Answer:
T² = (π² / 2G M_{s}) (R₁ + R₂)³
Step-by-step explanation:
Let's use Newton's second law where the force is the universal gravitational force
F = m a
the acceleration is centripetal
a = v² / r
the out of gravitational universal is
F = G
m / r²
we substitute
G M_{s} m / r² = m v² / r
G M_{s} / r = v²
the planet's speed in orbit is
v = d / T
v = 2π r / T
we substitute
G M_{s}/ r = 4π² r² / T²
T² = 4π² / G M_{s} r³
In the case of an elliptical orbit the distance r is the length of the semi-major axis, see attached for the nomenclature
T² = 4π² / G M_{s} a³
indicates that the minimum distance is R₁ = a -c and the maximum distance is R₂ = a + c, let's add these two expressions
R₁ + R₂ = 2 a
we substitute in the equation of the period
T² = 4π²/ G M_{s} (R₁ + R₂)³/2³
T² = (π² / 2G M_{s}) (R₁ + R₂)³