Rotation of a planet, kinetic energy

Click For Summary

Homework Help Overview

The problem involves calculating the ratio of the kinetic energy of rotation of a planet to the kinetic energy of its center of mass as it orbits around its sun. The planet's mass, radius, and orbital distance are provided, along with the duration of its day and year.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the formulas for kinetic energy, questioning the need to square the radius in certain calculations. There are attempts to clarify the correct application of the kinetic energy equations for both rotational and orbital motion.

Discussion Status

The discussion is ongoing, with participants providing guidance on the calculations and questioning the original poster's approach. There is recognition of potential errors in the division of the kinetic energy values, and some participants express uncertainty about the correctness of their results.

Contextual Notes

Participants are working within the constraints of homework rules and are encouraged to show their full calculations. There is mention of a previous answer that was deemed incorrect, leading to further exploration of the problem.

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

  • · Replies 21 ·
Replies
21
Views
5K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
6
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 9 ·
Replies
9
Views
3K