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Rotation of a planet, kinetic energy

 
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Mar5-10, 04:20 PM   #1
 

Rotation of a planet, kinetic energy


1. The problem statement, all variables and given/known data

Calculate the ratio between the kinetic energy of rotation of a planet (mass=4.30E+24 kg, radius=7.60E+6 m) to the kinetic energy of its center of mass orbiting around its sun at a distance of 1.20E+12 m. Like the Earth, it has a day lasting 24 hours and a year lasting 365.25 days.


2. Relevant equations

KErot= (1/2)Iw^2 and I believe KE=(1/2)mv^2

3. The attempt at a solution

For the KErot:
KErot= (1/2)((2/5)(4.3E24kg)(7.6E6m)(2pi/(24hr*60min*60sec))^2
and I got 2.62698301E29 J

for the other one:
KE=(1/2)(4.30E24kg)(2pi*1.2E12m/(365.25days*24hrs*60min*60sec))
and got 1.22730571E35 J

I devided KE by KErot and got 467192 for the ratio. but it's wrong and I know I'm doing something wrong but can't figure it out. any help will be great!
 
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Mar5-10, 04:50 PM   #2
 
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Hi Dotty21690! Welcome to PF!

(have an omega: ω and a pi: π and try using the X2 tag just above the Reply box )
Quote by Dotty21690 View Post
Calculate the ratio between the kinetic energy of rotation of a planet (mass=4.30E+24 kg, radius=7.60E+6 m) to the kinetic energy of its center of mass orbiting around its sun at a distance of 1.20E+12 m. Like the Earth, it has a day lasting 24 hours and a year lasting 365.25 days.

For the KErot:
KErot= (1/2)((2/5)(4.3E24kg)(7.6E6m)(2pi/(24hr*60min*60sec))^2
and I got 2.62698301E29 J

for the other one:
KE=(1/2)(4.30E24kg)(2pi*1.2E12m/(365.25days*24hrs*60min*60sec))
and got 1.22730571E35 J
Did you remember to square the radius in both cases?
 
Mar5-10, 05:08 PM   #3
 
Why would you square the radius in the second part? because isn't the V=Rw?? so the radius wouldn't be squared?
 
Mar5-10, 05:13 PM   #4
 
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Rotation of a planet, kinetic energy


It's still mv2, which is mR2w2.
 
Mar5-10, 05:16 PM   #5
 
i'm still getting the wrong answer and I don't know why.
 
Mar5-10, 05:25 PM   #6
 
am I using the wrong equation to find the kinetic energy of the center of mass for the orbit??
 
Mar5-10, 05:34 PM   #7
 
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You should be using 1/2 mv2 which is the same as 1/2 m (2πr/T)2
 
Mar5-10, 05:48 PM   #8
 
so when I plug in my information it will look like this?:

(1/2)(4.3E24kg)((2π*1.2E12m)/31557600sec)^2)

I get 1.22730571E35. but when I do the ratio, I'm getting the answer wrong.
 
Mar6-10, 03:53 AM   #9
 
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Hi Dotty21690!

(please use the X2 tag just above the Reply box )

Show us your full calculations (and cancel out the common factors before you start).
 
Mar6-10, 10:26 AM   #10
 
well I showed the work earlier to get the answers for the KErot and KE, then I divided. I'll post it again though:

For the KErot:
KErot= (1/2)((2/5)(4.3E24kg)(7.6E6m)(2pi/(24hr*60min*60sec))^2
and I got 2.62698301E29 J

for the other one:
KE=(1/2)(4.30E24kg)(2pi*1.2E12m/(365.25days*24hrs*60min*60sec))
and got 1.22730571E35 J

then I divided 1.22730571E35 J/2.62698301E29 J and got 467192.1. I don't even know if I am doing the problem right.
 
Mar6-10, 11:05 AM   #11
 
turns out the answer was 2.140*10-6, which I had got at one time, but it said it was wrong :( but I don't know how to get that answer again. so can someone help me with this one?
 
Mar6-10, 04:10 PM   #12
 
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I think you divided your ratio the wrong way. After all, the answer's power is suspiciously close to your answer's power.
 
Mar6-10, 04:35 PM   #13
 
turns out I did divide the wrong way!! but I wonder why I was told the answer was wrong when I did put in 2.140*10-6, hmm. But thanks guys!!
 
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