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Compare centripetal acceleration of Mars and Earth

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  1. Nov 7, 2011 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant 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.

    3. 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.
     
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  3. Nov 7, 2011 #2

    rock.freak667

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    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?)
     
  4. Nov 7, 2011 #3
    So with Newton's law of universal gravitation it's: F = [itex]\frac{G(m_{Earth})(m_{Sun})}{r^{2}}[/itex]?
     
  5. Nov 7, 2011 #4

    rock.freak667

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

    Do the same with Mars now.
     
  6. Nov 7, 2011 #5
    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]
     
  7. Nov 7, 2011 #6

    rock.freak667

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    Right, so you can see that if you take aearth/amars then GMsun will cancel out.
     
  8. Nov 7, 2011 #7
    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?
     
  9. Nov 7, 2011 #8

    rock.freak667

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    That should be correct.
     
  10. Nov 7, 2011 #9
    Thanks, I really appreciate all of the help :smile:. I just thought that the solution would involve more than one variable.
     
  11. Mar 5, 2012 #10
    wrong if radius is larger acceleration is less
     
  12. Mar 6, 2012 #11

    rock.freak667

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    Wrong about what?
     
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