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How to calculate mass of fuel needed to escape Earth gravity vs. Mars gravity

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  1. Sep 28, 2011 #1
    I am trying to determine approximately how much fuel it would take to launch a bi-propellant rocket from Mars and Earth such that they escape each planet's respective gravity. Unfortunately, my orbital mechanics is a little rusty, and the solution I'm getting does not make much sense. Here's what I've been trying - hopefully someone can point out my error:

    Assumptions:
    Earth escape velocity (Vesc,e) = 11 200 m/s
    Mars escape velocity (Vesc,m) = 5 027 m/s
    Gravity of Earth (ge) = 9.81 m/s2
    Gravity of Mars (gm) = 3.71 m/s2
    Specific Impulse of a bipropellant liquid rocket (Isp) = 450 s
    Dry mass of rocket (structure + payload) = Mdry
    Mass of rocket fuel = Mfuel
    Wet mass of rocket = Mwet

    Okay, using the Rocket Equation we have

    deltaV = Vexhaust * ln(Mwet/Mdry) = Vexhaust * ln((Mdry + Mfuel)/Mdry)

    Therefore, deltaV/Vexhaust = ln(Mdry + Mfuel)/Mdry)

    And, e(deltaV/Vexhaust) = e(ln(Mdry + Mfuel)/Mdry)) = (Mdry + Mfuel)/Mdry)

    Mdry * e(deltaV/Vexhaust) - Mdry = Mfuel

    Therefore, Mfuel = Mdry(e(deltaV/Vexhaust) - 1)

    If Vexhaust = Isp * g then Mfuel = Mdry(e(deltaV/Isp*g) - 1)

    Assuming that the deltav equals the escape velocity, the fuel required to escape Earth gravity is

    Mfuel,earth = Mdry(e(11200/(450*9.81)) - 1) = 5.8965 * Mdry

    and,

    Mfuel,mars = Mdry(e(5027/(450*3.71)) - 1) = 7.1828 * Mdry

    This would seem to say that more fuel is required to escape Mars than is needed to escape Earth, which seems completely incorrect. Can someone please point out to me where I've gone wrong?
     
  2. jcsd
  3. Sep 28, 2011 #2

    BobG

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    Your specific impulse for both rockets was determined using Earth's gravity. Then you used your fuel rate using Earth's gravity for the Earth rocket and Mar's gravity for the Mars rocket.

    The units for specific impulse (sec) and fuel rate (N/sec or lbs/sec) are totally artificial and are obtained by tossing 'g' into the equations for each. (Actually, it's just a unit conversion.)

    In reality, rockets are based on the simple principle of conservation of momentum. The mass times velocity of the fuel tossed out the back equals the change in momentum of the rocket.

    Mass doesn't change regardless of where you are. If you realize its the mass that really matters and that the 'g' tossed into each equation should cancel out, then you'll realize you need to use the same 'g' no matter which planet you're talking about.

    It does seem a strange way of doing things. But specific impulse is more a measure of how efficient the fuel process is than anything else. In the end, you do get a force from your thrusters measured in the proper units for force (Newtons or pounds).

    And I realize you didn't calculate the specific impulse - that it was given to you, but I guarantee 'g' was used in coming up with the specific impulse and the 'g' was 9.8 m/sec^2. (Which is why "If Vexhaust = Isp * g...." is true; you're pulling the 'g' back out of your specific impulse. You need to make sure you're pulling out the same 'g' that originally went in.) And, in reality, that specific impulse will be constantly changing, so be glad they just gave it to you, even if it leads to some confusion (as the pressure in the fuel tanks decrease, the specific impulse will decrease while, conversely, as the temperature builds up due to the chemical reactions, the specific impulse will increase - think PV=NRT).

    The working versions of equations for things are always the easiest for doing calculations, but tend to use some shortcuts that obscure what's really happening, so I can see why you would make the assumptions you did.

    Specific impulse is the velocity of the fuel coming out of the thruster divided by g. Fuel rate is the mass of the fuel being used per second times g. The g just cancels out, leaving mv. Of course, the rate you're using up fuel also determines the rate that the rocket is losing mass, which affects the acceleration from your thrusters, hence some of the other parameters in your equations. A lot of things are being rolled into just a few equations.
     
    Last edited: Sep 28, 2011
  4. Sep 29, 2011 #3
    Thank you, BobG. I appreciate the clarification.
     
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