User blog:Uninown/The Sun's rotational energy



On this occasion, I'll try calculating the Sun's rotational energy. Shouldn't be too hard.

The formula for rotational energy is 1/2* Ι*ω^2, where I equals to moment of inertia and ω to angular velocity.

Since the Sun is a near-perfect sphere, we'll use the moment of inertia of one, which is 2/5*m*r^2, where m is equal to mass and r to radius.

According to this site, the Sun's mass is 1.989 x 10^30 kilograms, whereas the radius is equivalent to 696,000 kilometers, or 696,000,000 meters.

Angular velocity, retrieving it from this site (don't have a single clue how to properly calculate that nor I've found any other sources, please correct me if faulty) is 2.865329607243705e-06 rad/s.

Now, the equation for the inertia:

2/5*1.989 x 10^30*69600000^2 = 3.8540137e+45

Having those values ready, we're now able to finish the last equation:

1/2*(3.8540137e+45*2.865329607243705e-06^2) = 1.5820945e+34 J

1.5820945e+34/4.184 = 3.7812966e+33/10^33 = 3.7812966 Yottatons, or near-baseline Large Planet level.