- #1

David Baker

- 2

- 1

(*) v

^{2}= 2GM/r

where r is the distance between the body and the escaping projectile.

it doesn’t seem to me that (*) can be right except as an approximation when M>>m. For let w be the escape velocity of the mass-M body in the center-of-mass coordinate system. w is

-(m/M)v,

so this would give

w

^{2}= 2Gm

^{2}/Mr, which isn’t of the same form as (*). Am I making a mistake here?

When I try to derive the escape velocity equation, what I get is put most naturally just using the variable r. Letting z be the rate of change of r, I get

(a) z

^{2}= 2G(M+m)/r.

Put in terms of the mixed variables, that’s

(b) v

^{2}= (2GM

^{2})/(M+m)r.

It is (b) (and hence (a)) rather than (*) that the energy argument seems to yield. The kinetic energy at escape velocity is half of mv

^{2}+ Mw

^{2}, i.e. half of mv

^{2}+ M(mv/M)

^{2}, i.e. half of v

^{2}times (M+m)m/M. This must be the negative of the potential energy, i.e. GMm/r. This yields (b), hence its equivalent (a).

Another reason for suspecting that it’s (a)/(b) rather than (*) that’s correct is that in the r formulation M and m should appear only in the combination M+m. For the 2-body problem is equivalent to a 1-body problem with reduced mass Mm/(M+m):

[Mm/(M+m)]dz/dt = -GMm/r

^{2},

i.e. dz/dt = -G(M+m)/r

^{2}.

So in the variable r, there’s no separate role of M and m besides in M+m.

Is this right? I've tried checking some orbital mechanics books, and none of them mention that the escape velocity equation is an approximation. But I really don't think I'm making a mistake here.