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 Be the first pioneers to continue the Astronomy Discussions at our new Astronomy meeting place...The Space and Astronomy Agora Uhmm Forum List | Follow Ups | Post Message | Back to Thread Topics | In Response ToPosted by Tim on April 30, 2003 22:15:45 UTC

let us look at it another way:

we know (1) F=MA
and (2) F=G(M1xM2)/R^2

also for case of M1=M2 we can say F=GM^2/R^2 holds....
notice in this case it is easy to see that we can equate from (1)&(2) A=GM/R^2

also if M1>M2 we can invision two cases:

(3) F=M2x(GM1/R^2) force of gravity exerted on M2 by M1
&
(4) F=M1x(GM2/R^2) force of gravity exerted on M1 by M2
and yes both (3) and (4) are the same thing but we can interpret the meaning by how we group the terms F[of M1 or (3)] = -F[of M2 or (4)]

again because of equation (1)
we can equate as follows A = GM1/R^2 for case (3) which is the case we are really interested in!
note that it matters not what value M2 holds in this case the acceleration shall allways be GM1/R^2 . the gravitation force may be larger or smaller depending on the size of M2 when M1>M2 but the acceleration of M2 remains the same no matter what the size of M2.

hence masses in the neighborhood of a more massive object will fall towards that object with the same acceleration regardless of the size of their mass..

regards tim