Rotational Kinetic Energy of Planet

1. Nov 16, 2013

rrfergus

1. The problem statement, all variables and given/known data
This question has two parts; the first part I understand but the second part I do not.
Part 1: What is the rotational energy of a planet about its spin axis? Model the planet as a uniform sphere of radius 6420 km, and mass 5.59x10^24 kg. Assume it has a rotational period of 24.0 h. The answer to this is 2.44x10^29J. I understand this part.
Part 2: Suppose the rotational kinetic energy of the planet is decreasing steadily because of tidal friction. Assuming the rotational period increases 15.0 μs each year, find the change in one day. (Let 1 yr = 365 d.). The answer to this is -2.32x10^17 J/day, but I do not understand how to get it. Thank you! :)

2. Relevant equations
KE = 1/2Iω^2
I = 2/5MR^2
ω = 2π/T (T is period)

3. The attempt at a solution
If T increases by 15μs each year, then at the end of 1 year the period will be 24h(3600s/h) + 15x10^-6s. So, I use that to find the rotational kinetic energy of the planet at the end of the year and I get 2.4369x10^29J. Then, I subtract the original rotational kinetic energy (the answer from part 1) and divide the result by 365 (since there are 365 days/year). The result is about -8.39x10^23J/day, which is way off from the correct answer.

2. Nov 16, 2013

lucasem_

Make sure your units are consistent between problems. I recommend converting everything into units of days when determining the kinetic energy. Simply diving by 365 in the very end may not give you what you're looking for, try calculating $\Delta E = \frac{1}{2}I(\omega_2^2-\omega_1^2)$ after your units of $\omega$ are consistent. It's likely the error was made with unit conversion.

Once all your units are for a single day, you won't need to divide by anything in the end. Your answer will be the change in energy from one day to the next, so it would be equivalent to Joules/Day

3. Nov 16, 2013

rrfergus

Great! I got the right answer! Thank you so much!