How much dV to set Minmus straight?

[airelibre]

Firstly, I know this is impossible as planets and moons are on rails, so this is theoretical. I think Scott Manley once did a video on how much dV it would took to deorbit the moon or something along those lines. It has often annoyed me how Minmus's orbit is slanted. I'm just wondering how simple/hard it would be to calculate how much dV it would take to "correct" it, and if that would be physical possible, even if it's unfeasible. (I seem to remember the moon deorbit would be nigh-on impossible since the mass of fuel tanks would be around 1/4 or 1/3 the mass of the moon itself, this would require less I think)

[Superfluous J]

dV does not take mass into account, so it would take as much dV to "correct" Minmus' orbit as it would to correct a probe's orbit that was similar to Minmus'.

Now the amount of FUEL it would take is a whole other matter. I don't know but it's a lot. Like, more than has ever been ever.

[maltesh]

Minmus is in a circular orbit, whose orbital velocity of Minmus is 274.1 m/s. Minmus has an orbital inclinationof 6°.

As a result, the delta-v neccesary to take that orbit equatorial is 2 * ( 274.1 m/s) *sin (6°/2) = 28.7 m/s.

[airelibre]

Oh of course, I knew that!! It's just so common to refer to how much dV a craft has, when really you mean how much dV it's fuel provides.

[airelibre]

Minmus is in a circular orbit, whose orbital velocity of Minmus is 274.1 m/s. Minmus has an orbital inclinationof 6°.

As a result, the delta-v neccesary to take that orbit equatorial is 2 * ( 274.1 m/s) *sin (6°/2) = 28.7 m/s.

Ok thanks, I'll have to trust you on that. What's the next stage, ie. dV and mass -> fuel mass required?

[cantab]

The most efficient engine in stock is the PB-ION, with an Isp of 4200 s. To get 29 m/s of delta-V out of it requires a mass fraction of 1.0007.

That may seem hilariously small, but Minmus weighs 2.6 x 10¹⁹ kg. 0.0007 times that is still around 18 trillion tonnes of fuel. If you had a xenon version of the giant Kerbodyne tank, with the same wet and dry masses, you'd need 250 billion of them.

[YukiCruentus]

Soooooo... Lotsa kethane mining and Extra planetary launchpads connected to the surface via KAS?

[No one]

Assumptions:

We do not care about how much time this would take.

Mass of anything not a fuel tank/minmus is negligible.

We are perfect and infinitely precise at rocket science.

dv=g*Isp*ln(mi/mf)

Without staging, if we are using ion thrusters, 28.7=9.8(4200)*ln((2.65*10^16+.12x) / (2.65*10^16+.05x))

x = approximately 2.6*10^14 canisters of xenon, for approximately 2.6*10^13 tonnes.

With staging, we can do better.

dv = sum(k=1 to x, 28.7=9.8(4200)*ln((2.65*10^16+.12k) / (2.65*10^16+.12k-.07)

Wolfram alpha refuses to do this. So let's get a lower bound and assume that all of the xenon canisters are massless.

28.7 = 9.8×4200 log((2.65×10^16+0.12 x)/(2.65×10^16)) It would take approximately 1.54036x10^14 xenon canisters.

Is there a better way?

Yes.

We've been assuming that we're grabbing minmus and thrusting. What if we just hit it with something?

With the right gravitational assists we can get things to go really fast. Let's assume we have infinite time and infinitely precise rocket science. We could probably use gravitational assists to get an orbit with the periapsis at Kerbin's periapsis and the apoapsis very very far away. The escape velocity of Kerbol at the height of Kerbin's periapsis, that is, the maximum speed we can be going at before we stop orbiting, is 37,420 m/s. At the apoapsis it would be very easy (Require almost no delta-v) to turn around and get the orbit to be at the right angle to hit minmus to knock it into a less inclined orbit. I'm not sure what the angle would be, but for now I'll just assume it's perpendicular to Kerbin's orbit (Thus, the relative velocity is the entire velocity of the thing we're hitting minmus with).

If we did that, we would only need an object with a mass of only...

MmVmr/(Vmr+Vr)=2.6457897×28.7*37420/(28.7+37420)=2*10^13 tons!

Extremely elaborate plans shave off about .65*10^13 tons.

On the other hand, performing a similarly elaborate impact on Gilly would take less mass, and if we did that then we could (Using infinitely precise rocket science and infinite time) get Gilly into the right path, and then use Gilly to knock Minmus into a less inclined orbit.

[Jesrad]

It has often annoyed me how Minmus's orbit is slanted.

It sure has me NOT annoyed at all ! Whenever I look up to the night sky, and see how disorderly and plain chaotic it all is everywhere in the Universe, I'm reassured about the mess I leave at home. Good enough for the Pleiades, good enough for anyone, I say !

Just learn to embrace the chaos

[Skorpychan]

Soooooo... Lotsa kethane mining and Extra planetary launchpads connected to the surface via KAS?

Nah, just a kethane mining rig and a big Kerbodyne engine. Naturally, you'd have to kill Minimus' spin first, but then you could just burn at the ascending/descending nodes every time, and keep repositioning to sit on a deposit.

Or keep refueling from Kerbin.

[Oafman]

The most efficient engine in stock is the PB-ION, with an Isp of 4200 s. To get 29 m/s of delta-V out of it requires a mass fraction of 1.0007.

That may seem hilariously small, but Minmus weighs 2.6 x 10¹⁹ kg. 0.0007 times that is still around 18 trillion tonnes of fuel. If you had a xenon version of the giant Kerbodyne tank, with the same wet and dry masses, you'd need 250 billion of them.

I foresee a slow frame rate.....

[Javster]

More fuel than all of Whackjob's ships combined.

[cantab]

Soooooo... Lotsa kethane mining and Extra planetary launchpads connected to the surface via KAS?

I don't know exactly how much kethane to expect on Minmus, but I know it's nowhere near enough. Even if you stripped the whole system of its kethane you'd still be many orders of magnitude short.

And if you tried to make the kethane from Kerbals, recording the names and stats of each one would probably make your persistent.sfs fill up your hard drive.

[legodragonxp]

Just grab a bunch of class E asteroids and put them in to position so that you could gravity tug Minmus slowly in to the proper orbit. No smashing required.

http://en.wikipedia.org/wiki/Gravity_tractor

[weezl]

The most efficient engine in stock is the PB-ION, with an Isp of 4200 s. To get 29 m/s of delta-V out of it requires a mass fraction of 1.0007.

That may seem hilariously small, but Minmus weighs 2.6 x 10¹⁹ kg. 0.0007 times that is still around 18 trillion tonnes of fuel. If you had a xenon version of the giant Kerbodyne tank, with the same wet and dry masses, you'd need 250 billion of them.

This begs another question.. since you need this much fuel.. how much of the Kerbian economy is totally consumed by production/transport of this fuel in preparation for use? And for how long? The biggest energy consumption I can think of on earth is oil.. which is produced at the rate of about 6.5 billion barrels/year... or about 1.044 trillion liters of oil. And the dollars required to just extract that from the ground are extreme.. but we are talking LOX and something like hydrazine.. which is MUCH more complex to make, and MUCH more dangerous to store. So if we go with 1000:1 efficiency difference (probably still too low), that gives us 1.044 billion liters of fuel per year...

I don't have the calcs on hand for fuel density.. but it wouldn't surprise me if its at least century or more of production to get 250 billion tanks. And that is assuming you commit 30% of an earth based economy to production of just that one item.

Of course.. no one has mentioned nukes.. I know they aren't in the game.. but for pure matter to energy conversion, you just can't beat it.. although the site where you are applying the force might have to be abandoned as a landing site for some decades after the adjustment event.

Edited June 12, 2014 by weezl

[Pipcard]

It's much, much easier to just make a mid-course correction on your way to Minmus.

[Cmdr. Arn1e]

All theoretical anyways; Minmus is 'on rails' and cannot be moved from its orbit! much the same as all the other planets and moons... Interesting seeing people work this out though!

[Taki117]

This is the video you are looking for. (And the one you reference) Basically to figure out how much fuel you need simply work the dV equation backwards.