- #1
David Baker
- 2
- 1
Text books ordinarily give the escape velocity of a mass-M body (in the center of mass frame of the system of the body and the escaping projectile, whose mass I'll label m) as
(*) v2 = 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
w2 = 2Gm2/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) z2 = 2G(M+m)/r.
Put in terms of the mixed variables, that’s
(b) v2 = (2GM2)/(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 mv2 + Mw2, i.e. half of mv2 + M(mv/M)2, i.e. half of v2 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/r2,
i.e. dz/dt = -G(M+m)/r2.
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.
(*) v2 = 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
w2 = 2Gm2/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) z2 = 2G(M+m)/r.
Put in terms of the mixed variables, that’s
(b) v2 = (2GM2)/(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 mv2 + Mw2, i.e. half of mv2 + M(mv/M)2, i.e. half of v2 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/r2,
i.e. dz/dt = -G(M+m)/r2.
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.