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Homework Help: Determine Jupiter radius based on graviational acceleration at surface

  1. Jun 26, 2010 #1
    1. The problem:
    The moon Europa, of the planet Jupiter, has an orbital period of 3.55 days and
    an average distance from the center of the planet equal to 671,000 km. If the
    magnitude of the gravitational acceleration at the surface of Jupiter is 2.36
    times greater than that on the surface of the Earth, what is the radius of
    Jupiter? (Hint: begin by calculating the rotation speed.)

    2. Relevant equations


    3. The attempt at a solution

    Not sure at all. I guess, the centripital force of the mon Europa has to be equal to the gravitational force. MeV^2/R=GMeMj/R^2, but then I am not sure what to do with the given gravitational acceleration of Jupiter and how to get the r-radius of jupiter.

    I appreciate any help.
  2. jcsd
  3. Jun 26, 2010 #2
    Re: Gravitation

    This is sort of a two step problem.

    Step 1.)
    Look at the first relevant equation (which appears to have a typo of one too many masses, but anyway...)
    The right side of that equation has Jupiter's radius in it, which is the thing you want to solve for. Given that on Jupiter's surface the acceleration is 2.36 g, what can you write the left side of the equation as? (Hint: If you were on the surface of the earth, what would you write it as? How does this change on Jupiter?)

    Step 2.) Hint: You'll need to use the info given about Jupiter's moon to calculate the piece of information you aren't directly given for the resulting equation in step 1. That's where you use the info from your suggested solution...
  4. Jun 26, 2010 #3
    Re: Gravitation

    Thank you pastro.
    Step 1. 2.36g=G/r^2 ?

    I found a solution to this problem in one of the previous posts, but there are few things I do not understand.

    radius of orbit of Europa, rm = 671000km
    rotational period of Europa, T = 3.55 days
    Acceleration due to gravity on Jupiter = GM/r² = 2.36g
    radius of Jupiter, r = ?

    2. Relevant equations
    Rotational velocity of Europa, ω = 2π/(3.55*86400) radians/second - WHY IS THERE NO 'r' v=2πr/T
    Centripetal acceleration, a =rmω² - ISN'T IT a=v^2/r?
    3. The attempt at a solution

    Gravitational acceleration on Europa

    Equating gravitational acceleration with centripetal acceleration,
    (2.36g)(r/rm)² = rmω²
    =74,038 km

    Could somebody please explain. I am trying to understand. Any help.
    Last edited: Jun 26, 2010
  5. Jun 27, 2010 #4
    Re: Gravitation

    Perform dimensional analysis and you'll see you're still missing something. Should be:
    2.36g = GMJ/r2

    So you need to know the mass of Jupiter, which is the unknown quantity I referred to before.

    ω is the angular velocity, not the linear velocity. In this case, you can think of angular velocity as the angle through which an object rotates divided by the time it takes to rotate through that angle. Since you are given the period (which means a rotation by 2π = 180 degrees), then ω is as you show.

    Note that linear velocity is related to angular velocity by the relationship v = ω r, which I use to answer your next question.

    This is a bit confusing, be cause you use rm, which appears to stand for "Jupiter-moon distance," not r m = distance times mass. Anyway, I'll make this distance rm to avoid confusion.

    v = ω rm, so a = v2/rm = (ω rm)2/rm = ω2rm
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