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Force of gravity if the Earth's radius, r = r/2

  1. Dec 10, 2006 #1
    Problem: Suppose that the Earth retained its present mass but was somehow compressed to half its present radius. What would be the value of g at the surface of this new, compact planet?

    My work: So, this seems pretty simple, and I get the right answer, but I seem to be off by a lot of decimal places. Can anyone tell me what is wrong here with my calculations?

    g = (GMe)/(Re)² where Me = mass of the Earth, and Re = radius of the Earth.

    So if Re is compressed to half its present radius, then:

    g = (GMe)/(Re/2)² = (6.67E-11*5.98E24)/(6370/2)² = 39344273 m/s²

    The answer in my text gives 39.2 m/s².

    To try and check what I was doing wrong, I tried to calculate for the known value of g = 9.81 m/s² and I got 9836068.3 m/s². So it looks as if the correct numbers are there, but I am somehow messing this up.

    Thanks for any help!
  2. jcsd
  3. Dec 10, 2006 #2


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    Staff: Mentor

    Check the units of the quantities that you're calculating with, and make sure that they're consistent with the units of the answer that you're supposed to get.
  4. Dec 10, 2006 #3
    A simpler way to do this, without going through all the numbers, is to notice that the new value of g will be 4 times the original value. (the 2^2 goes to the numerator and 4x9.8 = 39.2 ms^-2)

    The only problem with your calculation could be that you've not written down the radius in metres.
  5. Dec 10, 2006 #4
    Oh my... so simple huh?! ^_^ Thank you. It looks like I calculated the radius of the Earth in meters when it should have been kilometers!! Thanks for seeing that for me.
  6. Dec 10, 2006 #5
    Oh cool, thanks neutrino, that way is much easier. Thanks!
  7. Dec 10, 2006 #6
    No, it's the other way. :smile:
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