Minimus dV to to a 75x75 LKO orbit

[Nich]

I read in a post that the minimum dV required to get to a 75x75 LKO orbit is 6350 m/s. I have thought about this problem ever since I went to school for aerospace engineering. I was thinking of coming up with the prof of this my self but I though it might be more fun if I let everyone who is interested in helping help and we may get a better answer. If you have read or written a research paper on the topic please don't give away the answer.

I would like to take a step wise approach and move from the impossibly low but bare minimum dV required and add in more complexity until we get to a realistic minimum dV requirement.

Step 1 remove air friction. Do an instant horizontal burn at KSP and a second instant burn at AP to circularize.

A: 2389.7 (Orbital Vagabond)

Step 2 add in air friction. simple launch, vertical with infinite TWR, second burn at AP to circularize.

Step 3 Gravity turn with no air friction TWR limited to 2 at launch (increases as vehicle burns off fuel)

Step 4 Gravity turn with air friction TWR limited to 2 at launch (increases as vehicle burns off fuel)

Step 5 Variations from a gravity turn (does getting to lower density air sooner improve required dV)

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Step 1

I believe the easiest way to calculate this is conservation of energy where E = 1/2mV^2 + GMm/r

E_{1 =} E₂

V_{1 = ?}

V_{2 = ?}

V_{Start = 174.6 m/s}

V_{Final = 2287 m/s}

1/2V₁² + GM/r₁= 1/2V₂² + GM/r₂

(V₂² - V₁²) = 2GM(r_{2 -} r₁)/r₂r₁

V₁² = 2GM(r_{2 -} r₁)/r₂r_{1 -} V₂²

r₁ = 600,050 m (Orbital Vagabond)

r₂ = 675,000 m (Orbital Vagabond)

GM = 3.5316 x 10^12 m³s^-2 (Orbital Vagabond)

Edited June 21, 2015 by Nich

[Orbital Vagabond]

First, the minimum dV to get to a circular 75 km orbit around Kerbin from the surface is certainly not 6350... Is this a typo for 3650?

Second, you aren't going to find a proof due to the complexity of the math involved, especially concerning drag. Your second step will fail apart.

Finally, with fuel and mass held constant, a vessel's dV changes as a function of altitude due to ISP changes.

I can tell you that a Hohmann transfer from 70 m (about where KSC is) on the equator to a 75 Km circular orbit would take 2389.9 m/s, assuming infinite TWR and no velocity losses due to drag.

That's about the extent of what you can get for a theoretical minimum based on math (and even then, its not a proof). Beyond this, you'll need approximations and numerical integration methods, which don't qualify as mathematical proofs.

Addendum: It's Kerbin. Not Kirbin.

[Prepper-Jack]

6350 m/s? That's quite high, unless you're flying a parking lot into space for some reason. Or is that 75 million meters?

Absolute minimum would, in a vacuum, be slightly higher than orbital velocity at the desired orbit.

Radius of Kerbin is 600km

[Orbital Vagabond]

My only problems are I am not sure if GM = 9.81 or just G = 9.81 and I am not sure or the radius ® of Kirbin. So I have some more research to do.

GM_(Ker) = 3.5316 x 10^12 m³s^-2

r(Ker) = 600 km

All these values are available on the official wiki page.

[Red Iron Crown]

g₀ is 9.81 m/s². G is 6.674x10^-11 Nm²/kg².

[Admac]

Just off the top of my head I'd say it's a lot lower.

3500 DV for a fast ascent to 75x75 LKO

950 DV for Minimus transfer

(50 or less DV insertion burn if you take a non optimal launch window)

200 DV circ burn once you get there

(50 or less DV to correct it to a 0 inclination orbit, if that's a thing that you do)

I'd be pretty confident I could get there with 4600-4700 M/s DV just flying by sight.

If that 6350 number was from the days of souposphere, and a very slow launch vehicle I could see that.

Or if you wanted like 650 M/S to land and return in the days of 4500 M/S of DV to get a 75x75 LKO.

Even then 650m/s is pretty generous to land and return home.

[Orbital Vagabond]

Just off the top of my head I'd say it's a lot lower.

3500 DV for a fast ascent to 75x75 LKO

950 DV for Minimus transfer

Kerbin's more distant orbiting body is "Minmus", not "MinImus".

I think the OP was using the term "minimus" to refer to a least amount, i.e. the least amount of dV needed to reach 75km x 75km orbit. I don't believe he was referring to the satellite.

I had to read it a few times to figure it out as well.

[Admac]

Well that's terrifyingly vague.

I retract my advice!

[beabop]

I think you could get a rough estimate if you break the trajectory into linear segments with constant everything - air pressure, Isp. Drag is still a pain, but you can look up the drag cubes of your parts and guesstimate drag coefficient and area, maybe linearize the drag force. Gravity loss varies with flight angle (gravity turning v.s. slowing you), but since your segments are straight lines, cosine and sine of flight path angle are constant over time. Linear segments of trajectories got us to the moon, maybe you can use them.

You'd get the best answer using numerical integration and an optimization algorithm. Which I think Mechjeb does.

[heng]

I can tell you that a Hohmann transfer from 70 m (about where KSC is) on the equator to a 75 Km circular orbit would take 2389.9 m/s, assuming infinite TWR and no velocity losses due to drag.

But you do not have a 70mx70m orbit at launch time.

KSC is not moving at orbital velocity

So you need to add the circularization burn to your dv requirements. :-P

[Nich]

So after thinking about it I realized that the energy equation alone does not define the motion of the ship. You could be passing through Kerbin for all the energy equation cares. Then I was thinking conservation of angular momentum. Please correct me if I am wrong but if you had 2 roughly equally massive planets and you applied a force to one planet you would get a change in velocity and angular momentum based off the moment of inertia and distance of the tangent force to the center of mass of the two bodies. However because the the center of our two body problem is essentially the center of Kerbin the angular momentum of our ship should remain constant. this should give us a second equation to help solve our first one.

Thus r₁V_1/r₂ = V₂

(V₂² - V₁²) = 2GM(r_{2 -} r₁)/r₂r₁

r₁²V₁²_/r₂² - V₁^{2 =} 2GM(r_{2 -} r₁)/r₂r₁

((r₁² - r₂²)V₁²)/r₂^{2 =} 2GM(r_{2 -} r₁)/r₂r₁

(r₁² - r₂²)V₁^{2 =} 2GM(r_{2 -} r₁)r₂/r₁

(r₁ - r₂)(r₁ + r₂)V₁^{2 =} 2GM(r_{2 -} r₁)r₂/r₁

-(r₂ - r₁)(r₁ + r₂)V₁^{2 =} 2GM(r_{2 -} r₁)r₂/r₁

V₁^{2 = -}2GMr₂/(r₁₍r₁ + r₂))

_{Since all values are positive that leaves us with an imaginary number. Perhaps my assumption of constant angular momentum was not valid or hopefully I forgot a negative sign somewhere.}

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And heng I was wondering about that which was why I was trying to do the math my self.

Equation for circular orbit speed at 70m is when centripetal acceleration cancels out gravity

F = mV²/r

Where F=ma

a= 9.81

r = 600,070

V = 2426.25 m/s

Then subtract your orbital speed at KSP 174.9 and your at 2251.35 dV already i suspect getting to 675,000 takes a bit more then 138.55 dV but I will do the math.

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beabop

Originally my plan was to attempt to define everything in terms of r (isp, air density, flight angle, velocity, thrust, acceleration of gravity) Then integrate that from 600,070 to 675,000 with an initial condition of V₀ = 0 and V_f = 2287 in the horizontal direction. The big problem comes from drag which is not directly related to r and the fact that I do not believe the wiki has been updated for how the new aero model works

Edited June 17, 2015 by Nich

[Laie]

But you do not have a 70mx70m orbit at launch time.

KSC is not moving at orbital velocity

So you need to add the circularization burn to your dv requirements. :-P

While standing on the pad, the vessel is constantly held at the apoapsis of a very eccentric orbit. Methinks that those ~2400m/s would be the hohmann transfer from that ground-level apoapsis to LKO.

[Orbital Vagabond]

But you do not have a 70mx70m orbit at launch time.

KSC is not moving at orbital velocity

So you need to add the circularization burn to your dv requirements. :-P

The dV estimate I provided assumes a prograde velocity of 174.6 m/s, which was calculated based on Kerbin's radius, rotational period and KSC's elevation. The speed can be confirmed by parking a vehicle on the pad and switching the navball to "orbital" mode.

Edit: A now removed post brought it to my attention the original remainder of this message (which I have voluntarily removed) may have been unnecessarily harsh. Frankly, I found the rebuttal cited above to be quite rude and snide, as it not only criticizes from what appears to be ignorance of the calculations I performed (i.e. no clarification was requested, nor was any evidence provided for a counter-point), but the smilely and tongue smileys were also quite disrespectful given the calculations were accurate given the cited assumptions.

When criticisms, such as is quoted above, are made without full understanding of what was presented, it doesn't help the conversation. It only muddies the waters and confuses less well-informed readers.

I would really prefer it if, in the future, you could just ask about what was done before posting what you think were needed corrections.

/soapbox

Edited June 19, 2015 by Orbital Vagabond

[Lynch]

The dV required is 3900m/s averagely, can go down to 3300 with PERFECT gravity turn.

Say its 4000 with errors, and then you ll be spending 2750 for getting up to apoapsis with a horizontal velocity of 1000s and burn 1250 to circularise at horizontal velocity of 2250s, thus making 4000 but this is like the worst case scenario

[Orbital Vagabond]

Originally my plan was to attempt to define everything in terms of r (isp, air density, flight angle, velocity, thrust, acceleration of gravity) Then integrate that from 600,070 to 675,000 with an initial condition of V₀ = 0 and V_f = 2287 in the horizontal direction. The big problem comes from drag which is not directly related to r and the fact that I do not believe the wiki has been updated for how the new aero model works

Might I suggest you try modeling everything in terms of time? In my experience, it works much better since you can model r in terms of t, and then all r-dependent factors like drag and ISP in terms of r.

[Rickenbacker]

The dV required is 3900m/s averagely, can go down to 3300 with PERFECT gravity turn.

Say its 4000 with errors, and then you ll be spending 2750 for getting up to apoapsis with a horizontal velocity of 1000s and burn 1250 to circularise at horizontal velocity of 2250s, thus making 4000 but this is like the worst case scenario

Yep. So if you have 6350 m/s Delta-V on board, you can get to orbit, go to Duna and return with some fuel left!

[heng]

but the smilely and tongue smileys were also quite disrespectful given the calculations were accurate given the cited assumptions.

My apologies. I really did not want to sound disrespectful.

As you have propably gathered by the lack of numbers in my post, i have absolutely no graps on the underlying mathmatics and are always in awe if someone actually knows what they are talking about. The smileys were intended to clarify a tounge-in-cheek silly comment, i am sorry for my mishap there!

[Warzouz]

I read in a post that the minimum dV required to get to a 75x75 LKO orbit is 3650 m/s. I have thought about this problem ever since I went to school for aerospace engineering. I was thinking of coming up with the prof of this my self but I though it might be more fun if I let everyone who is interested in helping help and we may get a better answer. If you have read or written a research paper on the topic please don't give away the answer.

...

From experience, I usually pack 3650m/s for my LKO stage. BUT they are self deorbiting and landing rocket SSTO.

Depending on the payload, I did a LKO with 270m/s left. So I used less than 3400m/s. Further more, this is VAC dV. On early acceleration, ISP is lower then "real" dV is even less.

From intuition, I would say that minimal dV would be around 3200m/s, but as KER or MJ don't give "real dV", but only full VAC or full ASL (which either of them aren't realistic - except VAc for vacuum worlds), it's hard to tell.

Rocket shape impact your drag and your flight plan. This has a huge impact on dV as well (probably even bigger that ISP variation).

[WeirdCulture]

From experience, I usually pack 3650m/s for my LKO stage. BUT they are self deorbiting and landing rocket SSTO..

are you using kOS?

[Warzouz]

No, I fly back the launch stage myself to near KSC. I left the payload in orbit and deal with it at next orbit.

Here are my rockets, including flight plan recommendations.

http://forum.kerbalspaceprogram.com/threads/123195-Cygnus-family-Recoverable-SSTO-Rockets-from-12-to-240-tons

[SAI Peregrinus]

Minimum delta-V to get to 70x70km orbit with 1.0.2 stock aero is on the order of 2900m/s. See

for a demonstration.

Orbital speed at 75km is about 2426 m/s (sqrt(GM/a) where GM is the standard gravitational parameter (3.5316000×10¹² m³/s² for Kerbin) and a is the semi-major axis (for a circular orbit of Kerbin, orbital height + radius of kerbin = 600070m). As losses (drag and gravitational during ascent) approach 0 the delta-V required should approach the orbital speed. And you get 174.53m/s of that for free from Kerbin's rotation.

[Nich]

Very impressive demonstration video. Although that craft had an ungodly TWR. Also you forgot to add the 75km when I run the numbers at 75km i get 2287.2379584301482994121048375395 and 2425.9664837144090973858235072786 at 70m

[SAI Peregrinus]

Aah, yes, accidentally ran 70. It's not a huge delta-V difference though. 140m/s. And you should use significant figures, all those extra decimal places have no meaning.

[MarvinKitFox]

i get 2287.2379584301482994121048375395 and 2425.9664837144090973858235072786

I beg to differ! You have an error in the 17th digit. Or was that the 24th?

And I am 100.0000000000000000000000000000000000000000000000000000000000000000000000000000% sure of that!

[Orbital Vagabond]

Step 1 remove air friction. Do an instant horizontal burn at KSP and a second instant burn at AP to circularize.

A: 4641.55 m/s (Orbital Vagabond)

I never said anything like this!!! The only dV estimate I provide was half this value!

I can tell you that a Hohmann transfer from 70 m (about where KSC is) on the equator to a 75 Km circular orbit would take 2389.9 m/s, assuming infinite TWR and no velocity losses due to drag.

I'm really kinda starting to regret even commenting on this thread...

Edited June 20, 2015 by Orbital Vagabond