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# Capture Delta-V

## Question

Hello Everyone,

I've been wondering for a little over an hour about this now, how can you calculate the delta-v from needed to get into orbit of a body once you enter its Sphere of Influence?

On many delta-v maps like this one there is a delta-v needed to get into orbit (mun: 310m/s). I understand the Hohmann transfer which gives the other values but I don't know how to get the delta-v needed to get into orbit once you're intercepted by a body. So can anyone help please?

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Well, take a look at the moment of capture into the Mun's SOI.

You know the Mun's velocity vector, and can make a good estimate of your vessel's velocity vector: approximately parallel to the Mun's trajectory, at your apoapsis velocity.

Thus, your Mun capture velocity is approximately the Mun's velocity, minus your transfer orbit's apoapsis velocity. In practice, it's probably going to be a bit more, because you're probably not going to be at exact apoapsis and you're probably not going to be exactly parallel to the Mun's orbit at the moment of capture.

Thus, you now have a Mun-relative V, and you already know G*m for the Mun, and your altitude is the Mun's SOI (be sure you're using SOI radius from the exact center of the Mun, not from the Mun's surface).

An example, assuming you started from 100 km periapsis of Kerbin, and will capture 20 km above the Mun:

Note that the vis-viva equation simplifies to V^2 = mu/a when you are in a circular orbit (r == a)

Vpark = sqrt(mu(Kerbin) / rpark) = 2246.139 m/sec at parking orbit

Next, your transfer orbit apoapsis will be about at the Mun's SMA of 12,000,000m, so your transfer SMA will be 0.5*(700,000 + 12,000,000), or 6,350,000m.

Vtrans = sqrt(mu(Kerbin) * ((2/rpark) - (1/atrans))) = 3,087.738 m/sec

So, your Mun injection burn requires (3087-2246) = 841.6 m/sec

Next up is calculating your Mun-relative velocity at capture.

Vapokerbin = sqrt(mu(Kerbin) * ((2/apo) - (1/atrans))) = 180.1 m/sec

We know the Mun's orbital velocity, 542.5 m/sec. So, at capture, your Mun-relative velocity will be approximately 362.4 m/sec.

We can now plug this in to determine our Mun-capture SMA using the Mun's SOI radius as our altitude.

V(mun)^2 = mu(Mun) * ((2/SOI) - (1/a))

Rearrange, and you get a = 1 / ( (2/SOI) - (V^2/mu(Mun)))

We now have a semi-major axis of -838387 meters. Yes, I know a negative SMA seems nonsensical, but it's how SMA works for hyperbolic orbits.

We can then plug that in to determine our velocity at peri-Mun of 220000 meters

Vperi = sqrt(mu(Mun) * (2/220000 - 1/a)) = 818.45 m/sec

We can then calculate the velocity of our destination, circular 20 km (220 km from the center) orbit

Vdest = sqrt(mu(Mun)/220000) = 544.14 m/sec

From that, we know the delta-V of our insertion burn: 818.45 - 544.14 m/sec = 274.315 m/sec

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It is easiest (for me) to do with the vis-viva equation, which relies on conservation of orbital energy. You have a certain amount of relative velocity incoming to the Mun, plus gravitational potential energy.

The vis-viva equation is: V^2=G*m*((2/r) - (1/a)), where:

V is velocity, G is the gravitational constant, m is the mass of the Mun or whatever body you're transferring into, r is current altitude, and a is semi-major axis. You will often see the Greek letter mu substituted for G*m; that product is the body's standard gravitational parameter, and due to not having a very precise estimate of G, real world astronomers usually use mu directly.

So, if you know what Mun-relative velocity you capture with, and the SOI radius, you can solve for a, and plug that back in with an r much closer to the surface, and now you know your velocity at periapsis. From there, calculate the velocity you will have in the target orbit, and subtract.

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The delta-v required to capture depends on your trajectory and  speed through the target SOI.  As you mention, the delta-v maps you see online give you numbers, based on a fairly close to ideal Hohman transfer.  But sometimes you'll come in at a less than ideal approach and it will take more.  Conversely, if you spend more delta-v before hitting the SOI getting your orbit closer to the target's, it can take less once you're in the target's SOI.

The easiest way to see what you'll need to capture is generally with maneuver nodes.  If you can get your nodes lined up close enough that you'll hit your target's SOI, then you can put another node at your periapisis vs. the target, and just see how big of a retrograde burn it takes to capture (or circularize, if you prefer).

One other thought: in order to capture cheaply, you generally want your periapsis to be as close to the target as possible (to take advantage of the Oberth effect).  This is what those delta-v maps assume as well.  So it's worth it to fine-tune your maneuvers for a close approach.

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@Aegolius13 Thank you for your input, but I was looking for a way using maths to get the delta-v cost of a capture, assuming an ideal Hohmann transfer, without having to rely on maneuver nodes. I'm sorry if I did not make it clear enough.

@Starman4308 I was trying to use the vis-viva equation but I'm not sure how. If we keep the example of the mun with the same delta-v map, we expect to need about 310m/s to get into orbit right?

First we need to know what's our transfer maneuver looks like. Let's see how the math checks out:

Spoiler

I already did the math and noticed that what is used for that map is Hohmann transfer with a periapsis of 80km above Kerbin (680 000 m from the center) and with an apoapsis of 11 400 000 above Kerbin (12 000 000 m from the center, which is the radius of the mun's circular orbit). So we get:

Dv  = sqrt ( muKerbin / rinitial ) * ( sqrt ( 2*rfinal / ( rinitial + rfinal ) ) - 1 )   ←   Let's replace with values

Dv = sqrt ( 3,5315984 * 10^12 / 680000 ) * ( sqrt ( 2 * 12000000 / (680000 + 12000000 ) ) -1)

Dv = 856,355 m/s ~ 860 m/s   ←  So it checks out so far

Now we get to the encounter ans this is the part where I am lost and I'm no how to get the delta-v to get into orbit. I'm not even sure how to use the vis-viva equation since we're not in orbit and therefore the semi-major axis "a" is the one of a hyperbole. So I'm kind of lost. Any help?

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@Starman4308 Amazing! That's exactly what I was looking for! Thanks a lot

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