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Homework Help: Pendulum swinging above earth's surface

  1. Nov 26, 2011 #1
    1. The problem statement, all variables and given/known data
    Assume all oscillators are frictionless.
    a) A pendulum consists of point mass Mo swinging on a massless string of length Lo, with period To = 9.42 s on the surface of the Earth (at RE, the radius of the Earth). Find T1, the period of the same pendulum if it swings on a horizontal platform of height H = 2.96RE (stationary, not orbiting) above the Earth's surface.

    2. Relevant equations
    T = 2∏√(L/g) , g = GM/R2 , Radius of earth = 6.37e6m, mass of earth = 5.98e24kg

    3. The attempt at a solution
    g(T/2∏)2 = L
    9.81m/s2√(9.42s/2∏)2 = 22.05012604m

    g = GM/R2
    (6.67e-11 * 5.98e24kg)/(2.96 * 6.37e6)2 = g1
    3.98866e14/3.55518567e14 = 1.121927339m/s2 = g1

    T1 = 2∏√(L/g1)
    T1 = 2∏√(22.05012604/1.121927339)
    T1 = 27.85499219s

    The answer i calculated is wrong, any help with figuring out why it's wrong would be greatly appreciated:smile:
  2. jcsd
  3. Nov 27, 2011 #2


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    The R in the formula g = GM/R2 is the distance from the centre of the Earth.
    This pendulum is set up 2.96RE above the surface, so 3.96RE from the centre.
  4. Nov 27, 2011 #3
    Thank you for the response Peter. Are you saying that the R in the equation should be RE + 2.96RE?
  5. Nov 27, 2011 #4


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    I think yes. After you have worked it out, I will show you how I would have done the question - in a very simply way.
  6. Nov 27, 2011 #5
    Ok so I redid the problem with the correction of the R. g = (6.67e-11 * 5.98e24kg)/((2.96*6.37e6)+6.37e6)2 and got a g = .626841558. Then T = 2∏√(22.05012604/.626841558) and got T = 37.26s. I think that period should be right, I'm interested in your alternate method to solving this type of problem:smile:
  7. Nov 27, 2011 #6


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    OK it amounts to using variation.

    You already knew

    T = 2∏√L/g and g = GM/R2

    Substitute the expression for g into the first expression and we get

    T = 2∏√(LR2/GM)

    most of those terms don't change, so replacing 2∏√(L/GM) with something simple like k we get

    T = kR

    Now the new R value is 3.96 times the R value at the surface, so simple arithmetic →

    T = 9.42 x 3.96 → 37.3072s or 37.3 s when respecting the significant figures in supplied data.

    A couple of things to note:

    1: This type of comparative approach is very powerful and leads to simple calculation.

    2: Giving the distance above the Earth Surface is a common "tester" used in problems.
    It is always fun to see people calculate that gravity 600km above the surface is way stronger than at the surface, because someone calculates with R = 600 000 m rather than 600 000 m MORE than the Radius of the Earth.

    Another example:
    Compare the centripetal acceleration of a satellite orbiting a distance RE above the surface of the earth, to one orbiting at a distance 2RE above the surface.

    The centripetal acceleration is simple g at the point.

    Knowing the formula g = GM/R2 leads some to assume an answer 4:1, but of course it is actually 9:4 since the R values to use in the formula are not RE & 2RE but 2RE & 3RE
  8. Nov 27, 2011 #7


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    Note: I think your answer of 37.26 varies from my 37.3026 due to the rounded off values of Earth mass and Radius you have used.

    had you expressed your answer to 3 figures you would be giving the same answer as me.
  9. Nov 27, 2011 #8
    Thank you for your help. I have a habit of not rounding, I'm glad my professor isn't too harsh about it. I found that your alternate way of doing this calculation to be much more efficient. I think the extra example you mentioned is cleverly worded and is very possible for my professor to put something similar on the next exam. I appreciate your time and patience spent helping me.
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