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Absolute Motion

Started by neilski_2003 4 replies 1k views
Original Post
neilski_2003
neilski_2003
Cheers for any help that can be given. Neil [Edited by - neilski_2003 on April 25, 2006 6:11:12 AM]
rhinovirus
rhinovirus
Quote:
Original post by neilski_2003

As the point be is connected to the earth at a distance of 6378000 metres then v = 637800 * 7.27 *10^-5
= 463.68k m/s (which seems quite high to be honest - though checking it against the circumference seems to make it correct)


Check your units again, I'm getting 463.8 m/s, which doesn't seem unreasonable.
rhinovirus
rhinovirus
As for the acceleration - there is no acceleration tangential to the orbit path, but there is a centripetal acceleration, calculated as a = v^2/r. Hope that helps.
neilski_2003
neilski_2003
Hi,

Thanks very much for that - i'll go back over my workings - i may well have over generalised some numbers somewhere.

I'll look into the other thing about the acceleration see what i can mak of it.

Oh just to check - would you agree that the velocity of A follows the same as that above just with the extra 240km added to the radius?

Cheers fella

Neil
rhinovirus
rhinovirus
Quote:
Original post by neilski_2003
Oh just to check - would you agree that the velocity of A follows the same as that above just with the extra 240km added to the radius?


Yeah, same method different radius.

rhinovirus

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