# Escape Velocity and the Motion of Two Massive Bodies

• Conor Smith
In summary, the escape velocity equation, V_e = (2GM1/r)**.5, assumes that the escaping body is less massive than the body it is escaping from. When dealing with two more comparable masses, the general formula V_e= sqrt(2G(m1+m2)/r) should be used. Additionally, the Kepler method, which assumes one body to be of negligible mass, may not accurately calculate escape velocity in this scenario.
Conor Smith
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).

Conor Smith said:
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.

Conor Smith said:
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!

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

## 1. What is escape velocity confusion?

Escape velocity confusion refers to the confusion surrounding the concept of escape velocity, which is the minimum velocity required for an object to escape the gravitational pull of a celestial body. This confusion often arises due to misconceptions about the factors that affect escape velocity and the difference between escape velocity and orbital velocity.

## 2. How is escape velocity calculated?

Escape velocity is calculated using the formula v = √(2GM/r), where v is the escape velocity, G is the universal gravitational constant, M is the mass of the celestial body, and r is the distance between the object and the center of the celestial body. It is important to note that this formula assumes a point mass and does not take into account the effects of atmospheric drag.

## 3. What factors affect escape velocity?

The factors that affect escape velocity include the mass and radius of the celestial body and the distance between the object and the center of the celestial body. A larger mass or radius will result in a higher escape velocity, while a greater distance will result in a lower escape velocity. The shape of the celestial body and the presence of an atmosphere can also affect the escape velocity.

## 4. Can escape velocity be exceeded?

Yes, escape velocity can be exceeded. When an object is given a velocity greater than the escape velocity, it will achieve escape velocity and leave the gravitational field of the celestial body. This is how spacecraft are able to leave Earth's orbit and travel to other celestial bodies.

## 5. What is the difference between escape velocity and orbital velocity?

The main difference between escape velocity and orbital velocity is that escape velocity is the minimum velocity required to escape the gravitational pull of a celestial body, while orbital velocity is the velocity required for an object to maintain a stable orbit around a celestial body. Escape velocity is higher than orbital velocity, as it requires enough energy to overcome the gravitational pull completely, while orbital velocity only needs enough energy to counteract the pull and maintain a circular or elliptical orbit.

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