Calculate the rotational kinetic energy of Earth on its axis. The Mass of the Earth is 5.98 x 1024 kg and the radius is 6.38 x 106 m.

1.22×10^38Joules

Explanation:

Rotational Kinetic energy is expressed as I = 1/2MR²

M is the mass of the earth

R is the radius of the earth

Given

M = 5.98 x 10^24 kg

R = 6.38 x 106 m

Substitute

I = 1/2(5.98 x 10^24)(6.38 x 10^6)²

I = 1/2(243.41)×10^{24+12}

I = 1/2(243.41)×10^36

I = 121.71×10^36

I = 1.22×10^38

Hence the rotational kinetic energy of Earth on its axis is 1.22×10^38Joules

i think the answer is a and b

i wish you luck