## Escape velocities

One of the concepts alluded to by Steve in his Black Holes for Beginners series was *escape velocity*. However, it’s worthy of a more full explanation in my opinion. We’ll explore the mathematics behind escape velocities, and show how they can be calculated. We’ll end with a generalised formula for calculating the escape velocity from any object.

### Assumptions

Before we start we need to assume some things to make our calculations easier. First, we’ll ignore air resistance as modelling it is a real pain. Secondly, we’ll ignore the prescence of other big masses in the solar system like the Sun. Lastly, we’ll ignore tiny little problems like gravity waves and the quantisation of time and distance (the Planck time and length).

### Non-constant gravity

The first thing you need to know that contradicts with normal school physics is that the *strength of gravity depends on where you are*. While you might have used the value of *g* of 9.8 m s^{-2}, this is actually totally dependant on how far away from the Earth you are. This makes sense: if you’re further away from the earth, you might expect th Earth to pull with less force than if you were standing right next to it. The pull of the earth of you is conventially called your *weight* and as it is a force, should be expressed in newtons (i.e., saying “My weight is 70 kg” is technically wrong!)

Isaac Newton, being quite a clever cookie, came up with a law that predicts how much your weight is at different distances from the Earth (in fact, it’s more general than that, but we’ll ignore the extra generality for now). It is:

Where R_{E} is the radius of the earth, d is your distance from the surface of the Earth, m is your mass, m_{E} is the mass of the Earth and G is Newton’s gravitational constant.

### Kinetic and potential energy

When you move an object (perhaps yourself) *against a force* (i.e., in a direction opposite to the direction of the force), you have to spend energy. Think jumping: you’re moving against your weight and so you have to expend energy. If you jumped and jumped for hours, you’d get tired, which is your body’s way of telling you that it’s out of energy.

How much energy do you spend, though? If the force is the same all the way through then it’s easy, and you’ve probably seen the formula before: E = F d, where E is the energy you expend, F is the strength of the force and d the distance you move the object through. However, if the force isn’t constant, you’ve got to get clever and use some calculus:

*
*Where the symbols mean the same as above, apart from this time F is a function, not a value.

But where does come from, and where does it go? Imagine projecting a ball from the Earth’s surface upwards. Initially, the ball has a velocity of u, and so a kinetic energy of 1/2 m u^{2}. When the ball comes to rest, it will have no kinetic energy. Through this time, the ball’s potential energy has been increasing as it’s been moving against its weight. So we can see that the ball’s kinetic energy has been converted into potential energy. As energy must be conserved, we can say that the amount of potential energy it gained must be equal to the amount of kinetic energy it lost.

### Getting to 100m

How does this discourse help us? Well, we might want to ask “How fast does the ball have to be going initially to get to 100m from the Earth’s surface?”. Lets work this out.

We know that the amount of kinetic energy it lost is the initial kinetic energy minus the final kinetic energy. The final point is when the ball is stopped, i.e. it has a velocity and kinetic energy of 0. So, if we let K_{E }be the amount of kinetic energy lost, then we can see:

The force we’re working against here is the ball’s weight, which we use Newton’s formula for. We choose r to represent the distance from the ball’s centre of mass to the Earth’s centre of mass. And we know that the amount of kinetic energy lost must be equal to the potential energy gained, so we have:

Which we can work out using the following values:

So, u = 4.43 m s^{-1}. You may have noticed that we haven’t used the mass of the ball. Indeed, the m cancelled out in our calculations above. What does this mean? The velocity needed to get to 100m is independant of your mass. It simply doesn’t matter.

### Escape velocities

But we don’t want to get to 100m. We want to get to infinity. That is, we want the speed such that we have enough kinetic energy to go on forever. This seems impossible, as no matter how far away you are, the Earth will still pull on you with some miniscule force. However, one of the most startling things about mathematics is that you can have an infinite amount of things which add up to a finite number.

Not convinced? Think about the series 1 + 1/2 + 1/4 + 1/8 + 1/16 + …. This gets closer and closer to 2, but never actually reaches it. You could keep going for as many terms as you like but you still would be some distance away from 2. As the number of terms approaches infinity, the sum approaches 2. Mathematicians call this this *convergence* and we say that series has a sum-to-infinity of 2, or converges to 2.

This applies with weight as well. Even though you always have some weight, its decreasing so fast that it actually adds up to a finite value. Your weight keeps sapping at your kinetic energy, decreasing it, but as your weight is decreasing as well your kinetic energy never actually reaches 0, so you never stop (given sufficient initial kinetic energy, that is). If we replace R_{E} + 100 with infinity in the above integration, we can calculate the initial velocity needed to ‘reach infinity’, i.e. never stop. This is the escape velocity of the Earth, and it turns out to be about 11200 m.

What does that mean? Quite simply, if you can start off moving at 11.2 km s^{-1}, you’ll never stop moving. You’ll get very, very slow, but not stop.

### A formula for escape velocities

What if we start from somewhere other than the Earth? For example, in Steve’s essays, we started in a black hole. Can we generalise our formula? It turns out this is quite easy. Use the generalised form of Newton’s law of gravitation:

Where G is the gravitational constant as before, m_{1} the mass of the first object (say, the ball), m_{2} the mass of the second (say, the black hole or Earth) and r the distance between their centre of masses.

Using that formula and integrating as above, we can obtain a generalised formula for escape velocities.

Where r_{0} is the initial distance between the two centre of masses. Beforehand, we started from the Earth’s surface and so r_{0} = R_{E}, but this needn’t be true any more with our new general formula. Again, this is independant of the mass of the projectile, like it was for our 100m example.

The Wikipedia page on escape velocities contains a table of escape velocities calculated for various masses around the solar system. You should check that out.

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