Question

The planet in a distant solar system has a diameter equal to 1.50×107 m and a free-fall acceleration on the surface that is equal to 13.1 m/s2 . . The planet orbits a distance 2.20×1011 m from its star (center-to-center distance) with a period of 402 earth days. Part AWhat is the mass of the planet?Part BWhat is the mass of the star?

Answer 1

Part A: 1.47 * 10^ 20 kg

Part B: 5.22 * 10^30 kg.

Step-by-step explanation:

The acceleration due to gravity on a planet of mass M and radius r is given by

`a=G\frac{M_{}}{r^2}`

where G = 6.67 *10^-11 m^3 / kg *s^2 = gravitational constant.

Now solving for M gives

`M=(ar^2)/(G)`

Putting in a = 13.1 m /s^2 and r = (1.5 * 10^9 / 2) gives

`M=(13.1*(1.5*10^9)/2)/(6.67*10^(-11))`

which evaluates to give

which is the mass of the planet.

Part B.

Here we use Kepler's third law, which says

where

T = orbital period of the planet

M = mass of the star

a = distance between the star and the planet.

G = gravitational constant

Now solving for M gives

`M=(4\pi^2)/(GT^2)a^3`

Now putting a = 2.20 * 10^11 , G = 6.67 * 10^-11, and T = 402 * 24 * 60 *60 gives

`M=(4\pi^2)/((6.67*10^(-11))*(402\cdot24\cdot60\cdot60)^2)\cdot(2.20*10^(11))^3`

Hence, the mass of the star is 5.22 * 10^30 kg.