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”# Tag: Gravity

# 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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