Escape Velocity |

Calculating Escape Velocity |

First of all, what is escape velocity anyway? Escape velocity is the velocity that an object must have to escape the gravitational field of a body like the earth, or another planet, or any mass. That's my definition, by the way.

Why would you be interested, anyway? Well, you need to know this if you want to calculate how to fly to the moon, for example. Or maybe it would be interesting for other things as well...

So, let's assume we can neglect air resistance, and let's assume we are directing our object vertically, i.e. "straight up". What is the speed the object needs to have to escape from the 'gravity well'?

There are two things going on here. First, our object has a certain velocity upwards, away from the planet. Note that the object is not being propelled. Second, the gravity field of the planet is pulling our object down. Which of the two will win?

The first reaction is to think that the object will always come down eventually, since the planet keeps on pulling on it, and the object is not propelled. So, eventually, gravity will win, right?

Wrong. The thing I didn't mention yet, is that the higher our object is, the less gravity will be. Gravity diminishes with the square of the distance to the center of the planet. This is a consequence of Newton's Law of Gravity. We need more equations here, so let's see.

In English: two bodies with masses M and m will attract each other with a force that is proportional to the masses of the bodies, and inversely proportional to the square of the distance between the centers of the bodies. There is a constant in there as well.

**F** is the
attracting (gravitational) force.

**M** is de mass of the first body,

**m** is the mass of the second body,

**r** is the distance between the centers,

**G** is the gravitational constant.

We have another equation that describes the motion of an object.

This describes the position of an
object at a certain time **t**, given a start
position (**s(0)**), a start velocity (**v(0)**,
in our case the escape velocity) and an accelleration **a**
(in our case, the acceleration - or deceleration - caused by
gravitation).

One more equation, Newton's Second Law of Motion (see text box later on, though)

Combining [1] and [3] (they are equal) we find that:

So, we could substitute [4] in [2]
and try to find a solution (we need to fiddle with signs, since
the direction of **v(0)** is opposite to that of **a**).
However, this becomes quite complex, since there are dependencies
between the value of **s(t)** and **a**.
Maybe it is solvable (at the very least you can feed the problem
to a computer and approximate the solution numerically), but it
becomes too complex for me, I'm afraid.

Fortunately, there is a better
way. We can calculate the potential *energy* of the object
in the gravitational field. Some very basic physics state that:

Where **W**_{AB} is the amount of work (energy) required
when moving an object with force **F(r)** over a
distance **dr**, from point A to point B. Turns out
that using [1] for the expression of force yields an integral
that can be solved easily.

Now, if we choose our point B to
be infinitely far away (no gravity there), and our point A to be
on the surface of our planet, introduce **U** for
the potential energy, we find that

So, to escape the gravity of our
planet, our object must have a kinetic energy that is equal to
this potential energy. Since we know that the kinetic energy **E**
of an object with mass **m** and velocity **v**
is gven by

we find that

or

so

That's it. We're done. Just for fun, let's fill in some values and do a sanity check.

For Earth, we have

and

so,

which agrees nicely with the known
value. Just for kicks, let's do the same thing for Mars. Here we
know that **r**_{m }= 3397
km, and **M** = 6.4185 * 10^{23} kg, so we
find that the escape velocity = 5.02 km/s, which also agrees
nicely with literature.

I'm always
somewhat confused by Newton's Second Law of Motion. You
usually find it stated (as I did here) as:
where with So, using the normal mathematical rules of differentiation, we find that:
Note that The term
However, in the case of
spaceships and rockets and all that, |

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