What is the rotational kinetic energy of a planet about its spin axis? Model the planet as a uniform sphere of radius 6000 km, and mass 6.21 x 10^24 kg. Assume it has a rotational period of 24.0 h.

I=2mR²/5

ω=2π/T
KE= Iω²/2

KE= m(R^2)ω²/5

To find the rotational kinetic energy of a planet about its spin axis, you can use the following formula:

Rotational Kinetic Energy (KE) = (1/2) * I * ω^2

Where:
- KE is the rotational kinetic energy
- I is the moment of inertia of the planet
- ω is the angular velocity of the planet

First, let's calculate the moment of inertia (I) of the planet. For a uniform sphere, the moment of inertia is given by the formula:

I = (2/5) * m * r^2

Where:
- m is the mass of the planet
- r is the radius of the planet

Given values:
- Mass (m) = 6.21 * 10^24 kg
- Radius (r) = 6000 km = 6 * 10^6 m

Plugging these values into the formula, we get:

I = (2/5) * (6.21 * 10^24 kg) * (6 * 10^6 m)^2

Simplifying this expression, we find the moment of inertia of the planet.

Next, let's calculate the angular velocity (ω) of the planet. The rotational period of the planet is given as 24.0 h. The angular velocity is the rotation angle per unit time. In this case, one rotation is equal to 2π radians, and the time is given in hours.

Angular velocity (ω) = (2π radians) / (time in seconds)

To convert the rotational period from hours to seconds, multiply it by 60 (minutes per hour), and again by 60 (seconds per minute). Then, calculate the angular velocity using the formula.

Finally, we can plug the values of the moment of inertia (I) and angular velocity (ω) into the formula for rotational kinetic energy (KE) to find the answer. Don't forget to multiply by (1/2) to get the correct result.