# How much fuel do I need to send cubesat 1kg from Earth orbit to Moon orbit? [closed]

How much fuel do I need to send cubesat 1kg from Earth orbit to Moon orbit? And how much stages? For i.e. C12H22O11 + KNO3, polyethylene + oxygen or kerosene + oxygen.

## closed as off-topic by Gert, ACuriousMind♦, user36790, CuriousOne, John DuffieldNov 1 '15 at 10:47

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First you want to find the average amount of energy you would need to send it to the moon. For the approximation we assume no atmosphere to the earth and that your making it go to the moon but it does not comme back. We find that the energy needed for the trip is the difference in potential energy between when the mass is at the surface of the earth and when, after the trip, it is on the surface of the moon. So you get $$E=\Delta U=(U_{moon})-(U_{earth})$$ $$E=(-\frac{GM_{moon}m}{r_{moon}}-\frac{GM_{earth}m}{D})-(-\frac{GM_{earth}m}{r_{earth}}-\frac{GM_{moon}m}{D})$$ where here $m$ is the object's mass (in this case 1kg), $r_{moon}$ and $r_{earth}$ are the moon's and earth's radii respectively, $D$ is the distance between the earth and the moon and where $M_{earth}$ and $M_{moon}$ are their masses.

We can see that because D has a very large value, we can put 0 where we had D on the denominator (indeed the potential energy from earth is going to be negligable on the moon and the potential energy from the moon is going to be negligable on the earth). We are then left with $$E=(-\frac{GM_{moon}m}{r_{moon}})-(-\frac{GM_{earth}m}{r_{earth}})$$ $$E=-\frac{GM_{moon}m}{r_{moon}}+\frac{GM_{earth}m}{r_{earth}}$$ so finding the values and putting

This gives you the energy you need for the trip from the earth to the moon, and it will be with that energy that you'll be able to calculate the ammount of fuel that you need, knowing the enthalpy of your combustion.

Hope this helps!

• M Earth is 10^24kg; D is 10^8m. G is common - so ignored M Moon is 10^22kg R moon is 10^6m => Me/D = 10^(24-8) = 10^16 => Mm/Rm = 10^(22-6) = 10^16 You can't consider MEarth/D insignificant. – UKMonkey Jul 30 '18 at 12:57

To get a spacecraft to the Moon we normally use a Hohmann transfer orbit. The fuel is used in two steps:

1. increase the velocity of the scapecraft to put it into an elliptical orbit with its apogee at the Moon.

2. when the spacecraft reaches the Moon increase its velocity again to match the velocity of the Moon.

The amount of fuel required is described by the Delta-V for the transfer. This in turn depends on exactly how the manouver is done. You can find figures for the Earth-Moon delta-V in the Wikipedia article on the Delta-V budget.