# Is the usual Escape Velocity eqn an approximation?

• David Baker
In summary, the conversation discusses the escape velocity equation for a mass-M body and the escaping projectile (m). The equation (*) v2 = 2GM/r is mentioned, but the speaker argues that it can only be an approximation when M>>m. Instead, they propose the equation (b) v2 = (2GM2)/(M+m)r, which is supported by an energy argument and is equivalent to (a) z2 = 2G(M+m)/r. The conversation also touches on the role of M and m in the 2-body problem and the assumption made in the usual escape velocity equation. Overall, the conversation presents (b) and (a) as the more accurate equations for escape velocity.
David Baker
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.

David Baker said:
(b) v2 = (2GM2)/(M+m)r.

That looks right for the velocity of one mass in the original rest frame of the two bodies.

The usual escape velocity equation is a very minor approximation based on the assumption that the larger body does not move. How much does the Earth move if a space probe is launched?

PeroK said:
The usual escape velocity equation is a very minor approximation based on the assumption that the larger body does not move.

Minor indeed! The ratio of the two velocities differs from 1 by only 1 part in 1016 when considering the escape velocity of a rocket 1000 times more massive than the Saturn V.

## 1. What is the usual Escape Velocity equation?

The usual Escape Velocity equation is a formula used to calculate the minimum speed an object needs to escape the gravitational pull of a planet or other celestial body. It is given by the equation: Ve = √(2GM/R) where Ve is the escape velocity, G is the gravitational constant, M is the mass of the planet, and R is the distance from the center of the planet to the object.

## 2. Is the usual Escape Velocity equation always accurate?

No, the usual Escape Velocity equation is an approximation and may not always give a completely accurate value for the escape velocity. It assumes a spherically symmetric planet and does not take into account factors such as air resistance or the rotation of the planet.

## 3. What are the limitations of the usual Escape Velocity equation?

The usual Escape Velocity equation is only applicable for objects escaping from the surface of a planet. It does not take into account objects that are already in orbit around a planet or objects that are escaping from other types of gravitational fields.

## 4. Are there alternative equations for calculating escape velocity?

Yes, there are alternative equations such as the Tsiolkovsky rocket equation which takes into account the mass and velocity of the object. There is also the General Relativity equation which is a more accurate and complex equation that takes into account the curvature of spacetime.

## 5. How important is it to accurately calculate escape velocity?

Accurately calculating escape velocity is crucial for space travel and understanding the behavior of objects in space. However, for most practical purposes, the usual Escape Velocity equation is accurate enough and can provide a good estimate of the minimum speed needed for an object to escape a planet's gravitational pull.

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