Question

On october 15, 2001, a planet was discovered orbiting around the star hd68988. its orbital distance was measured to be 10.5 million kilometers from the center of the star, and its orbital period was estimated at 6.3 days. what is the mass of hd68988

Answer 1

We can solve the problem by using Netwon's third law:
$\frac{T^2}{R^3} = \frac{4 \pi^2}{GM}$
where:
T is the orbital period of the planet
R is the average radius of the orbit (so, the average distance of the planet from the star)
$G=6.67 \cdot 10^{-11} m^3 kg^{-1}s^{-2}$
M is the mass of the star.

First, we need to convert the period and the radius into SI units:
$T=6.3 days = 5.44 \cdot 10^5 s$
$R=10.5 \cdot 10^6 km=10.5 \cdot 10^9m$
And then, re-arranging the formula and substituting the numbers we find
$M= \frac{4 \pi^2 R^3}{GT^2}=2.31 \cdot 10^{30}kg$