Question

A planet of mass m = 4.25 x 1024 kg orbits a star of mass M = 6.75 x 1029 kg in a circular path. The radius of the orbits R = 8.85 x 107 km. What is the orbital period Tplanet of the planet in Earth days? ​

Answer 1

285.3 days

Explanation:

The centripetal force $F_c$ experienced by the planet is the same as the gravitational force $F_G$ so we can write

$F_c = F_G$

or

$m\dfrac{v^2}{R} = G\dfrac{mM}{R^2}$

where M is the mass of the star and R is the orbital radius around the star. We know that

$v = \dfrac{C}{T} = \dfrac{2\pi R}{T}$

where C is the orbital circumference and T is orbital period. We can then write

$\dfrac{4\pi^2R}{T^2} = G\dfrac{M}{R^2}$

Isolating $T^2$, we get

$T^2 = \dfrac{4\pi^2R^3}{GM}$

Taking the square root of the expression above, we get

$T = 2\pi \sqrt{\dfrac{R^3}{GM}}$

which turns out to be $T = 2.47×10^7\:\text{s}$. We can convert this into earth days as

$T = 2.47×10^7\:\text{s}×\dfrac{1\:\text{hr}}{3600\:\text{s}}×\dfrac{1\:\text{day}}{24\:\text{hr}}$

$\:\:\:\:\:= 285.3\:\text{days}$