I know it's on rails, but how much would it take to de-orbit the Mun?

[Souper]

Title says it all. Whoever solves it first gets a E-cookie. Also, you just lost The Game.

[LLlAMnYP]

Just under 1.43*10^26 Joules

[theend3r]

About one impossibillion of jumbo tanks.

[Redjoker]

In the words of Scott Manley

[Ralathon]

The mun has a circular orbit with a SMA of 12Mm.

To deorbit the mun we need to lower the (SMA to 12Mm+0.8Mm)/2 = 6.4Mm. (0.8Mm is the combined radii of Kerbin and the Mun)

Vis-viva equation tells us that the mun needs to move at 191.8 m/s at apoapsis. It currently moves at 542.5 m/s so it takes 350.7 m/s of dV.

If we use a NERVA to deorbit the mun (over the course of many many orbits) we're going to need:

350.7 = 800*9.81*ln(M_wet/M_dry)

M_wet = 1.0457*M_dry

Wet to dry ratio of tanks in Ksp is 9 to 1.

9*M_dry+M_mun = 1.0457*M_dry+1.0457*M_mun)

M_dry=(0.0457/7.9743)*M_mun=5.6e15 tons = 1.4e15 Jumbo tanks.

So just 1.4 Petatanks!

[SunJumper]

The energy requirements are already known, but fuel requirements depend on what propulsion you use, and on whether it is permitted to use resources from the mun itself.

Using ion propulsion however, you may need less than a petatank (here 32000kg of xenon)

[Souper]

Exactly how long would a jumbo with infinite fuel and a mainsail take?

[Souper]

also, Ralathon wins.

[Souper]

*gives you a E-cookie*

[Ralathon]

Exactly how long would a jumbo with infinite fuel and a mainsail take?

Depends on how you burn and how many engines you use. Each engine you add increases the mass and thrust. So you have less dV but faster burns. If you want to do an optimal burn you need to burn retrograde while the mun is at apoapsis and then wait for it to come back around. If you just don't care and go for a spiral orbit you need twice the dV I calculated earlier, so about 700m/s. Unless you plaster half the mun in Nerva engines it is going to take a long ass time though.

If you know your flightplan it isn't too hard to calculate it exactly, just calculate your acceleration as a function of time and integrate it to get your dV. Then you solve for the given dV.

[K^2]

You can't burn continuously. Continuous burn would actually cause you to gently spiral down, requiring a much longer delta-V. Of course, if you are only time-limited and not fuel-limited, then that doesn't matter. But if fuel is a concern, you can only burn at apoapsis, once per revolution. That's going to take a very, very long time.

[Souper]

How about an engine that was half the size of the mun? (I wonder if i could scale a part / engine up to that size without getting into trouble with the 2.5 km distance limit :|

And then, how powerful would it need to be to completely cancel it's velocity relative to Kerbin in just under 1 minute?

(Sorry for being 1+ years late.)

[-Velocity-]

Also keep in mind that sources like the KSP wiki arbitrarily assumed that the gravitational constant in the KSP universe is the same as ours, and that objects like Mun or Kerbin are made of some ridiculously dense substance that does not exist in our universe and is not at all similar to the substance that rockets in KSP are made out of. A simpler explanation is that the gravitational constant in the KSP universe is higher and that rockets are made out of the same form of matter that makes up planets (as in real life). This second set of assumptions is simpler, but under it, we don't know what the masses of the objects in KSP are, as the estimated masses presented in the Wiki are from the objects' gravitational effects. Since the planets are on rails, it is impossible to determine what the actual gravitational constant is (as we can't apply a force to a gravitating body and determine how much it accelerates to determine its inertial mass), and as such, we cannot answer how much it would take to de-orbit the Mun.

There is another explanation though- if we take the "planets are on rails" part of the engine as an actual physical trait of the KSP universe. Since no force we apply to the planets appears to change their courses, one could conclude that the masses of gravitating bodies in KSP is extremely high, so high that any delta-V we impart on them is below the measurement floor. This would mean that gravitating bodies (even Gilly) are nearly infinitely massive and gravity is nearly infinitely weak to compensate. So the amount of impulse it would take to de-orbit the Mun is nearly infinite.

Edited April 2, 2015 by |Velocity|