Question

8 × 107 m and its rotation period to be 22.3 hours. You have previously determined that the planet orbits 4.6 × 1011 m from its star with a period of 402 earth days. Once on the surface you find that the acceleration due to gravity is 22.4 m/s2. What are the mass of (a) the planet and (b) the star? (G = 6.67 × 10-11 N · m2/kg2) 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 × 107 m and its rotation period to be 22.3 hours. You have previously determined that the planet orbits 4.6 × 1011 m from its star with a period of 402 earth days. Once on the surface you find that the acceleration due to gravity is 22.4 m/s2. What are the mass of (a) the planet and (b) the star? (G = 6.67 × 10-11 N · m2/kg2) (a) 2.7 × 1025 kg, (b) 2.8 × 1031 kg (a) 5.0 × 1025 kg, (b) 2.8 × 1031 kg (a) 5.0 × 1025 kg, (b) 4.8 × 1031 kg (a) 2.7 × 1025 kg, (b) 4.8 × 1031 kg

Answer 1

(a) $2.7\cdot 10^{25} kg$

The acceleration due to gravity on the surface of the planet is given by

$g=\frac{GM}{R^2}$ (1)

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

Here we know:

$g=22.4 m/s^2$

$d=1.8\cdot 10^7 m$ is the diameter, so the radius is

$R=\frac{d}{2}=\frac{1.8\cdot 10^7 m}{2}=9\cdot 10^6 m$

So we can re-arrange eq.(1) to find M, the mass of the planet:

$M=\frac{gR^2}{G}=\frac{(22.4 m/s^2)(9\cdot 10^6 m)^2}{6.67\cdot 10^{-11}}=2.7\cdot 10^{25} kg$

(b) $4.8\cdot 10^{31}kg$

The planet is orbiting the star, so the centripetal force is equal to the gravitational attraction between the planet and the star:

$m\frac{v^2}{r}=\frac{GMm}{r^2}$ (1)

where

m is the mass of the planet

M is the mass of the star

v is the orbital speed of the planet

r is the radius of the orbit

The orbital speed is equal to the ratio between the circumference of the orbit and the period, T:

$v=\frac{2\pi r}{T}$

where

$T=402 days = 3.47\cdot 10^7 s$

Substituting into (1) and re-arranging the equation

$m\frac{4\pi r^2}{rT^2}=\frac{GMm}{r^2}\\\frac{4\pi r}{T^2}=\frac{GM}{r^2}\\M=\frac{4\pi r^3}{GMT^2}$

And substituting the numbers, we find the mass of the star:

$M=\frac{4\pi^2 (4.6\cdot 10^{11} m)^3}{(6.67\cdot 10^{-11})(3.47\cdot 10^7 s)^2}=4.8\cdot 10^{31}kg$