Previously we calculated the fuel requirements of rockets to reach escape velocity. In that calculation, we did not take into account the effect of gravity during the fuel burn and underestimated the fuel requirements. We improved on this through a simulation better approximating reality. This simulation takes into account the force of gravity during the burn, as well as that force weakening as the rocket increases its distance from Earth, and takes into account the necessary escape velocity decreasing as the rocket’s distance to Earth grows. On this second post, Basker V commented that a closed-form solution can be found when the burn time is sufficiently small, allowing the rocket’s distance to Earth to be considered constant during the burn. In effect, this enables us to ignore both the diminishing force of gravity and decreasing escape velocity. Let’s find this closed-form.

Continue reading “Closed-Form Rocket Fuel Requirements”# Category: Physics

# Rocket Fuel Requirements Revisited

In a previous post we looked at the fuel requirements for rockets to reach escape velocity. We calculated the fuel requirements using the rocket equation. This equation takes into account the conservation of momentum. However, momentum is not the only property influencing the velocity of the rocket during a launch.

Rockets expel their fuel over time. During this time, the rocket is pulled back due to gravity. Only if a rocket could instantaneously expel all of its fuel, and when ignoring atmospheric drag, the escape velocity would be reached instantaneously and the equation would hold.

Taking the burn-time and gravity into account yields a difficult differential equation. We can implement that equation in a computer program to simulate the launch.

# Black Holes

A black hole is an object from which nothing, including light, is able to escape. As nothing can go faster than light, this can be more formally defined as an object for which the escape velocity is greater than the speed of light. In my post about escape velocity we found an equation relating the velocity required to escape from the gravitational pull of an object and that object’s mass.

v = \sqrt{2G \frac{M}{r}}

In this equation, v is the escape velocity, M is the mass of the object we want to escape from and r is the distance from the center of mass of the object we’re escaping from. G is the gravitational constant.

If the required velocity v is greater than the speed of light, c, even light will not able to escape the object, making it a black hole. We have two variables, M and r, so we can derive two equations from this. One equation gives the distance from the center of mass required to make escaping from the object impossible given the mass, and the other gives the mass required given the distance from the center of mass.

\begin{aligned}

\sqrt{2G \frac{M}{r}} &= c \\

2G \frac{M}{r} &= c^2 \\

\frac{M}{r} &= \frac{c^2}{2G} \\

M &= \frac{r c^2}{2G} \\

r &= \frac{2GM}{c^2}

\end{aligned}

The latter equation, r = \frac{2GM}{c^2}, is known as the *Schwarzschild radius*. It is the radius of the perfect sphere around the center of mass of the object, such that if all the mass is within that sphere the resulting escape velocity is equal to the speed of light. In other words, if the object were smaller than this, it would become a black hole. For Earth, the radius is slightly surprising:

\begin{aligned}

r &= \frac{2G M_{\oplus}}{c^2} \\

&= \frac{2 \cdot 6.67 \cdot 10^{-11} \cdot 5.97 \cdot 10^{24}}{3.00 \cdot 10^8} \approx 8.87 \text{ mm}

\end{aligned}

So, Earth would only become a black hole if it was compressed to the size of a marble. A black hole can be smaller than its Swartchzschild radius, however. In this case, the radius acts as the *event horizon* of the black hole: matter, or information, inside the radius would not necessarily be inside the black hole itself, but it would no longer be able to escape to outside the event horizon. In other words: everything that happens inside the event horizon of a black hole, is invisible to outside observers.

# Escape Velocity (And: How Much Fuel Do Rockets Need?)

During the launch of a rocket, the Earth’s gravitational field is pulling the rocket back. The rocket needs a certain speed to be able to escape from the Earth’s gravitational field, such that it won’t fall back to Earth nor get into an orbit around it. Escape velocity is the speed a rocket requires to be able to escape from a body without having to burn more fuel later during the maneuver. For a body as massive as Earth, the required velocity is relatively high, and this is why rockets literally need tonnes of fuel.

In this post, by making a few simplifications and using the rocket equation that we found earlier, we will derive an equation to calculate the amount of propellant needed to escape from Earth.

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# 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!

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# 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.

Continue reading “Relative Velocity (Or: The Velocity-Addition Formula)”

# 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.

Continue reading “The Kessel Run in Less than Twelve Parsecs”

# 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.