I'm working on a hypothetical constellation of orbiters at Venus for a school project, and I am currently trying to figure out how much fuel I need in the transfer stage. I have already calculated that the total mass of the spacecrafts and dry mass of the transfer stage is 6900 KG. The propellant I am using is Liquid Hydrogen + Liquid Oxygen, and the total d/V is around 3.9 km/s. I used the formula for Kinetic Energy, where I got 20 Tons of propellant. I can tell that this isn't possible, and I do not understand exactly how Isp fits in, and how to calculate it. What process should I use to be able to calculate the weight of the spacecraft?

  • $\begingroup$ What's wrong with the numbers you got? 6.9 tonnes dry, 3.9 km/s, 20 tonnes propellant. Seems reasonable to a first order of approximation (it's wrong, but not impossibly so). Also, what formula did you use, exactly? $\endgroup$ Apr 9, 2018 at 16:56
  • 2
    $\begingroup$ Look up the rocket equation. A single kinetic energy is not the way to calculate this. $\endgroup$
    – Mark Adler
    Apr 9, 2018 at 17:00
  • $\begingroup$ I used the formula Ek=1/2*m*v^2, where m was the dry mass, and v was the d/V. $\endgroup$ Apr 9, 2018 at 17:37
  • $\begingroup$ You have everything you need for the rocket equation except the specific impulse. This number characterizes the engine efficiency. An RL-10 or two should be good for your application, look them up to find out its specific impulse and as @MarkAdler suggests, apply your numbers to the rocket equation. $\endgroup$ Apr 9, 2018 at 18:25

1 Answer 1


From Tsiolkovsky's rocket equation (see this Wikipedia article) you can find out the mass of the fuel used during the shuttle's launch.

The equation is stated as it follows

$${\Delta v=v_{\text{e}}\ln {\frac {m_{0}}{m_{f}}}}$$

Where ${\displaystyle \Delta v}$ is the maximum change of velocity of the vehicle (with no external forces acting), $m_{0}$ is the initial total mass including propellant, ${\displaystyle m_{f}}$ is the final total mass without propellant or also known as dry mass, and ${\displaystyle v_{\text{e}}}$ is the effective exhaust velocity.

By rearranging, we can find out for the final mass of the shuttle after consuming all the fuel.

$${{m_{f}}={m_{0}\over e^\frac {\Delta v}{v_{\text{e}}}}}$$

Furthermore, we know that the initial mass minus the final mass equals the mass of the fuel. Thus, now we plug in the values that you provided us in an attempt to find for the final mass.

$${{m_{f}}={{6900\,\mathrm{kg}}\over e^\frac{3.9\,\mathrm{km/s}}{4.44\,\mathrm{km/s}}}}={\frac {6900\,\mathrm{kg}}{2.40699}}={2866\,\mathrm{kg}}$$

In this last set of calculations, I use the effective exhaust velocity value of the engine SSME. You might want to check this table for another value of $v_{\text{e}}$.

The mass of the fuel required is ${6900\,\mathrm{kg}-2866\,\mathrm{kg}=4034\,\mathrm{kg}}$

  • $\begingroup$ Please if anyone knows how to put the divide sign with the math code for Mf equals, you are more than welcome to modify my post. $\endgroup$
    – Matthew
    Apr 9, 2018 at 18:37
  • $\begingroup$ I think this is backwards: the final mass should be 6900kg. At least that's how I'd understand the question. Then the fuel mass would be approx 6900kg * (2.4 - 1) $\endgroup$ May 4, 2018 at 21:46
  • $\begingroup$ @DanielJour As he stated in his question the mass of the spacecraft and the dry mass equals to 6900kg, which is equal to the total mass of the spacecraft. After my calculations, we found that the final mass of the spacecraft after using all the fuel is equal to 2866kg. Therefore the mass of the fuel required is the mass of the rocket before launch minus the mass of the rocket after it has consumed all its fuel, this equals to the mass of the fuel required which is 4034kg. Hoped that clarified things. $\endgroup$
    – Matthew
    May 8, 2018 at 6:03
  • $\begingroup$ the spacecrafts (plural, OP refers to the orbiters, not the whole thing) are the payload which get delivered by a transfer stage whose dry mass is known. So we know payload mass and dry mass. Now the question is how much fuel do we need to add to that? $\endgroup$ May 8, 2018 at 7:52

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