# 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.

First, we will find the amount of work the Earth’s gravitational field puts into pulling a rocket back during a successful escape. This work is what has to be overcome by the work of the thruster of the rocket. When saying a rocket escapes from Earth (without having to burn any more fuel), we can say it moves from some distance from the center of the Earth, r = r_0, to infinitely far away from Earth, r = \infty (given enough time, of course).

The Earth’s gravitational field acts as a force on the rocket:

F(r) = G \frac{Mm}{r^2}

Here, G \approx 6.674 \cdot 10^{-11} \text{ N} \text{m}^2 \text{kg}^{-2} is the gravitational constant, M \approx 5.972 \cdot 10^{24} \text{ kg} is the mass of the Earth and m is the mass of the rocket, which we’ll leave unspecified for now. Of course, r is the distance from the rocket to the center of the Earth.

The work done by gravity is equal to the force with which gravity acts upon the rocket multiplied by the displacement of the rocket (relative to the force’s direction). If the force applied by gravity was constant, we could calculate the work done by gravity as W = F (r_\text{end} – r_\text{start}). However, the magnitude of the force of gravity depends on the distance between the two objects, and as such we’ll have to perform a more delicate calculation.

We could try to calculate the work done by multiplying the force of gravity F(r) at a point in the path by a very small distance traveled at that point, \Delta r, and add all those “pieces of work” together to get the total work done. If we take this just one step further and say we do this for infinitesimally small pieces of the path, we arrive at the precise amount of work required through taking the integral.

\begin{aligned}
W &= \int_{r_0}^\infty F(r) dr \\
&= \int_{r_0}^\infty G \frac{Mm}{r^2} dr \\
&= \int_{r_0}^\infty G Mm \frac{1}{r^2} dr \\
&= \int_{r_0}^\infty G Mm r^{-2} dr \\
&= \frac{1}{-1} G Mm r^{-1} \Bigr\rvert_{r_0}^{\infty} \\
&= -G Mm \frac{1}{r} \Bigr\rvert_{r_0}^{\infty}\\
&= -G \frac{Mm}{r} \Bigr\rvert_{r_0}^{\infty} \\
&= (-G \frac{Mm}{\infty}) – (-G \frac{Mm}{r_0}) \\
&= (-0) + G \frac{Mm}{r_0} \\
&= G \frac{Mm}{r_0}
\end{aligned}

So, we found that W = G \frac{Mm}{r_0} is the work done by the Earth’s gravitational field on a rocket traveling from a distance of r_0 to a point infinitely far away. This is the work the rocket has to overcome by its speed.

An object with a certain speed has a kinetic energy, which is the “energy of movement” that object has. This energy is equal to E_k = \frac{1}{2} m v^2 with m the mass of the object and v the speed. To escape the Earth’s gravitational field, our rocket’s kinetic energy has to, at least, be equal to the work done by the gravitational field.

\begin{aligned}
E_k = \frac{1}{2} m v^2 &= G \frac{Mm}{r_0} \\
v^2 &= 2 G \frac{Mm}{mr_0} \\
v^2 &= 2 G \frac{M}{r_0} \\
v &= \sqrt{2 G \frac{M}{r_0}}
\end{aligned}

We can now calculate the speed required to escape from the Earth! If we assume the rocket starts to accelerate at sea-level (r_0 \approx 6.376 \cdot 10^6 \text{ m}), and if we ignore atmospheric drag, we get:

v \approx \sqrt{2 \cdot 6.674 \cdot 10^{-11} \frac{5.972 \cdot 10^{24}}{6.376 \cdot 10^6}} \approx 1.118 \cdot 10^4 \text{ m/s} = 11.18 \text{ km/s}

Now, we use the rocket equation to calculate the mass of the propellant needed to make this happen!

\Delta v = v_e \ln \left(\frac{m_i}{m_f} \right)

Here, \Delta v is “delta-v”, the required change in speed, m_i is the initial mass of the rocket and m_f is the final mass of the rocket. v_e is the exhaust velocity of the propellant used. If we say that all fuel is used during the maneuver, m_f is equal to the payload of the rocket. A relatively modest rocket could have a payload of 2,000 \text{ kg}. A pretty good exhaust velocity is v_e = 5,000 \text{ m/s}. We require \Delta v = 1.118 \cdot 10^4 \text{ m/s}.

We want to know the initial mass m_i of the rocket, and specifically we want to know the mass of the propellant, which is equal to m_i – m_f.

\begin{aligned}
\Delta v &= v_e \ln \left(\frac{m_i}{m_f} \right) \\
\frac{\Delta v}{v_e} &= \ln \left(\frac{m_i}{m_f} \right) \\
e^{\left(\frac{\Delta v}{v_e}\right)} &= \frac{m_i}{m_f} \\
m_i &= m_f e^{\left(\frac{\Delta v}{v_e}\right)}
\end{aligned}

If we plug in our values for m_f = 2,000 \text{ kg}, \Delta v = 1.118 \cdot 10^4 \text{ m/s} and v_e = 5,000 \text{ m/s}, we get:

m_i = 2,000 \cdot e^{\left(\frac{1.118 \cdot 10^4}{5,000}\right)} \approx 1.871 \cdot 10^4 \text{ kg} \approx 19 \text{ tonnes}

So, for a rocket with a payload of 2 tonnes to escape from Earth with an exhaust velocity of 5 km/s, we require m_i – m_f = 1.871 \cdot 10^4 – 2,000 = 1.671 \cdot 10^4 \approx 17 \text{ tonnes} of fuel. That is roughly 90\% of the total initial mass of the rocket!

## 5 thoughts on “Escape Velocity (And: How Much Fuel Do Rockets Need?)”

1. Vipul Patel says:

This is very good calculations to understand basic science of rockets. I am interested in understanding the cumulative fuel usage vs height gained by rocket. At first instant rocket will have zero initial velocity. This means it will have to generate enough force by burning fuel to overcome inertia & gravitational pull. As it slowly starts to move up and accelerates, it loses its mass due to fuel burn, that means for same rate of fuel burn the rocket can gain greater acceleration. Of Course now it will have higher drag as its velocity increases but I think the drag is not very significant portion of the forces acting on rockets. I think, since the rocket has to accelerate from zero velocity and fight against the gravity and drag from the air, the amount of ‘fuel used/height gain’ is greater initially and gradually it will be decreased. The reason for this is the weight loss in terms of mass of fuel used, reduction in gravitational pull due to increasing distance from the center of the earth and momentum gain and requirement of lower acceleration. I think a plot of cumulative fuel usage vs height gained by rocket would make a sense. I am trying to figure out, instead of starting the rocket engine at zero velocity, how much fuel we can save if we provide it some initial velocity. This will result in significant savings of fuel, can make the size of the rocket smaller by reducing the size of fuel tank. That way we can reduce the overall size of rockets to carry the same payload.