Rotation of a planet, kinetic energy

AI Thread Summary
The discussion centers on calculating the ratio of kinetic energy of a planet's rotation to the kinetic energy of its center of mass orbiting its sun. The user initially calculated the rotational kinetic energy (KErot) and the orbital kinetic energy (KE) but received an incorrect ratio. It was pointed out that the user mistakenly divided the values in the wrong order, leading to confusion about the correct answer. After clarification, the user realized the correct ratio should be approximately 2.140E-6, although there was uncertainty about previous feedback on this answer. The conversation highlights the importance of careful calculations and understanding the formulas used in physics.
Dotty21690
Messages
13
Reaction score
0

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!
 
Physics news on Phys.org
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:)
Dotty21690 said:
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:
 
Why would you square the radius in the second part? because isn't the V=Rw?? so the radius wouldn't be squared?
 
It's still mv2, which is mR2w2.
 
i'm still getting the wrong answer and I don't know why.
 
am I using the wrong equation to find the kinetic energy of the center of mass for the orbit??
 
You should be using 1/2 mv2 which is the same as 1/2 m (2πr/T)2 :smile:
 
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.
 
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
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
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
I think you divided your ratio the wrong way. After all, the answer's power is suspiciously close to your answer's power.
 
  • #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!
 

Similar threads

Back
Top