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## by Kevin Wise

Submitted : Spring 2012

In order to find out how much fuel it would take to land on Mars and return to Earth I had to begin by assuming a few things. I assumed there was no air resistance in Earth’s atmosphere or in space and I assumed Earth’s and Mars’s gravity remained constant until entering space, which at that point became 0. Therefore while in space the spaceship would never slow down, and it wouldn’t require any more thrust to reach Mars once it left Earth’s atmosphere.

As for the rocket I implemented a two stage scenario comprised of a primary and secondary engine. When first fuel tank, which carried the fuel for the primary engine weighing approximately 35,000kg, ran out, it would detached itself from the rocket, which weighs approximately 150,000kg, and the secondary engines would take over. However, the primary engines were the only engines that would be used while taking off from Earth, and the secondary engines would be the only engines used when leaving mars. In this case I assumed that there would be just enough fuel in the primary engine to leave Earth and just enough fuel in the secondary engine to leave Mars. Any extra fuel would be a waste.

The primary engine is driven by liquid fuel so the thrusters burn with an exhaust velocity of approximately 4.5 km/s while the secondary engines are driven by solid propellant which produces exhaust velocity of around 2.5 km/s. The velocity that is attainable by the primary engines is approximately 7 km/s which is the fastest space shuttle humans have produced to this day. This engine produces thrust for 480s until it runs out of fuel. However, the secondary engine isn’t as powerful because they are driven by solid propellant and therefore can only reach velocities up to 1.5 km/s. This engine runs out of fuel in 120s.

For the calculations the first thing I had to do was come up with an equation using calculus and physics. I started my calculations by summing all the forces acting on the rocket, excluding air resistance. By using derivative notation and integrals I found that change in velocity is equal to the exhaust velocity times the natural log of the initial mass divided by the final mass minus gravity times time. I then manipulated the equation algebraically and solved for the mass of the propellant for the two stage rocket. From my calculations I found that the rocket would need to be carrying 3,630,000 kg of fuel on takeoff which is roughly 96% of the total weight of the aircraft.