Interplanetary injection dv

[Crixomix]

Let's say I want to park a satellite in a huge orbit (40Mm) around Duna. 

 

Is it more efficient to plot an intercept that has a Duna periaps of 40Mm, and then circularize at that point? Or to plot an intercept that gets as low as possible (60km) and then burn into a elliptical orbit which has an apoaps of 40Mm, then circularize at the 40Mm. 

 

Obviously the second one takes more time but I don't care about that, I'm wanting to know more about the Delta v. My initial brain logic (very fallable) says that by burning close to Duna, you take advantage of the oberth effect, shedding your speed faster, and it will end up being a cheaper overall dv cost. 

 Am I right? Also, if it is more efficient, is it significant at all? Or is it something stupidly low like .5% cheaper dv. 

[OhioBob]

1 hour ago, Crixomix said:

Is it more efficient to plot an intercept that has a Duna periaps of 40Mm, and then circularize at that point? Or to plot an intercept that gets as low as possible (60km) and then burn into a elliptical orbit which has an apoaps of 40Mm, then circularize at the 40Mm.

The latter is better, by probably about 400 m/s.

[DerekL1963]

55 minutes ago, OhioBob said:

The latter is better, by probably about 400 m/s.

But risks having Ike wander by and try to join in the fun.

[Archgeek]

3 hours ago, DerekL1963 said:

But risks having Ike wander by and try to join in the fun.

Heheh, adjust your timing just right and you make Ike help you with the task a bit.  Get an encounter where Ike'll be a decent ways ahead of your pe, set your burn so you fly out in front of Ike for a gravity desist, adjust further with a burn at Ike to get that ap right where you want it, and re-inflate your orbit into a circle at ap.  Ike and Tylo just want to help, you see.

[Crixomix]

4 hours ago, OhioBob said:

The latter is better, by probably about 400 m/s.

I did the math. All the ugly math. I wasn't able to take any of the gravity assist stuff into account, so I'm not sure how big of a deal that is. But it looks like you save about 240dv by making use of the oberth effect. I did it all on paper but I'll double check it when I get back to an excel spreadsheet at home. 

[OhioBob]

19 hours ago, Crixomix said:

I did the math. All the ugly math. I wasn't able to take any of the gravity assist stuff into account, so I'm not sure how big of a deal that is. But it looks like you save about 240dv by making use of the oberth effect. I did it all on paper but I'll double check it when I get back to an excel spreadsheet at home. 

Here my math, please correct me if I'm wrong...

The ∆V depends on the spacecraft's speed entering Duna's space.  I always measure this using something called hyperbolic excess velocity.  The equation used is,

V² = V_esc² + V_∞²

where V_esc is escape velocity and V_∞ is hyperbolic excess velocity.

A good encounter with Duna will generally have V_∞ = 700 to 900 m/s (could be a lot more if the encounter is suboptimal), so I took the average and said V_∞ = 800 m/s.

Let's first consider the scenario where we insert direct into a 40 Mm orbit.

We first need the escape velocity at an altitude of 40 Mm above Duna, which is

V_esc = SQRT(2*μ/r) = SQRT(2*3.013632E+11/40320000) = 122 m/s

We can now compute the velocity of the approaching spacecraft when reaching this altitude,

V = SQRT(V_esc²+V_∞²) = SQRT(122²+800²) = 809 m/s

Orbit insertion requires that we slow down from 809 m/s to the orbital velocity at 40 Mm.  The orbital velocity is,

V_orb = SQRT(μ/r) = SQRT(3.013632E+11/40320000) = 86 m/s

(where for Duna μ = 3.013632E+11 m³/s², and radius = 320 km.)

So the required ∆V is,

∆V = 809 - 86 = 723 m/s

Let's now consider the second scenario where we insert into a highly elliptical orbit, then circularize.

The escape velocity at an altitude of 60 km is,

V_esc = SQRT(2*3.013632E+11/380000) = 1259 m/s

So the spacecraft's velocity at this altitude is,

V = SQRT(1259²+800²) = 1492 m/s

We're assuming a 60 x 40,000 km orbit, which has a semimajor axis of,

a = (60+40000)/2+320 = 20,350 km 

Using the vis-viva equation, the velocity at periapsis is,

V = SQRT(μ*(2/r-1/a)) = SQRT(3.013632E+11*(2/380000-1/20350000)) = 1254 m/s

Therefore the orbit insertion ∆V is,

∆V (orbit insertion) = 1492 - 1254 = 238 m/s

To this we must add the ∆V needed to circularize the orbit at 40 Mm when reaching apoapsis.  The velocity at apoapsis is,

V = SQRT(3.013632E+11*(2/40320000-1/20350000)) = 12 m/s

We've already computed that the orbital velocity of a 40 Mm circular orbit is 86 m/s.  Therefore the circularization ∆V is,

∆V (circularization) = 86 - 12 = 74 m/s

For a total ∆V of,

∆V (total) = 238 + 74 = 312 m/s

That's a savings of,

723 - 312 = 411 m/s.
 

Edited August 15, 2017 by OhioBob

[Vanamonde]

This question about how to play the game has been moved to Gameplay Questions. 

[Crixomix]

3 hours ago, OhioBob said:

Here my math, please correct me if I'm wrong...

The ∆V depends on the spacecraft's speed entering Duna's space.  I always measure this using something called hyperbolic excess velocity.  The equation used is,

V² = V_esc² + V_∞²

where V_esc is escape velocity and V_∞ is hyperbolic excess velocity.

A good encounter with Duna will generally have V_∞ = 700 to 900 m/s (could be a lot more if the encounter is suboptimal), so I took the average and said V_∞ = 800 m/s.

Let's first consider the scenario where we insert direct into a 40 Mm orbit.

We first need the escape velocity at an altitude of 40 Mm above Duna, which is

V_esc = SQRT(2*μ/r) = SQRT(2*3.013632E+11/40320000) = 122 m/s

We can now compute the velocity of the approaching spacecraft when reaching this altitude,

V = SQRT(V_esc²+V_∞²) = SQRT(122²+800²) = 809 m/s

Orbit insertion requires that we slow down from 809 m/s to the orbital velocity at 40 Mm.  The orbital velocity is,

V_orb = SQRT(μ/r) = SQRT(3.013632E+11/40320000) = 86 m/s

(where for Duna μ = 3.013632E+11 m³/s², and radius = 320 km.)

So the required ∆V is,

∆V = 809 - 86 = 723 m/s

Let's now consider the second scenario where we insert into a highly elliptical orbit, then circularize.

The escape velocity at an altitude of 60 km is,

V_esc = SQRT(2*3.013632E+11/380000) = 1259 m/s

So the spacecraft's velocity at this altitude is,

V = SQRT(1259²+800²) = 1492 m/s

We're assuming a 60 x 40,000 km orbit, which has a semimajor axis of,

a = (60+40000)/2+320 = 20,350 km 

Using the vis-viva equation, the velocity at periapsis is,

V = SQRT(μ*(2/r-1/a) = SQRT(3.013632E+11*(2/380000-1/20350000) = 1254 m/s

Therefore the orbit insertion ∆V is,

∆V (orbit insertion) = 1492 - 1254 = 238 m/s

To this we must add the ∆V needed to circularize the orbit at 40 Mm when reaching apoapsis.  The velocity at apoapsis is,

V = SQRT(3.013632E+11*(2/40320000-1/20350000) = 12 m/s

We've already computed that the orbital velocity of a 40 Mm circular orbit is 86 m/s.  Therefore the circularization ∆V is,

∆V (circularization) = 86 - 12 = 74 m/s

For a total ∆V of,

∆V (total) = 238 + 74 = 312 m/s

That's a savings of,

723 - 312 = 411 m/s.
 

Beautiful answer. I had done the hyperbolic excess velocity part wrong. Thanks for the help! 

[Jas0n]

Wow that's a lot of math.... I just give it as much delta V as delta V maps tell me, with 1000m/s or so of extra.

[OhioBob]

1 hour ago, Crixomix said:

Beautiful answer. I had done the hyperbolic excess velocity part wrong. Thanks for the help! 

There are three equations that if you can master, you can do a heck of a lot in KSP.  They are Tsiolkovsky rocket equation, vis-viva equation, and hyperbolic excess velocity.
 

1 hour ago, Jas0n said:

Wow that's a lot of math.... I just give it as much delta V as delta V maps tell me, with 1000m/s or so of extra.

I usually use Transfer Window Planner, which gives the ejection and insertion delta-v, then add made 200-300 m/s margin.  I only did the math this time because that's the only way I could answer Crixomix's question.

[Foxster]

Wouldn't it be most efficient to aerobrake to establish an orbit and then tweak it afterwards?

[WanderingKid]

3 hours ago, OhioBob said:

There are three equations that if you can master, you can do a heck of a lot in KSP.  They are Tsiolkovsky rocket equation, vis-viva equation, and hyperbolic excess velocity.

Is there a good link you know of for an introduction to vis-viva equation?  I've never heard the term before and would like to get acquainted.

[ThreePounds]

56 minutes ago, WanderingKid said:

Is there a good link you know of for an introduction to vis-viva equation?  I've never heard the term before and would like to get acquainted.

It's nothing fancy really. It's this one here:

v: relative speed; μ = GM = G*M: standard gravitational parameter; r: distance from elliptical focus, i.e. radius of central body + altitude; a = SMA: semi-major axis

EDIT: For those who think I butchered the equation, please square both sides.

Edited August 15, 2017 by Three_Pounds

[Spricigo]

2 hours ago, Foxster said:

Wouldn't it be most efficient to aerobrake to establish an orbit and then tweak it afterwards?

I think that's a particular case of capturing with low pe. Or rather a potential added advantage.

[Crixomix]

3 hours ago, Foxster said:

Wouldn't it be most efficient to aerobrake to establish an orbit and then tweak it afterwards?

Most definitely. Because it's essentially the same as the second case, but you get to slow down for free! 

[OhioBob]

5 hours ago, WanderingKid said:

Is there a good link you know of for an introduction to vis-viva equation?  I've never heard the term before and would like to get acquainted.

@Three_Pounds covered it well in his post.  The vis-viva equation is used to calculate velocity at any point in a satellite's orbit, provided the radius is know.  It's often used to calculate periapsis and apoapsis velocities because those are the points where is it most efficient to perform maneuvers, but we can really calculate the velocity anywhere.  Note that if the orbit is circular, the vis-viva equation simplifies to,

v = SQRT( μ / r)

While we'll never have a perfectly circular orbit, for many calculation the orbit is likely close enough that we can assume circular to simplify the calculations.

To compute escape velocity, the equation becomes,

v_esc = SQRT( 2 * μ / r )

From this we can see that escape velocity is always SQRT(2) times orbital velocity.
 

Edited August 15, 2017 by OhioBob

[OhioBob]

There's something else I better explain that will help complete this discussion.  I earlier said that a spacecraft approaching Duna on a good Kerbin-Duna trajectory will typically have a hyperbolic excess velocity of about 700-900 m/s.  But how do I know that?  The math to figure that out from scratch is rather complex and more than we need to get into, however there's an easier way.  I use the Transfer Window Planner mod, or the web version Launch Window Planner.

Using the web version, I enter origin Kerbin, initial orbit 80 km, destination Duna, final orbit 60 km, earliest departure year 1 day 1, and transfer type ballistic.  Plotting that (and refining transfer) we see that we have a launch window with a departure date of year 1, day 236.  The ejection ∆v is 1059 m/s and the insertion ∆v is 652 m/s.  From those numbers we can compute hyperbolic excess velocity using the equation,

V² = V_esc² + V_∞²

We'll compute it for Duna, but the method is exactly the same for Kerbin.  We first need to know the orbital and escape velocities at an altitude of 60 km,

V_orb = SQRT(3.013632E+11/380000) = 890.5 m/s

V_esc = SQRT(2*3.013632E+11/380000) = 1259.4 m/s

V is the velocity of the spacecraft prior to the orbit insertion burn, i.e. it is orbital velocity plus the insertion ∆v,

V = 890.5 + 652 = 1542.5 m/s

So now we just rearrange the first equation and solve for V_∞,

V_∞ = SQRT(V²-V_esc²) = SQRT(1542.5²-1259.4²) = 890.6 m/s

If we were to do the math for different transfer windows, we'd find that for Duna V_∞ usually varies between about 700-900 m/s.  (It's just coincidence that in this example V_orb and V_∞ are nearly identical.)

I've also found that using Transfer/Launch Window Planner, the computed value of V_∞ varies slightly depending on the orbit altitude entered.  For instance, if we enter a final orbit altitude of 200 km, we get an insertion ∆v of 634 m/s.  Recomputing V_∞ using these values we get 887.6 m/s.  I'm pretty certain that the reason for this difference is because of the way KSP uses patched conics.  In real life this wouldn't be the case.
 

Edited August 15, 2017 by OhioBob

[Crixomix]

8 hours ago, OhioBob said:

There's something else I better explain that will help complete this discussion.  I earlier said that a spacecraft approaching Duna on a good Kerbin-Duna trajectory will typically have a hyperbolic excess velocity of about 700-900 m/s.  But how do I know that?  The math to figure that out from scratch is rather complex and more than we need to get into, however there's an easier way.  I use the Transfer Window Planner mod, or the web version Launch Window Planner.

Using the web version, I enter origin Kerbin, initial orbit 80 km, destination Duna, final orbit 60 km, earliest departure year 1 day 1, and transfer type ballistic.  Plotting that (and refining transfer) we see that we have a launch window with a departure date of year 1, day 236.  The ejection ∆v is 1059 m/s and the insertion ∆v is 652 m/s.  From those numbers we can compute hyperbolic excess velocity using the equation,

V² = V_esc² + V_∞²

We'll compute it for Duna, but the method is exactly the same for Kerbin.  We first need to know the orbital and escape velocities at an altitude of 60 km,

V_orb = SQRT(3.013632E+11/380000) = 890.5 m/s

V_esc = SQRT(2*3.013632E+11/380000) = 1259.4 m/s

V is the velocity of the spacecraft prior to the orbit insertion burn, i.e. it is orbital velocity plus the insertion ∆v,

V = 890.5 + 652 = 1542.5 m/s

So now we just rearrange the first equation and solve for V_∞,

V_∞ = SQRT(V²-V_esc²) = SQRT(1542.5²-1259.4²) = 890.6 m/s

If we were to do the math for different transfer windows, we'd find that for Duna V_∞ usually varies between about 700-900 m/s.  (It's just coincidence that in this example V_orb and V_∞ are nearly identical.)

I've also found that using Transfer/Launch Window Planner, the computed value of V_∞ varies slightly depending on the orbit altitude entered.  For instance, if we enter a final orbit altitude of 200 km, we get an insertion ∆v of 634 m/s.  Recomputing V_∞ using these values we get 887.6 m/s.  I'm pretty certain that the reason for this difference is because of the way KSP uses patched conics.  In real life this wouldn't be the case.
 

I might be wrong here, 

 

wouldn't excess hyperbolic velocity be, roughly, equal to the difference in speed between your vessel and the body that you're encountering?

 

I.e. if you go to duna, your velocity at apoaps (around kerbol) versus Duna's velocity. Relative to your target, duna, it's going about 800m/s. Since you encounter the SOI of duna far away, isn't your velocity at that exact moment roughly equal to your "velocity at infinity"? Like if you were already at duna, and you accelerated to exactly escape velocity, your velocity at the edge of the SOI would be close to zero (since the SOI isn't infinite, it would be greater than zero, but probably still close). So since your velocity at infinity is zero, your velocity at the edge of SOI is (almost) zero. 

 

Is this making any sense? It seems to me like for the most part, hyperbolic excess velocity is going to be pretty much the same as whatever your relative target speed (orbital speed) is when you leave the SOI

Edited August 16, 2017 by Crixomix

[OhioBob]

55 minutes ago, Crixomix said:

wouldn't excess hyperbolic velocity be, roughly, equal to the difference in speed between your vessel and the body that you're encountering?

Hyperbolic excess velocity is equal to the relative velocity, which is the difference between the velocity vectors.  It is important to take the direction into account, not just the difference in speed.  Maybe that's what you meant, but I just wanted to be clear.  Other than that clarification, you are correct.  In practice we have to figure out the velocity vector of each body at the intersection point and subtract the vectors.  This makes the problem a lot more difficult then something like an idealistic Hohmann transfer.
 

 

Edited August 16, 2017 by OhioBob

[Crixomix]

3 minutes ago, OhioBob said:

Hyperbolic excess velocity is equal to the relative velocity, which is the difference between the velocity vectors.  It is important to take the direction into account, not just the difference in speed.  Maybe that's what you meant, but I just wanted to be clear.  Other than that clarification, you are correct.  In practice we have to figure out the velocity vector of each body at the intersection point and subtract the vectors.  This make the problem a lot more difficult then something like an idealistic Hohmann transfer.

 

Okay, so that makes the calculations simpler because you could find the hyperbolic excess without needing to know the injection dv. 

 

And I think in most cases your velocity vector is fairly close to the planets at the intercept point, or at least that's always my goal. Obviously with the highly eccentric or the very inclined orbits this may not be possible. I'll always use the transfer window planner mod. But it's fun to be able to do all these calculations myself too. 

[OhioBob]

@Crixomix, if you want to get serious about studying the math, you can look at this:  http://www.braeunig.us/space/interpl.htm.