Disclaimer: rocket science isn't exactly the easiest thing to understand, but I'm going to try to make this as simple as I possibly can without dumbing things down. This *will* still contain math. Considering, proceed at your own risk. I am not responsible for any exploding minds caused by advanced mathematics.

 

The acceleration due to gravity is typically treated as a constant number, 9.81 m/s². This is really just an approximation, though—the Earth's gravitational pull is going to be less farther away from Earth, and more closer to the center.

 

More specifically, the formula for universal gravitational attraction is:

 

 

Where a is the acceleration, G is the universal gravitational constant, M is the mass of the object, and r is the distance from the object. G, the constant, is always 6.67384 × 10^-11 m³ kg^-1 s^-2.

 

If we model an object that has a constant speed, guessing how far it'll go in any period of time is pretty easy. If we have an object that has a changing speed, we can still guess how far it'll go based on the rate at which it's changing... but this is much more difficult. The problem here is that we don't just have a changing speed, but a changing speed that changes.

 

So, let's say that we have a spaceship above Earth. Using the formula and the known mass of Earth, we can know the acceleration (a) if we know the distance ® from Earth. If we're farther away from Earth, the pull is less, and so we accelerate less. If we're closer to Earth, the pull is greater, and so we accelerate more.

 

As our object gets closer to the other object it will be pulled more, so the acceleration changes. The acceleration is the change in speed, and the speed is the change in position. So, here's the thing: can we model the change in position based off of a speed that changes with a variable rate of change?

 

Yes, we can. It's just a little difficult.

 

We know that acceleration is the rate of change of velocity, and velocity is the rate of change of position, so we can write this as:

 

 

 

 

Where dv/dt means change in velocity per change in unit of time, and d²r/dt² refers to the change in the rate of change of the position, per unit of time. Rearranging this, we have:

 

 

It looks more complicated than it really is. What this means is that the acceleration towards earth multiplied by the distance from earth squared is always going to be equal to the gravitational constant times Earth's mass. This is a differential equation, but unfortunately, it is not one that I know how to solve. Even using a computer algebra system (Maple 13 for anyone curious), I wasn't able to find the equation for the distance from Earth as a function of time.

 

However, computers are really good at doing a lot of simple calculations very quickly. So, the next best thing is to use computer software to approximate any solution to this equation. So, we can plug in some numbers and see if we can find some solutions.

 

Let's pretend we have a spaceship in orbit around Earth. Our orbit is at the same distance from Earth as the moon—384,400 km. We fire our engines to kill our velocity, so that we're not moving at all. Relative to Earth, we're now a static object. Our initial distance from Earth is 363,295 km, and now it's pulling us in.

 

Since we know the Earth's mass is 5.972e24 kg, we can plug this in and see where the computer takes us.

 

 

r(0) = d means that the initial starting distance at t = 0 is d, which is the value set in the parameter box. r'(0) = v means the initial speed is v, which is also the value given in the parameter window.

 

 

 

 

 

Now, I can give any value of t, and find the corresponding values of r (distance) and r' (change in distance) through numerical approximation. So at t = 86400 seconds, or 1 day, the distance away from Earth is r = 374,243 km. We're moving toward Earth at a speed of 237 meters per second.

 

After 2 days:

 

After 3 days:

 

After 4 days:

 

After 5 days:

 

 

Hmm...

 

 

So, let's look at what happened here. After t = 419307.05 seconds, the computer can no longer solve the equation due to a "singularity"... whatever the heck that is. That's roughly 4.85 days, so something must have happened at that point.

 

We were getting closer and closer to Earth, and as we were getting closer, our speed was increasing. We inevitably got so close to Earth that our distance reached zero, and our acceleration reached infinity. We got so close to the Earth, that the gravitational pull was too strong to escape. In essence, we crashed into the Earth.

 

See, at exactly 419,307 seconds, our distance was 16.793 km from the center of Earth. At that time, our velocity was 217.860 kilometers per second.

 

At 419,306 seconds, just one second prior to that, our distance was 125.617 km from Earth's center.

 

At 116 hours, the distance is 17,197 km from the center of the Earth and the velocity is 6.654 km/s. In the next hour it goes over the point of singularity, so we can guesstimate that we'll be hitting the Earth's atmosphere between hour 116 and hour 117.

 

At that point, we'll need a completely different set of equations to model the trajectory. I'm getting lazy and my eyes are starting to hurt, so that's a subject for some other time.

 

 

Before I'm totally done yet, we need some way to check and make sure these numbers are correct. There is another way to look at this same problem—using the law of conservation of energy.

 

The law states that the total energy in any given system doesn't change. Normally we could model kinetic energy as:

 

 

and potential energy as:

 

 

Where g is the acceleration due to gravity. But that's not going to work in this case, because as previously stated, the acceleration changes. So, we're going to need a different equation for that one.

 

Work is a form of energy. For a variable force, work is computed as the integral of the force times the change in position. By evaluating the integral, we have:

 

 

Where d is the starting distance before the descent, and m is the mass of the spaceship.

 

Since the total energy is just the sum of the work and kinetic energy, we can just add the two. And since there can be no increase or decrease in energy, the sum should be equal to zero.

 

So, all the work done should be converted into kinetic energy for every frame of time. So, this means:

 

 

Since m appears on both sides of the equation, we can cross that out. As such, the mass of the ship is unimportant, and has no effect on the calculation. This leaves us with:

 

 

Using the same numbers calculated previously, we can say that the position at 116 hours is 17,197 km from Earth's center. So, we can plug all these numbers into the equation and solve for v to get 6.654 km/s. Everything checks out, and it's all good. These numbers make sense, so the calculations are correct.

 

Perhaps some other time I'll also look at modeling orbits in two dimensions, but that gets even more complicated. I need to figure that one out first. You see, this is what I do when I don't have homework...