## Saturday, May 5, 2007

### It's Just Rocket Science: Part 1

People seem to have a conception that rocket science is very difficult; in effort to show that it really isn't, I'm going to try to explain its basic concepts and equations. I'll start with delta v. (Note that I copied this from my old blog.)

Delta v means change in velocity; velocity is basically the quantity that represents both speed of an object and the direction it is moving. Delta is the science/engineering shorthand for change, and thus delta v is the change in the speed and direction of an object.

For our purposes, that object is a spacecraft. How does a spacecraft move? Well, with a rocket, of course. A rocket is a simply a device that tosses material in one direction, causing whatever it is attached to move in the opposite direction. The reason for this is what Sir Issac Newton called "Conservation of Momentum". Momentum is the product of the velocity and the mass of an object, and for an object not affected by external forces, is a constant. Mathematically, this is expressed as:

m1 * v1 = -m2 * v2

Where m1 and v1 are the mass and velocity of the spacecraft and m2 and v2 and the mass and velocity of the exhaust. The negative means that the directions of the velocities are opposite. Let's look at an example:

Suppose a spacecraft that has a mass of 30,000 kg is at rest (not moving) fires a rocket that instantaneously shoots 10,000 kg of its mass out its back at a speed of 3000 m/s. How fast is the spacecraft moving now? Well, its mass after the burn is going to be 30,000 kg minus the 10,000 kg that was fired, leaving 20,000 kg. The equation above can thus be written as:

20,000 kg * v1 = -10,000 kg * -3000 m/s

Solving for v1, we get:

v1 = 1500 m/s

Which converts to about 5400 km/hr or 3355 mph. Note that the exhaust velocity is negative, meaning it is in the opposite direction than the spacecraft is pointing.

The problem with this example,though, is the word instantanously. In reality, things never move instantanously, and so we need to take into account the time that it takes for the exhaust to get out of the rocket. To do this takes calculus, so I won't go into the details, but the end result, as found by the great Konstantin Tsiolkovsky, is called the rocket equation and can be written:

delta v = -ve * ln( mi / mf )

Where ve is the exhaust velocity, mi is the mass of the spacecraft before the burn, mf is its mass after the burn, and ln is the natural logarithm.

Let's return to our previous example and see what we get when apply the rocket equation to that situation:

delta v = -(-3000 m/s) * ln( 30,000 kg / 20,000 kg )

Solving, we get:

delta v = 1216.4 m/s

Or 4379 km/hr or 2721 mph, which is just 81% of our previous value.

Now, if you read the specifications of a rocket, you typicaly will not see the exhaust velocity, but you will probably see something called "specific impulse" or Isp. Specific impulse is simply the exhaust velocity divided by the accelleration due to gravity at the Earth's surface, called one g, or 9.8 m/s2.

Isp = -ve / g

Thus, we can write the rocket equation as:

delta v = (Isp * g) * ln( mi / mf)

And there you have it. With the above equation, you can calculate the delta v of a spacecraft maneuver with just some basic information about the spacecraft. But what is delta v useful for? Well, that's what the next installment is for...

### I'm going to Arizona

I've just been accepted into the Astrophysics PhD program at Arizona State University. It's a new department, less than a year old, so getting in on the ground floor should be fun... Also, I'll be at JPL this summer again, hopefully bringing MSCL to the point that we can get real funding...