Re: Is escape velocity dependent on mass of captive object?

Date: Mon Mar 2 13:08:25 1998
Posted By: Dan Berger, Faculty Chemistry/Science, Bluffton College
Area of science: Physics
ID: 888287280.Ph

Message:

My neighbor says that the earth's escape velocity is a constant 11 km/s, regardless of the mass of the body escaping, at least up to the point where the captive is equal in mass to the captor. To prove his point he asks, assuming the moon was at rest on the surface of the earth, "Is the mass of the moon significant to change the velocity required for escape from the earths' gravity well to <.05 or <.01?"

Since I believe the moon's gravity would increase the velocity necessary to separate the two, my question is "Is the escape velocity for any two bodies the sum of their separate escape velocities, computed assuming the escape of an object with an insignificant gravitational force, i.e. in this case, earth 11, moon 2 for a total of 13 km/s?

Though I am really rusty on my algebra, my guess is that an object 1/100th the size of the earth or larger would increase the escape velocity by 1% or more - Anything smaller than that would be insignificant.

Your neighbor is correct, but for the wrong reasons. The mass of the escaping body does not matter at all! The velocity required for any single body to "escape" the gravitational influence of any other body is dependent only on the mass of the second body, that is, the body being "left behind."

To understand why this is so, why escape velocity is dependent only on the mass of the body defined as being at rest, let's investigate how escape velocity is calculated.

In this discussion we will use the following conventions:

A subscript letter g indicates something due to gravity.
Upper-case M indicates the parent body, the one "being escaped from," which is defined as being at rest.
Lower-case m indicates the subordinate body, the one "doing the escaping," which is in motion relative to the parent body.

1. The gravitational force between two objects is proportional to the product of their masses and inversely proportional to the square of the distance between them.

This is pretty well-understood and seems to be the source of your difficulty. Newton's Third Law says that both masses (M and m) will feel the same gravitational force. However, in order to produce motion the force must act on a mass, so that one of the masses (m) gets re-involved in the calculation. This will ultimately lead to the counterintuitive conclusion that the value of m doesn't effect the result!

As an example, consider how we calculate the acceleration of any object in a gravitational field. Since force is mass times acceleration, we need to divide by the mass of the object (m) to get the acceleration. This leads us to the conclusion that the acceleration experienced by any object in the gravitational field of another object is independent of the mass of the subordinate object:

Note that this says, for example, that the Earth experiences the same acceleration due to the mass of the Moon as any other object 380,000 km from the Moon! This is actually observable: the Earth-Moon system revolves around its center of mass, which is some 4700 km, or 3/4 of the Earth's radius, from the center of the Earth, so that the Earth wobbles in its orbit around the Sun (the center-of-mass follows a smooth orbit around the Sun, though).

We see this in any situation involving orbiting objects for which the mass of one is a significant fraction of the mass of the other. Jupiter makes the Sun wobble considerably; their common center of mass (neglecting the other planets, which is a reasonable approximation since Jupiter contains 90+% of the total non-solar mass of the system) is about 2/3 of the way out from the center of the Sun.

So far it looks good for your side -- see, this is a function of the substantial size of the second body! -- but perhaps we should keep going.

2. All potential energy problems involving objects in a force field are referenced to the removal of the body to infinite distance.

This is a standard physical convention. The energy of the body at infinite distance is taken to be zero. Since (for an attractive force like gravity) all bodies at less than infinite distance will have lower energy than a body at infinite distance, gravitational potential energy is always negative.

Now, since energy is defined as force times distance, and attractive forces are always defined as negative, the gravitational potential energy of any body in a gravitational field is given by

3. Escape velocity (v_e) is defined as the velocity required to remove a body, in a gravitational field, from distance r to an infinite distance.

Another way of putting this is to say that the total energy of the body must be zero relative to the "parent" mass. This requires that the body's kinetic energy be equal and opposite to its gravitational energy.

Now we know the expression for kinetic energy: ½mv². Thus, if the "escape" kinetic energy equals the gravitational potential energy,

Notice that the mass of the subordinate, escaping body (m) does not appear in the final expression.

The only assumption here is that the inertial mass (involved in the expression for kinetic energy) is equal to the gravitational mass, or in other words, the m on the left side of the equation is equal to the m on the right side. This assertion has been tested many times and never found to be false.

Thus, the velocity required for the Moon to escape from the Earth is dependent only on the mass of the Earth; the velocity required for the Earth to escape from the Moon is dependent only on the mass of the Moon; but the kinetic energy required by either body to escape from the other will be the same.

Or, to answer your question more directly, it is not true that
the escape velocity for any two bodies the sum of their separate escape velocities.

	Dan Berger
	Bluffton College
	http://cs.bluffton.edu/~berger

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