Question

You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 1.8×107m and its rotation period to be 22.3 hours . You have previously determined that the planet orbits 2.2×1011m from its star with a period of 407 earth days. Once on the surface you find that the free-fall acceleration is 12.2m/s2.
What is the mass of the Planet?What is the mass of the star?

Answer 1

Answer:

Mass of the planet = 1.48 × 10²⁵ Kg

Mass of the star = 5.09 × 10³⁰ kg

Explanation:

Given;

Diameter = 1.8 × 10⁷ m

Therefore,

Radius = $\frac{\textup{Diameter}}{\textup{2}}$  = $\frac{\textup{1.8}\times10^7}{\textup{2}}$

or

Radius of the planet = 0.9 × 10⁷ m

Rotation period = 22.3 hours

Radius of star = 2.2 × 10¹¹ m

Orbit period = 407 earth days = 407 × 24 × 60 × 60 seconds = 35164800 s

free-fall acceleration = 12.2 m/s²

Now,

we have the relation

g = $\frac{\textup{GM}}{\textup{R}^2}$

g is the free fall acceleration

G is the gravitational force constant

M is the mass of the planet

on substituting the respective values, we get

12.2 = $\frac{6.67\times10^{-11}\times M}{(0.9\times10^7)^2}$

or

M = 1.48 × 10²⁵ Kg

From the Kepler's law we have

T² = $\frac{\textup{4}\pi^2}{\textup{G}M_{star}}(R_{star})^3$

on substituting the respective values, we get

35164800² = $\frac{\textup{4}\pi^2}{6.67\times10^{-11}\timesM_{star}}(2.2\times10^{11})^3$

or

$M_{star}$ = 5.09 × 10³⁰ kg