Calculating interplanetary delta-v

[tomf]

Someone was asking about how to do calculations in another thread but having typed this up I thought it might deserve a thread of it's own. Corrections, suggestions etc are of course welcome.

This is how I would plan a mission to Moho. This is going to assume that you intercept Moho at its average distance from Kerbol. It ignores mid course transfers. I am assuming you can already work out the orbital speeds required for circular orbits and are familiar with Hohmann transfers.

You are going to be using the Vis-viva_equation

First your transfer orbit has apoapsis at Kerbin =1.36E+10

and periapsis at Moho = 5.26E+09

a (semi major axis) = 9.43E+09

so we plug into the equation r=1.36e10 and a= 9.45e9 and GM is Kerbols gravitational parameter 1.17E+18

Therefore when you leave kerbin soi your kerbol orbit speed needs to be 6935m/s.

By the same equation you speed when arriving at moho is 17921.72597

Kerbin is already moving at 9284.5 m/s so at the point you leave Kerbin you will need a kerbin speed of 2348m/s in a retrograde direction.

Using this as our V and the size of Kerbin's soi 8.42E+07 as r we can work out a, the sma of the escape orbit by rearranging the equation. If our parking orbit is 100km (r_PE_ = 7e5) we can work out that our speed at the end of an instantaneous burn is 3939m/s.

A 100km orbit of kerbin has a speed of 2246m/s so we need to do a burn of 1693 m/s. Yey, we have the same value as other people have got!

Delta-v to match inclination depends on where in your transfer orbit the ascending/descending node is but the worst case is that you have to correct immediately before your moho encounter. In that case you will be traveling at 18km/sec and you want to be traveling at 18 km/sec but at a 7 degree angle.

The delta v required for that can be worked out from trigonometry as 2*18*sin(7/2) = 2188m/s.

The best case is aligning at kerbin's orbit with a delta-v of 846.

Arriving at Moho we need to do a capture burn, using the same maths as the kerbin escape burn.

Our moho relative v at moho soi is 17921-14925 = 2997

The velocity at a moho pe of 30km is 3269. A 30km circular orbit is 935m/s so the capture+ circularize burn is 2333m/s

Next installment - calculating the delta-v required to get of a body with no atmosphere.

[tomf]

Lifting of from an airless world is easy, or at least calculating a minimum delta v is.

You first need to build up enough speed that you would be doing a circular orbit with an altitude of 0, then continue burning to raise your AP, and another burn to raise pe as a normal Hohmann transfer.

After our first burn we want an orbit with pe=2.5e5, ap=2.8e5 and sma = 2.65e5

At pe that orbit has a v of 1018m/s

on the equator of Moho you are moving at 13m/s so our first burn is 1016.

The Hohmann calculation will tell us the second burn is 27

So the total delta-v from the surface to a 30km orbit is 1043m/s

The ideal landing is the same process in reverse so has the same delta-v

WARNING - if you try taking of like this you are in danger of crashing into mountains, or even molehills.

And landing like this will likely lead to Kerbals splattered all over the landscape when you don't judge it perfectly.

[metaphor]

You can do a lot more to make a transfer to Moho or another planet more efficient. Since Moho is pretty highly inclined to the plane of Kerbin's orbit, there's a high delta-v requirement to change inclination. You can get rid of some of this delta-v by including it in the interplanetary transfer burn from Kerbin orbit. So instead of burning 2000 m/s forward to leave Kerbin and then 2000 m/s up or down to change inclination, you can burn 2800 m/s at a 45 degree angle and achieve the same final orbit. Of course this depends how close to the ascending/descending node you are when you leave Kerbin orbit, but you can always get some better efficiency unless the nodes are at exactly 90 degrees from you.

Another thing you can do is a gravity assist from Eve. This is what I always do for my Moho missions since it cuts down the amount of delta-v significantly. So you only need to burn about 1100 m/s from Kerbin for an Eve transfer, and then adjust your trajectory when you're close to Eve to get the Eve exit direction you want. Eve has enough energy so that with a gravity assist you can either change your inclination to Moho's inclination, or get your sun periapsis down to Moho's orbit, which both save a lot of delta-v at no propellant cost. Of course if you want to encounter Moho in the same orbit, you have to time your Kerbin launch right so Moho is in the right place, but otherwise if you don't worry about where Moho is in its orbit, you can get a Moho encounter in one or two more sun orbits at most after leaving Eve.

When you launch from an airless planet, there's no atmosphere drag, but there's still gravity drag. So unless your thrust to weight ratio is very high, you have to burn some of your delta-v vertically. So the total delta-v needed to orbit is a little higher based on what engine you have, probably around 1100-1200 m/s instead of 1043 m/s for Moho.

Other than that, the calculations are spot on. If you are actually making a Moho mission, you always want at least 10% more delta-v than required in case of errors and course correction burns.

[tavert]

tomf, this is great, thank you. I'd love to see all this broken down for the different planets and different burns in a spreadsheet, either on Google docs or an ods posted somewhere. The various delta-V maps try to show this type of data, but it's not as useful in my mind if I can't see the math behind it. Also very useful to tweak initial and final conditions, like parking and destination orbit altitudes, etc. And we could use this kind of thing to compare AN/DN inclination burns vs timing transfers to be at the nodes (not always possible, but worth aiming for).

Edited May 3, 2013 by tavert

[Tada]

When playing KSP I really hate myself for not paying atention during math and science classes

[Alfastar]

Well, the Delta-V needed to escape Kerbin is the same as the Delta-V needed for a short sub-orbital trip on earth

[John FX]

I`m currently trying to get a landing done on Moho and I`m noticing it`s harder than the other planets.

So much Delta-v...

There are some good tips here. I`ll try and lose some speed using Eve, that should help, also a single burn for deorbit and inclination. I would have thought that getting nearly 7000m/s into LKO would have done it with lots to spare but it seems tight.

[John FX]

I think I`ll try a combination of just getting out of Kerbin SOI, adjusting inclination, burn to eve intercept -> phasing orbit, intercept burn, orbit, land.

I was previously misunderstanding inclination adjustments and doing them on the smaller orbits when I should do them as far out as possible (makes sense)

1693+(846-2188)+2333+ about 1000 to land=5872-7214 Delta-V to land on Moho from Kerbol LKO

Hmmm, I can see how a poorly flown vehicle would not make it to Moho with 7000 Delta-V...

I`ll have to be careful.

EDIT: Realised I was wasting almost 400 Delta-V getting to an orbit I could warp at. I`ll leave my Moho lander at 75Km until I burn to leave SOI as this might make the difference...

Edited May 22, 2013 by John FX

[maltesh]

I think I`ll try a combination of just getting out of Kerbin SOI, adjusting inclination, burn to eve intercept -> phasing orbit, intercept burn, orbit, land.

I was previously misunderstanding inclination adjustments and doing them on the smaller orbits when I should do them as far out as possible (makes sense)

1693+(846-2188)+2333+ about 1000 to land=5872-7214 Delta-V to land on Moho from Kerbol LKO

Hmmm, I can see how a poorly flown vehicle would not make it to Moho with 7000 Delta-V...

I`ll have to be careful.

Based on the typical "all planets are in equatorial and circular orbits" and "My spaceccraft is capable of instantaneous impulse" assumptions, for planetary destinations, if your starting point is a 100km orbit over Kerbin the "Burn to barely escape Kerbin, followed by interplanetary transfer burn outside the SOI" costs about 75% more delta-V than a direct burn from 100kmLKO to destinations like Eve or Duna, up to more than 100% more when heading to distant destinations like Moho and Eeloo.

[John FX]

Based on the typical "all planets are in equatorial and circular orbits" and "My spaceccraft is capable of instantaneous impulse" assumptions, for planetary destinations, if your starting point is a 100km orbit over Kerbin the "Burn to barely escape Kerbin, followed by interplanetary transfer burn outside the SOI" costs about 75% more delta-V than a direct burn from 100kmLKO to destinations like Eve or Duna, up to more than 100% more when heading to distant destinations like Moho and Eeloo.

Yup. I figured as much. I wondered whether the saving in fuel doing a plane adjust at the higher orbit would outweigh the savings from a direct burn but they do not. I`ve added another set of boosters to my launch vehicle which gives me an extra 1000 Dv at LKO and shaved some weight off for another 500 which should make things easily possible rather than just possible...

[Amagi82]

Using this as our V and the size of Kerbin's soi 8.42E+07 as r we can work out a, the sma of the escape orbit by rearranging the equation. If our parking orbit is 100km (r_PE_ = 7e5) we can work out that our speed at the end of an instantaneous burn is 3939m/s.

A 100km orbit of kerbin has a speed of 2246m/s so we need to do a burn of 1693 m/s. Yey, we have the same value as other people have got!

What's the exact equation being used here? It's been a long time since math class. I managed to figure out everything before this part, but I've tried several different equations here and I keep getting wildly incorrect results.

Oh, and I know I'm dredging up an old thread, but thanks for posting this! I've been learning a lot.