Interplanetary Rocket Travel and the Rocket Equation

In rocket travel, one of the most essential elements is ∆v – the change in a ship’s velocity. The spaceship needs to accelerate to get out of the atmosphere, and then speed up to achieve orbit. If it’s going to another planet, it needs to achieve escape velocity from Earth, then speed up/slow down its orbit around the Sun to become intercepted by that other planet.

One of the most famous equations that governs this is (aptly) called the “rocket equation“: \Delta v = v_e ln\frac{m_0}{m_f} . This is complicated, so let’s break down each variable.

\Delta v in this equation is what we discussed above – the change in velocity.

v_e is the “effective exhaust velocity” – for our purposes, it just matters that this depends on the engine design. We’ll use SpaceX’s Raptor Engine, which has a “specific impulse” (or I_{sp} ) of 380s in a vacuum. You can get v_e by multiplying I_{sp} by the gravity at the Earth’s surface, which gives us v_e (Raptor Engine) = 380s \cdot 9.8m/s^2 = 3724m/s .

m_0 is the initial mass of the rocket, and m_f is the final mass of the rocket (after earlier stages have been discarded and fuel burned). The important thing to notice about these is they are within the logarithm ln\frac{m_0}{m_f} . That matters, because it means that as you want \Delta v to increase, the ratio \frac{m_0}{m_f} has to increase exponentially.

To give a bit of intuition for why that is: imagine you start with a probe. To get it to move, you only need a somewhat small engine – say we have a rocket with the same mass as the probe, and that gets us some acceleration. If we want it to go further, we need a rocket powerful enough to power both the probe and the engine we were using it before. So to get that same acceleration, we need to double the mass of the whole rocket again, leaving us with a new rocket four times as large. To get the same acceleration again, the rocket needs to be eight times as large. There’s the exponential climb.

Here’s a graph showing that exponential decline in more precision:

Source: Wikipedia

The interesting thing about this equation is that m_0 and m_f are the main things we control. Orbital mechanics gives us a required \Delta v if we want to orbit Earth, or go to Mars, or leave the solar system. The only thing we control is: how big is the probe we’re sending, and how big is the rocket we’re sending it on. So if we want to go to one of these places, we can find out exactly what percent of the mass of our first rocket we could theoretically keep. That’s what I calculated.

Below is the result of all of this. For each other world in our solar system, what percent of our mass can we keep? If we want to send 1 ton of stuff (cameras, sensors, food, people, etc.) to that world, how big of a rocket do we need?

WorldRequired ∆v% of Mass to Worldm_0 for m_f = 1t
Moon15.93 km/s1.392%72 tons
Mars21.3 km/s0.329%304 tons
Venus21.2 km/s0.338%296 tons
Mercury23.3 km/s0.193%519 tons
Juptier24 km/s0.160%627 tons
Saturn25 km/s0.122%819 tons
Neptune26 km/s0.093%1072 tons
Pluto26 km/s0.093%1072 tons
Escape Sun26.5 km/s0.082%1226 tons

Again, all of this assumes that we only use Raptor Engines to power our rocket, it ignores the different specific impulse in the atmosphere, and (most importantly) this is only enough to send your ship crashing into the other planet – not into orbit, and certainly not to land on it gently. All of those only take up more delta-v, which means you need a bigger rocket to start with.


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