Landing from Low vs High Orbit

[Rhomphaia]

To calculate a stop and drop landing I would stick with the Vis Viva equation.

if v = 0 then 2/r = 1/a

so you can get the SMA of the "orbit" after you stop. then calculate the velocity at surface altitude from that.

I get 315.22m/s from a 36km drop and 494.24m/s from the 120km drop. assuming a landing site at 0 elevation (200km radius) and an infinite TWR so no gravity drag during burns.

Edited March 26, 2014 by Rhomphaia

[Streetwind]

My math is rusty. But I'm thinking in low gravity bodies such as Gilly or Bop. If you're in a highly elliptic orbit (which is cheaper to achieve by burning retrograde at periapsis when you enter the moon's SOI), you will have a lot less horizontal speed at your Ap that at your Pe. So, since gravity isn't speed you too much, does it make sense in those bodies to kill horizontal speed at the Ap, making a slow vertical descend?

You always lose more to gravity than you save on your horizontal burn at high altitudes. Always. See the previous page(s).

Your zeroing out at the apopasis of an elliptical orbit only serves to make you subject to gravity even longer than you would if you were in a circular one of equal orbital energy. The actual gravity of the celestial body is irrelevant in this, because horizontal velocities go down together with gravity, so you save less by trying for the lowest horizontal speed (there's hardly any difference between lowest and highest).

In practice, this approach is actually worse than all other alternatives because you're actually spending dV with the burn at periapsis just to set yourself up for maximum dV losses to gravity later-on. Why not spend the dV at periapsis to decelerate instead of accelerating? You could be halfway into a landing already, with just a minimum of vertical distance to cover.

Edited March 26, 2014 by Streetwind

[ScottyDoesKnow]

Gravity drag isn't a force, of course gravity is, but gravity drag is simply a measurement of the reduction of net performance of a vessel from having to hold its own weight in a gravitational field. on a vessel which is in freefall there is no need for it to hold its own weight so the measurement which would be considered as gravity drag on a typical vertical accent is simply the increase in negative vertical speed due to gravity during a descent.

But we're not talking about freefall, we're talking about landing.

[FREEFALL1984]

But we're not talking about freefall, we're talking about landing.

But if you freefall to the surface and suicide burn until you land, is that not in fact a landing?

[EtherDragon]

Here are some math hints that will probably get you pointed in the right direction...

Even when you are landing, you are bound by Kepler's Laws of Planetary Motion, which Newton generalized into all orbiting objects.

At any point in your orbit, even during landing, you apply force to change the shape of your orbit - in the case of landing, it happens that you (eventually) change your orbit such that you end up with a highly eccentric (approaching 1.0) orbit with the apoapsis is at ground level, and the periapsis is at the center of the body you are landing on.

Let's look as some different circular orbits, and transfer orbits.

If you have a circular orbit of 300km around Mun - that is 100km above the surface - you end up with a 300x300km orbit. There are a number of ways to land, all of which actually take about the same energy - because the result is the same - you end up with a 0x200km orbit at touchdown.

First you can drop your periapsis from 300km to 210km, then drop your apoapsis from 300km to 0km (making in the new periapsis), and finally dropping your new 210km apoapsis to 200km for touchdown. Note, your surface velocity during the 210x300km phase will be quite high.

Second way, you can just hit the breaks and chance your 300x300km orbit to a 0x300km orbit, free fall for a while, and lastly slow down near the surface to make a 0x200km orbit. Note, this actually takes the same amount of energy as the scenario above! Also, you will be falling quite fast once you get closer to the surface.

Now, consider a 220x220km orbit. First off, it has less energy than a 300x300km orbit.

Phase one - change 220x220 to 210x220km for approach. Phase two - at periapsis hit the breaks and come to a 0x210km orbit. Phase three - settle into a 0x200km orbit at touch-down. Note, in phase two, at 210x220km your surface velocity will be much lower than before when you were at 210x300km.

Remember, I refer to these as "orbits" because if the Mun wasn't there, but it's gravity well was, you would still orbit about the center, just at high eccentricities. Since you are landing on the Mun, however, it ends up being a ballistic trajectory. Hope that helps get you pointed in the right direction for your calculations.

[ScottyDoesKnow]

But if you freefall to the surface and suicide burn until you land, is that not in fact a landing?

It is, one that tries to minimize gravity drag by burning for the least amount of time.

[FREEFALL1984]

Yes, I suppose when you phrase it like that it does make sense to apply it to landings, but of course the effect would be minimal almost negligible on a very short, powerful burn.

[Claw]

So I put an MPL into Mun orbit as a base for my Munar landing missions.

Now I was thinking about whether a high or a low orbit required more fuel for the lander; I'm pretty sure it's the low orbit but I'm stuck on the math.

The two factors for fuel usage in this case are the orbital velocity, which is higher for the low orbit - i.e. I need more m/s dV to kill the horizontal speed -, and the gravity from the Mun (which I see as the vertical speed component of the descent vector).

For the orbital velocity, I came up with this calculation:

since vo = sqr(gp / r) with gp = gravitational parameter (65,138,398 km/s^2 for the Mun); r = radius ~= 200km + orbital altitude,

for 36km orbit: vo = sqr(65,138,398 / 236) ~= 523,15 m/s.

for 120km orbit: vo = sqr(65,138,398 / 320) ~= 451,17 m/s

so I spend ~70 m/s more dV to kill my horizontal speed from the low orbit if I'm correct up to here (can only theorycraft right now because I'm not at my spacefaring computer).

But how do I calculate the dV I need to spend to fight Mun gravity for the orbits?

Basically: if I kill my orbital and surface velocity to 0m/s and then let go, gravity should pull me down and add up to a velocity I'll have to counter if I want to land safely...

So how high is that going to be? I know Mun gravity to be 1.63 m^2, but what formula to use to calculate my end speed, knowing the distance I'll be travelling?

Or am I looking at it entirely wrong?

The mass of my lander will be about 2t, 0.5t of which will be fuel (using the FL-T100 tank)... but that shouldn't matter for the dV, only for the fuel I need to reach it, no?

I didn't check your orbital velocities, but they look about right.

We don't need to get all fancy with orbital mechanics to answer what you're asking...

You ask about killing all surface velocity and dropping straight down. The equation for dropping straight down is: Velocity = square root (2 * distance fallen * g)

So falling straight down from 36km: V = sqrt (2 * 36000 * 1.63) = ~342 m/s.

Falling straight down from 120km: V = sqrt (2 * 120000 * 1.63) = ~625 m/s

So yes, it will take less dV to "stop" your orbital velocity at 120km (specifically 523 - 451 = 72 m/s less). However, it will take more dV to "instantly stop" your descent from 120km (specifically 625 - 342 = 283 m/s). The ship has a lot more opportunity to speed up during the fall, and the earlier you start fighting gravity with thrust, the more dV you will spend against gravity.

So it's "cheaper" overall to get your orbit as low as possible using all the usual cost saving tricks and transfers. Then once you are low, burn to kill your horizontal speed and spend as little time falling as necessary, allowing enough altitude for your skills and craft to readjust orientation and land safely.

Edited March 26, 2014 by Claw
typo

[ScottyDoesKnow]

Yes, I suppose when you phrase it like that it does make sense to apply it to landings, but of course the effect would be minimal almost negligible on a very short, powerful burn.

Exactly, and that's pretty much what we're discussing, the efficiency of a suicide burn vs slower deceleration.

[FREEFALL1984]

Exactly, and that's pretty much what we're discussing, the efficiency of a suicide burn vs slower deceleration.

Yeah, at first I thought we where comparing an orbital decent to a vertical descent, both with an applied suicide burn.

[Rhomphaia]

I didn't check your orbital velocities, but they look about right.

We don't need to get all fancy with orbital mechanics to answer what you're asking...

You ask about killing all surface velocity and dropping straight down. The equation for dropping straight down is: Velocity = square root (2 * distance fallen * g)

So falling straight down from 36km: V = sqrt (2 * 36000 * 1.63) = ~342 m/s.

Falling straight down from 120km: V = sqrt (2 * 120000 * 1.63) = ~625 m/s

So yes, it will take less dV to "stop" your orbital velocity at 120km (specifically 523 - 451 = 72 m/s less). However, it will take more dV to "instantly stop" your descent from 120km (specifically 625 - 342 = 283 m/s). The ship has a lot more opportunity to speed up during the fall, and the earlier you start fighting gravity with thrust, the more dV you will spend against gravity.

So it's "cheaper" overall to get your orbit as low as possible using all the usual cost saving tricks and transfers. Then once you are low, burn to kill your horizontal speed and spend as little time falling as necessary, allowing enough altitude for your skills and craft to readjust orientation and land safely.

Kind of do need some orbital mechanics. Gravity is not constant, but rather decreases with altitude. at the 36km altitude you would start accelerating at 1.17ms^-2 from 120km your initial acceleration would only be 0.64ms^-2

The values I calculated here are quite a bit lower especially for the 120km drop, but the method seems to correspond to experimental results after I accounted for the 4km altitude of the crash site.

Edited March 26, 2014 by Rhomphaia

[euphrates95]

To find the speed which you'll hit the ground at, use the formula

X_f=X_i+v(t)+1/2a(t²)

Assuming of course, that the height of your orbit at t is also your altitude, and that the Mun's gravity doesn't diminish with distance.

Solve for t to find the time until impact, then place that t into this formula:

V_f=V_i+a(t)

Also, keep in mind that since your velocity is tangential to your orbit the initial vertical velocity should (theoretically) be 0.

EDIT: Just saw the previous posts, I was always under the impression that in KSP the gravitation was modeled constantly(ish). Obviously not at extremely high altitudes, but this would situation would be close to the surface.

Edited March 26, 2014 by euphrates95

[maccollo]

To find the speed which you'll hit the ground at, use the formula

X_f=X_i+v(t)+1/2a(t²)

That doesn't work, since we're talking about ranges where the strength of gravity varies quite a bit.

One way to do it accurately is to use the delta in gravitational potential energy. Gravitational potential (U) is the change in potential energy per unit of mass when falling to the specified altitude from →∞.

U = -GMm/r

G is the gravitational constant, M is the mass of the planet, m is the mass of the space craft and r is the altitude. Since the mass of our space craft is insignificant we can assume that it's 1 kg

For Mun we can simplify it to 65138398000/r.

This means our final velocity upon impact will be

sqrt 2(ÃŽâ€U+v^2/2) = final velocity

So, falling from 20 000 meters at Mun ÃŽâ€U will be

-65138398000/220000-(-65138398000/200000) ≈ 29608

This means from a standstill we will hit the ground at 243 m/s

Let's say our initial velocity is 100 m/s (the vector does not matter). That means we add that as the initial velocity

29608+100^2/2=34608

thus our impact velocity is

sqrt 2x34608 ≈ 263 m/s.

Edited March 26, 2014 by maccollo
fixed calculation typo

[Xavven]

I love all the math in this post and it's very educational. Someone mentioned the reverse gravity turn as a way to think about the best way to land, and I found that the easiest way to intuit the answer, math notwithstanding. Assuming most of you know why we do a gravity turn on ascent, and that in KSP we do it at 10,000m, usually, because that's approximately the point at which air drag is smaller than gravity drag. On an airless Mun, the optimal ascent involves an immediate gravity turn as soon as possible (while avoiding tall mountains if present) because there is no air drag. The most fuel optimal landing is no different.

Put another way, an orbit is sometimes described by physicists as falling around the horizon. The longer you are allowing gravity to curve you around the horizon instead of directly towards the ground, the less delta-v you have to use to counter your velocity relative to the ground.

[Claw]

Kind of do need some orbital mechanics. Gravity is not constant, but rather decreases with altitude. at the 36km altitude you would start accelerating at 1.17ms^-2 from 120km your initial acceleration would only be 0.64ms^-2

The values I calculated here are quite a bit lower especially for the 120km drop, but the method seems to correspond to experimental results after I accounted for the 4km altitude of the crash site.

Yep, you got me on that one. I know gravity drops off, but I poorly assumed that it wasn't dropping off fast enough that it would matter much in the resulting difference (284 m/s). But it clearly did, because that's about 100 m/s too high. You are indeed correct, thanks.

Incidentally, the orbital mechanics I was referring to was people breaking out Oberth. Also, there are some discussions about converting potential to kinetic, and I think I recall seeing something about the energy always being the same, no matter how you get there. That all over complicates the question, but I tried to oversimplify...

[cantab]

The only reason I can see that a lower orbit might not be better is if it means you need to make a later deorbit burn in order to not hit terrain. From a given circular orbit, the ideal approach is a deorbit burn 180° from the landing site, then a suicide burn to land. Deorbiting closer to the landing site uses more dV overall. But the lower your starting orbit, the shallower your descent, and for any given landing site there's a limit on how shallow it can be.

Whether this effect is ever enough to offset the reduction in landing dV due to the lower orbit I don't know.

[magnemoe]

One downside with an too low orbit in real cases as then you want to land at an specific location is that the orbital plane shift will be more expensive. This can easy eat the dV you earned.

[Temstar]

You're suppose to line up your approach to your target LZ in orbit before you start your descent burn. When you're in shallow orbit near your landing site the only choice should be "land now" vs "land long further down range". Neil Armstrong ended up landing further down range in order to find a clear landing zone, he did it by thrusting upwards to cancel his downward motion. He didn't cancel out all horizontal motion and fly cross range to find a LZ.

[daniu]

Wow this got quite the traffic over night, thanks to everybody taking part in the discussion, everything said here is very enlightening

For a vertical landing, an instant velocity change at 0 altitude will require the dV you can get from a simple equation (e.g. I drop a ball from 10m, how fast will it be going when it hits the ground). Every extra second you take to land costs you an extra 9.8 m/s dV (or whatever the acceleration due to gravity is in your case).

That's what I initially assumed, but I talked to my coworker about it and he said it didn't matter. Note to self: Sheldon was right, physicists are not the same as rocket scientists

When you fall a certain distance you loose a certain amount of potential energy based on the difference in altitude. This potential energy goes into your kinetic energy, but kinetic energy is exponential, so that addition of energy results in different velocities depending on your initial velocity.

Lets say you fall 20 000 meters and this results in a kinetic energy gain of 23200 j/kg.

If your initial velocity was 600 m/s, that means you initial kinetic energy was 180000 j/kg, and your velocity after falling the 20 000 meters will be 637.5 m/s.

Wait wait wait where did you lose the mass and energy in the step from j/kg to m/s?

(I kind of have a lot of formulas in my head, but sometimes miss the connections between them).

To calculate a stop and drop landing I would stick with the Vis Viva equation.

if v = 0 then 2/r = 1/a

so you can get the SMA of the "orbit" after you stop. then calculate the velocity at surface altitude from that.

I get 315.22m/s from a 36km drop and 494.24m/s from the 120km drop. assuming a landing site at 0 elevation (200km radius) and an infinite TWR so no gravity drag during burns.

That's pretty much exactly what I'm looking for; hadn't come across the vis viva equation before, thanks.

So that's a difference of 180m/s for the vertical component versus the 70m/s I got for the horizontal one.

Accounting for it being better to not burn the component individually (as in, first horizontal then vertical), that should give me two vectors (vertical dV | horizontal dV) to impact:

36km: (315.22m/s | 523.15m/s); length = sqr(315,22^2 + 523.15^2) ~= 610.78m/s

120km: (494.24m/s | 451.17m/s); length = sqr(494.24^2 + 451.17^2) ~= 669.20m/s

With a total difference of ~60m/s (under ideal conditions and instantaneous burn to kill them).

Edited March 27, 2014 by daniu

[Rhomphaia]

That's pretty much exactly what I'm looking for; hadn't come across the vis viva equation before, thanks.

So that's a difference of 180m/s for the vertical component versus the 70m/s I got for the horizontal one.

Accounting for it being better to not burn the component individually (as in, first horizontal then vertical), that should give me two vectors (vertical dV | horizontal dV) to impact:

36km: (315.22m/s | 523.15m/s); length = sqr(315,22^2 + 523.15^2) ~= 610.78m/s

120km: (494.24m/s | 451.17m/s); length = sqr(494.24^2 + 451.17^2) ~= 669.20m/s

With a total difference of ~60m/s (under ideal conditions and instantaneous burn to kill them).

You can't really land from orbit in a single burn though. Most efficient method to land at zero altitude would be to drop periapsis to 0, then burn off all velocity at periapsis. giving me ideal values of 615.95m/s from a 36km orbit and 688.59 from a 120km orbit. (not accounting for Muns rotation, and of course most of Mun is higher than 0 altitude.)

[daniu]

You can't really land from orbit in a single burn though. Most efficient method to land at zero altitude would be to drop periapsis to 0, then burn off all velocity at periapsis. giving me ideal values of 615.95m/s from a 36km orbit and 688.59 from a 120km orbit. (not accounting for Muns rotation, and of course most of Mun is higher than 0 altitude.)

So about 70m/s difference... still a bit less than I expected actually.

Anyway, set this to Answered now; got a much better idea of the math involved and a feeling about the ballpark I need to expect.

[Lukaszenko]

While landing from the lowest possible orbit is theoretically best, practically that is not always the case. I landed on Tylo the other day, and found that due to insufficient TWR, I couldn't cancel all my velocity while burning retrograde in time before "reaching" the surface. I had to therefore retry, but with some thrust wasted for fighting gravity. I suspect that starting from a slightly higher orbit would be more efficient in this case.

[maccollo]

Wait wait wait where did you lose the mass and energy in the step from j/kg to m/s?

(I kind of have a lot of formulas in my head, but sometimes miss the connections between them).

I did skip over quite a few steps. Going from j/kg to m/s is just reverting the kinetic energy formula, so sqrt(2xKe) = velocity.

The reason for j/kg, or assuming that the mass of your space craft is 1 is that we're not interested in the total energy, but velocity, so we can simplify it by not having the mass in the equations.

The point was that if your initial velocity is higher you will gain less velocity as you fall towards the planet or moon. This means you want to do all your big burns as close to the planet or moon as you can.

[Xavven]

While landing from the lowest possible orbit is theoretically best, practically that is not always the case. I landed on Tylo the other day, and found that due to insufficient TWR, I couldn't cancel all my velocity while burning retrograde in time before "reaching" the surface. I had to therefore retry, but with some thrust wasted for fighting gravity. I suspect that starting from a slightly higher orbit would be more efficient in this case.

I had exactly the same experience. My first Tylo landing attempt was from a 20km or 30km orbit, and I ended up hitting a mountain at high horizontal velocity on my way down due to insufficient TWR.