Skip to 0 minutes and 3 seconds JASMINA LAZENDIC-GALLOWAY: Is it possible to send humans to Mars, even today or in the near future? Well, in theory, yes, but practically, there some difficulties we have to overcome. We have something called the rocket equation that allows us to relate the velocity at which we want our rocket to move, the type of fuel we use for the rocket, and also the mass of the rocket. So we say the final velocity minus the initial velocity is equal to J, which is basically exhaust speed, and then logarithm of your initial mass versus the final mass.

Skip to 0 minutes and 35 seconds So this is in units of kilometres per second, this is in units kilometres per second, this is in units kilometres per second, and this is in kilograms.

Skip to 0 minutes and 48 seconds So initial velocity will be 0 because this is where we’re starting from. And let’s say we want to get to 3 kilometres a second. The exhaust speed is dependent on what kind of fuel we use. And with the chemical fuel we use these days, basically, that’s 3. So for the final velocity, we take just the weight of the ship, of our rocket. So let’s say we take 10 to the 6 kilograms, 1,000 tonnes. So then this equation becomes 3 equals 3 logarithm of initial mass over the final mass, which is 10 to the 6. So when we turn this around, we get the initial mass has to be around 2.7 10 to the 6 kilograms.

Skip to 1 minute and 32 seconds What this means is basically that we have to have almost three times more initial mass, which will be just in fuel. So in respect to the weight of the rocket, three times that has to be fuel. This is how much it takes just to get to 3 kilometres a second. So as you can see, there are two possibilities to improve our space travel, either to improve J, which is basically moving away from chemical fuels or finding even more efficient chemical fuels, and also to change this ratio of initial mass with the final mass. That’s slightly harder to do because rocket ships have to be quite heavy. So then a lot of effort is put into finding more efficient fuel.

Skip to 2 minutes and 26 seconds But another way maybe of changing this ratio is not basically changing the type of the rocket, but actually allowing rockets to refuel in orbit.

# Rocket equation

Watch Jasmina describe how space travel is determined by the rocket equation.

The step-by-step calculations are:

\[3\,{\rm km/s} - 0\, {\rm km/s} = (3\, {\rm km/s}) \times ln \Big( \frac{M_i}{10^6\, {\rm kg}} \Big)\] \[\frac{3\, {\rm km/s}}{3\, {\rm km/s}} = ln \Big( \frac{M_i}{10^6\, {\rm kg}} \Big)\]\(ln\) is the natural logarithm (similar to log), which works in base \(e\), which is just a number (approximately 2.7). So we use \(e\) on both sides of the equation to get rid of \(ln\):

\[e^1 = \frac{M_i}{10^6\, kg}\]and finally rearrange to get \(M_i\):

\[M_i = 2.7 \times 10^6\, {\rm kg}\]Can we overcome this “rocket equation problem” to send humans to Mars? Let’s find out in the next step.

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