- #1
rrfergus
- 4
- 0
Homework Statement
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! :)
Homework Equations
KE = 1/2Iω^2
I = 2/5MR^2
ω = 2π/T (T is period)
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.