# Escape Velocity Confusion

• I
Hey there,

If body 1, mass M1 has escape velocity V_e1 = (2GM1/r)**.5 but M2 is more massive than M1 is this relation still valid? In this case, the subordinate body really isn't the subordinate body so does this still hold? And r (distance b/t the two) changes not only due to the motion of M2 but the motion of M1 being dragged by M2 and I'm not sure this equation accounts for that change.

I guess my question is whether or not escape velocity accounts for the acceleration of the body being escaped from?

If it makes any difference, this question arose as I'm programming a simulation which when using Cowell's method (which accounts for the acceleration of both masses) yields an escape velocity much higher than when using Kepler's (which yields the accepted escape velocity, but the Kepler method also assumes one body to be of negligible mass, namely that the body at the center doesn't move).

Hey there,

If body 1, mass M1 has escape velocity V_e1 = (2GM1/r)**.5 but M2 is more massive than M1 is this relation still valid? In this case, the subordinate body really isn't the subordinate body so does this still hold? And r (distance b/t the two) changes not only due to the motion of M2 but the motion of M1 being dragged by M2 and I'm not sure this equation accounts for that change.

I guess my question is whether or not escape velocity accounts for the acceleration of the body being escaped from?

If it makes any difference, this question arose as I'm programming a simulation which when using Cowell's method (which accounts for the acceleration of both masses) yields an escape velocity much higher than when using Kepler's (which yields the accepted escape velocity, but the Kepler method also assumes one body to be of negligible mass, namely that the body at the center doesn't move).
Yes, you're right. Kepler's method assumes that the body escaping is far less massive than the body it escapes. So, it appears you've answered your own question. Kepler's method is inappropriate because it assumes the lesser mass is the escaping mass. This is clearly stated in the assumptions of Kepler's laws. We don't say the earth orbits the moon but we say the moon orbits the earth, and the earth orbits the sun. So the velocity needed for the sun to escape the earth gravitation is definitely not equal to the above formula. wherever the sun goes, the earth follows.

Janus
Staff Emeritus
Gold Member
Hey there,

If body 1, mass M1 has escape velocity V_e1 = (2GM1/r)**.5 but M2 is more massive than M1 is this relation still valid? In this case, the subordinate body really isn't the subordinate body so does this still hold? And r (distance b/t the two) changes not only due to the motion of M2 but the motion of M1 being dragged by M2 and I'm not sure this equation accounts for that change.

I guess my question is whether or not escape velocity accounts for the acceleration of the body being escaped from?

If it makes any difference, this question arose as I'm programming a simulation which when using Cowell's method (which accounts for the acceleration of both masses) yields an escape velocity much higher than when using Kepler's (which yields the accepted escape velocity, but the Kepler method also assumes one body to be of negligible mass, namely that the body at the center doesn't move).

The more general escape velocity equation is

$$V_e= \sqrt{ \frac{2G(m_1+m_2){r}}$$

In most real problems m2 is really small compared to m1, and m1 dominates the equation.
If the two masses are more comparable in size, then you need to use the general equation and if m2 is much much more massive than m1 you can just use it in the equation. Ve is just the velocity that the two masses would have to be moving relative to each other.

Ah! That clears things up nicely. Thank you both!

FactChecker
Adding missing '}' to @Janus' LaTex: $$V_e= \sqrt{ \frac{2G(m_1+m_2)}{r} }$$