# The Rocket Equation

Rockets in space, like all other objects, have to accelerate to change velocity. But space is a vacuum, so there is nothing to push against to create force. Instead, rockets accelerate by using the conservation of momentum. The momentum of an object is equal to the object’s mass multiplied by the object’s velocity: \vec{p} = m \vec{v}. In a closed system, the total momentum remains constant: \vec{p}_{0} = \vec{p}_{t}.

A rocket carries propellant that it expels at high velocities to accelerate. Imagine a rocket moving in space; at first the rocket is not expelling propellant and so its momentum does not change. Then it expels a part of its propellant. That propellant’s momentum is equal to its mass multiplied by its velocity. The rocket and propellant are part of a closed system, so the momentum of the rocket has to change such that the total momentum (that of the rocket plus that of the propellant) is equal to the momentum of the rocket before it expelled the propellant. As a result, the rocket gains velocity in direction opposite to that of the propellant.

Let’s find out how much velocity the rocket gains!

# Shorts: Vertical Travel Distance of Bouncing Objects

If a ball is dropped from a height of 10 meters, and on each bounce it reaches a maximal height of 0.75 times the previous height, then what is the total distance traveled? If we use h_i as the maximal height reached on bounce i, h_0 as the initial height, and $latex b$ as the factor of the maximal height achieved relative to the previous bounce, we have h_i = b h_{i-1}.

Then, the total distance traveled is:

\begin{aligned}
d &= h_0 + 2 h_1 + 2 h_2 + 2 h_3 + ...
\\
&= h_0 + 2 b h_0 + 2 b h_1 + 2 b h_2 + ...
\\
&= h_0 + 2 b h_0 + 2 b b h_0 + 2 b b b h_0 + ...
\\
&= h_0 + 2 b h_0 + 2 b^2 h_0 + 2 b^3 h_0 + ...
\\
&= -h_0 + 2 (h_0 + b h_0 + b^2 h_0 + b^3 h_0 + ...)
\\
&= -h_0 + 2 h_0 (1 + b + b^2 + b^3 + ...)
\\
&= -h_0 + 2 h_0 \sum_i \left(b^i\right)
\end{aligned}

Using \sum_n \left(x^n\right) = \frac{1}{1-x} if 0 \le x < 1, we get:

d = -h_0 + 2 h_0 \frac{1}{1-b}

Now we plug our values of h_0 = 10 \text{ m} and b = 0.75 into this equation to find d = -10 + 20 \cdot \frac{1}{0.25} = 70 meters. So, even though our ball will bounce on for eternity, it will travel exactly 70 meters!

# Can an Object with Constant Speed Be Accelerating?

An interesting question is whether an object with a constant speed can still be accelerating, and intuitively the answer would be “no”: acceleration means the object is speeding up or slowing down, right? Apparently, this is not the case. It actually is possible to have a constant speed while still having an acceleration. In this post, we will look at how this is possible. Hint: circular motion!

# Relative Velocity (Or: The Velocity-Addition Formula)

With the knowledge that the speed of light is constant and the same for every reference frame, and that no object can travel at the speed of light or faster than the speed of light in any reference frame, we ask ourselves the question: what happens when two spaceships leave from a space station in opposite directions and both reach a constant speed of 0.95c (with c the speed of light) relative to that space station? In the space station’s reference frame, the two ships travel at 2 \cdot 0.95c = 1.9c relative to each other. This doesn’t violate relativity, as neither spaceship is actually going at or faster than the speed of light. But, ignoring the effects of relativity, in the reference frame of either spaceship, the other ship would appear to travel at 1.9c. This would violate relativity!

Applying the equations we found before, we will find out what actually happens.

# The Kessel Run in Less than Twelve Parsecs

If you’ve seen Star Wars: A New Hope, you’ll have heard the above quote. People often think Solo refers to his ship’s speed when he says this, but that is not the case. A parsec is a unit of length, not of time! In fact, it is defined as the distance from the sun to an astronomical object such that the parallax angle (the angle between lines from the Earth and the Sun to that object) is one second of arc.

Using this information, we will calculate the length of a parsec in meters.

# Deriving the Lorentz Factor

The Lorentz Factor is defined as \gamma = 1 / \sqrt{1 - v^2 / c^2} with v the relative velocity between two reference frames (or two objects, if you will) and c the speed of light. This is the factor with which length contracts and time dilates between two reference frames. For example, if you’d sit in a rocket that was traveling at 80% the speed of light relative to your family on Earth, the Lorentz Factor would be \gamma = 1 / \sqrt{1 - 0.8^2} = 1 \frac{2}{3}.

What this means is that your rocket and everything in it — including you — would be shortened along the direction of motion for your family relative to the length you observe the rocket to be. In fact, its length would be contracted with factor \gamma; if the rocket’s length is x as measured by you, your family would measure it to be x' = x / \gamma. In this case, they would measure the rocket to be 60% of the length your measure it to be!

Calculating the time dilation is a similar process. In your reference frame, a clock in the rocket would tick every second. If your family back on earth are able observe that clock, it would appear to tick slower. For one second to pass on your clock, they would have to wait 1 \cdot \gamma = 1 \frac{2}{3} seconds. In general, if a time of t has passed in your frame of reference, a time of t' = \gamma t has passed for your family.

In this post, I will follow Einstein’s steps in deriving this factor.