Question

The certain forest moon travels in an approximately circular orbit of radius

Answer 1

Answer:

Mass of Exoplanet =  0.58 kg

Explanation:

First, we will calculate the speed of the forest moon:

$speed = v = \frac{Circumference}{time}$

circumference = 2πr = 2π(14441566 m) = 90739035.3 m

time = 6 days 10 hr = (6 days)(24 h/1 day)(3600 s/1 h) + (10 h)(3600 s/1 h)

time = 554400 s

Therefore,

$v = \frac{90739035.3\ m}{554400\ s}\\\\v = 163.67\ m/s$

We know that the centripetal force on forest moon will be equal to the gravitational force given by Newton's Gravitational Law, as follows:

$Centripetal\ Force = Gravitational\ Force\\\frac{m_{moon}v^2}{r} = \frac{Gm_{moon}m_{exoplanet}}{r^2}\\\\m_{exoplanet} = \frac{v^2r}{G}\\\\m_{exoplanet} = \frac{(163.67\ m/s)^2(14441566)}{6.67\ x\ 10^{-11}\ N.m^2/kg^2}$

Mass of Exoplanet =  0.58 kg