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Rotation of a planet, kinetic energy

  • Thread starter Dotty21690
  • Start date
  • #1
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Homework Statement



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.


Homework Equations



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

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!
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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Welcome to PF!

Hi Dotty21690! Welcome to PF! :smile:

(have an omega: ω and a pi: π and try using the X2 tag just above the Reply box :wink:)
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? :wink:
 
  • #3
13
0
Why would you square the radius in the second part? because isn't the V=Rw?? so the radius wouldn't be squared?
 
  • #4
Char. Limit
Gold Member
1,204
14
It's still mv2, which is mR2w2.
 
  • #5
13
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i'm still getting the wrong answer and I don't know why.
 
  • #6
13
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am I using the wrong equation to find the kinetic energy of the center of mass for the orbit??
 
  • #7
tiny-tim
Science Advisor
Homework Helper
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You should be using 1/2 mv2 which is the same as 1/2 m (2πr/T)2 :smile:
 
  • #8
13
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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.
 
  • #9
tiny-tim
Science Advisor
Homework Helper
25,832
250
Hi Dotty21690!

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

Show us your full calculations (and cancel out the common factors before you start). :smile:
 
  • #10
13
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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.
 
  • #11
13
0
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?
 
  • #12
Char. Limit
Gold Member
1,204
14
I think you divided your ratio the wrong way. After all, the answer's power is suspiciously close to your answer's power.
 
  • #13
13
0
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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