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

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prace
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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!
 
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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.
 
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.
 
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.
 
neutrino said:
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.

Oh cool, thanks neutrino, that way is much easier. Thanks!
 
prace said:
It looks like I calculated the radius of the Earth in meters when it should have been kilometers!
No, it's the other way. :smile: