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Physics behind Halo, the game 
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#19
Mar1512, 07:43 AM

P: 44

Okay, I'll try putting h=1000 m (which is small compared to 10000 km) Putting R=10000000 m Putting G= the usual value Putting M=1.7x10^17 kg (from wikipedia) Hmmmm. This gives me 0.000275 as the answer. That's really less. 


#20
Mar1512, 08:42 AM

Sci Advisor
Thanks
PF Gold
P: 12,185

Here's a bit of arm waving.
Divide your ring into equal segments. (small as you like). With the test mass at the centre, they are all equidistant from the mass so no net force. Move towards the edge. Each segment on ' your side' of the the ring will have a corresponding piece on the other side of the ring. You will be nearer to each of the pieces on your side than to the mirror image segments on the other side. The effect of each segment will be proportional to 1/d^{2} so the sum of contributions from each of the nearer pieces will be far greater than the sum (same number) of all contributions from their equivalent pieces. But the number of segments involved is only a linear function of your position on the radius. This has to result in a finite sum and not zero. In the limit, for a large ring and a small measurement frame, you would have a 'flat earth' situation. This contrasts with the shell situation in which there are progressively more elements on the 'opposite wall'  a 'square law, in fact  which compensates the inverse square law for the force from each. 


#21
Mar2112, 07:20 PM

P: 7




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