

Dᵨ₁ = 12.0271323°

The spaceship's sidereal orbital period around the exoplanet

P = 108000 sec

The spaceship's altitude above the exoplanet's surface at periapsis

h₀ = 52568218 meters

Find...

The exoplanet's mass

The exoplanet's average density

The semimajor axis of the spaceship's orbit

The eccentricity of the spaceship's orbit

2 Answers

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  • The orbit's eccentricity

    e = [sin(½Dₑ₀)/sin(½Dₑ₁)−1] / [sin(½Dₑ₀)/sin(½Dₑ₁)+1]

    e = 0.1488

    The planet's equatorial radius

    Rₑ = 2h₀ / {(1−e)[csc(½Dₑ₀)+csc(½Dₑ₁)] − 2}

    Rₑ = 8687064 meters

    The orbit's periapsis radius

    r₀ = Rₑ+h₀

    r₀ = 61255282 meters

    The orbit's semimajor axis

    a = r₀/(1−e)

    a = 71963442 meters

    The planet's average radius

    R = Rₑ ∛{sin(½Dᵨ₀)/sin(½Dₑ₀)}

    R = 8678368 meters

    The planet's geometric volume

    V = (4π/3)R³

    V = 2.737807e+21 m³

    The planet's average density

    ρ = 3πa³/(GP²R³)

    ρ = 6903.026 kg m⁻³

    The planet's mass

    M = ρV

    M = 1.889915e+25 kg

    The orbit's apoapsis radius

    r₁ = a(1+e)

    r₁ = 82671602 meters

    Orbital speed at periapsis, relative to the planet's center

    v₀ = √[GM(2/r₀−1/a)]

    v₀ = 4863.786 m/s

    Orbital speed at apoapsis, relative to the planet's center

    v₁ = √[GM(2/r₁−1/a)]

    v₁ = 3603.808 m/s

    The planet's polar radius

    Rᵨ = R³/Rₑ²

    Rᵨ = 8661002 meters

    The planet's oblateness

    f = 1 − Rᵨ/Rₑ

    f = 0.003

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

    g = GM/Rₑ²

    g = 16.71483 m/s²

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

    v = √(2GM/Rₑ)

    v = 17041.292 m/s

  • Anonymous
    4 months ago

    can you just give a 5 star best answer please

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