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.

# Tag: Astrodynamics

# 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.

Continue reading “Escape Velocity (And: How Much Fuel Do Rockets Need?)”

# 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!