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Centripetal acceleration of Earth around Sun

  1. Feb 3, 2008 #1
    The earth's orbit (assumed circular) around the sun is 1.5e11 m in radius, and it makes this orbit once a year. What is the centripetal acceleration of the earth?

    a = (v^2) / r
    T = (2*pi*r) / v

    My work:
    T = (2*pi*r) / v;

    1 year = (365 days / 1 year)*(24 hours / 1 day)*(60 mins / 1 day)*(60 secs / 1 min) = 3.1536e7 s;

    3.1536e7 s = (2*pi*1.5e11 m) / v;
    algebraically rearranged is: v = (2*pi*1.5e11 m) / (3.1536e7 s)
    v = 29885.8 m/s

    a = (v^2) / r;

    a = ((29885.8 m/s)^2) / (1.5e11 m);
    a = 0.005954 m/s^2 MY ANSWER

    My question is if this is correct? I've been bombarded with tough questions up until this one and I am curious to know if I solved this correctly. It 'seemed' too easy. Confirmation on the answer would be appreciated since I can't find any information on presumed circular rotation around the sun. Thank you.
  2. jcsd
  3. Feb 3, 2008 #2

    Doc Al

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    Staff: Mentor

    Looks OK to me.
  4. Feb 3, 2008 #3


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    I haven't plugged the numbers in but your working is correct.
  5. Feb 13, 2008 #4
    The answer looks right. An alternative way of solving this question (to check your answer) would be to just ask what is the centripetal acceleration of earth around the sun, given the sun's gravitation force at our distance.


    G=gravitational constant=6.67E-11 m(3)kg(-1)s(-2)
    m1=mass of sun (1.00 E30) kg
    r=distance to the sun = 1.5E11 m

    Ie. acceleration = 6.67E-11 m(3)kg(-1)s(-2) * (1.00 E30) m / [ (1.5E11 m) * (1.5E11 m)]

    Answer = 5.8987E-03 ms(-2)

    My mass of distance were approximations, but the answer is very close indeed.
    Last edited: Feb 13, 2008

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