Question

2. A particular planet has a moment of inertia of 9.74 × 1037 kg•m2 and a mass of 5.98 × 1024 kg. Based on these values, what is the planet’s radius?

Answer 1

`6.38\cdot 10^6 m`

Step-by-step explanation:

The planet can be thought as a solid sphere rotating around its axis. The moment of inertia of a solid sphere rotating arount the axis is

`I=(2)/(5)MR^2`

where

M is the mass

R is the radius

For the planet in the problem, we have

`M=5.98\cdot 10^(24) kg`

`I=9.74\cdot 10^(37) kg\cdot m^2`

Solving the equation for R, we find the radius of the planet:

`R=\sqrt{(5I)/(2M)}=\sqrt{(5(9.74\cdot 10^(37))/(2(5.98\cdot 10^(24))}=6.38\cdot 10^6 m`