• Support PF! Buy your school textbooks, materials and every day products Here!

Compare centripetal acceleration of Mars and Earth

  • #1

Homework Statement



Is the centripetal acceleration of Mars in its orbit around the Sun larger or smaller than the centripetal acceleration of the Earth?

Homework Equations



I assume that the relevant equation is a=v[itex]^{2}[/itex]/r since you are trying to compare the centripetal accelerations of Mars and Earth, however it is supposed to be a concept question according to the book so no data or variables should be needed to answer the question.

The Attempt at a Solution



So i know that Mars is farther away from the Sun than the Earth is, so r is obviously greater for Mars. Therefore the perimeter of Mars' orbit around the sun is greater than the perimeter of the Earth's orbit around the Sun. What I am confused about is how to figure out the velocity of Mars, or Earth for that matter. Is the velocity not needed to answer the question and I am merely missing something, or can you just assume that because Mars is farther away from the Sun it has a greater centripetal acceleration based on a trend?

If I did anything against the forum rules/policies please forgive me as this is my first post on Physics Forums.
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31
Considering the Earth only. The gravitational force between the sun and the earth provides the centripetal force for the Earth. Can you find an expression for this acceleration?

(Hint: What does the law of universal gravitation say?)
 
  • #3
Considering the Earth only. The gravitational force between the sun and the earth provides the centripetal force for the Earth. Can you find an expression for this acceleration?

(Hint: What does the law of universal gravitation say?)
So with Newton's law of universal gravitation it's: F = [itex]\frac{G(m_{Earth})(m_{Sun})}{r^{2}}[/itex]?
 
  • #4
rock.freak667
Homework Helper
6,230
31
So with Newton's law of universal gravitation it's: F = [itex]\frac{G(m_{Earth})(m_{Sun})}{r^{2}}[/itex]?

Right, so the centripetal acceleration of the Earth is therefore? (Newton's second law)

Do the same with Mars now.
 
  • #5
Right, so the centripetal acceleration of the Earth is therefore? (Newton's second law)

Do the same with Mars now.
so: m[itex]_{earth}[/itex]a = [itex]\frac{G*m_{Earth}*m_{Sun}}{r^2}[/itex]


divide by m[itex]_{Earth}[/itex]: a = [itex]\frac{G*m_{Sun}}{r_{Earth's Orbit}^2}[/itex]


Therefore for Mars the equation will be: a = [itex]\frac{G*m_{Sun}}{r_{Mars' Orbit}^2}[/itex]
 
  • #6
rock.freak667
Homework Helper
6,230
31
so: m[itex]_{earth}[/itex]a = [itex]\frac{G*m_{Earth}*m_{Sun}}{r^2}[/itex]


divide by m[itex]_{Earth}[/itex]: a = [itex]\frac{G*m_{Sun}}{r_{Earth's Orbit}^2}[/itex]


Therefore for Mars the equation will be: a = [itex]\frac{G*m_{Sun}}{r_{Mars' Orbit}^2}[/itex]
Right, so you can see that if you take aearth/amars then GMsun will cancel out.
 
  • #7
Right, so you can see that if you take aearth/amars then GMsun will cancel out.
So the only factor that matters when comparing the two centripetal accelerations is the radius of their orbits. Therefore Mars has a larger centripetal acceleration because its radius is larger. Is that right?
 
  • #8
rock.freak667
Homework Helper
6,230
31
That should be correct.
 
  • #9
That should be correct.
Thanks, I really appreciate all of the help :smile:. I just thought that the solution would involve more than one variable.
 
  • #10
wrong if radius is larger acceleration is less
 
  • #11
rock.freak667
Homework Helper
6,230
31

Related Threads on Compare centripetal acceleration of Mars and Earth

Replies
29
Views
5K
  • Last Post
Replies
6
Views
4K
Replies
15
Views
2K
Replies
2
Views
4K
  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
6
Views
7K
Replies
7
Views
2K
  • Last Post
Replies
7
Views
14K
Replies
3
Views
8K
Replies
1
Views
2K
Top