# 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

Relevance
• The orbit's eccentricity

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

e = 0.1488

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

Rₑ = 8687064 meters

r₀ = Rₑ+h₀

r₀ = 61255282 meters

The orbit's semimajor axis

a = r₀/(1−e)

a = 71963442 meters

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

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

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