Landing from Low vs High Orbit

[daniu]

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?

Edited March 27, 2014 by daniu

[MalfunctionM1Ke]

I cant help you with the match but I want to Support your Notion that a low parking Orbit would be the best way to go for a landing.

IMO when you do your deorbit burn, the time that you are travelling from your apoapsis to the ground Level will accelerate you by the ground/gravity of the given Body.

This effect is not to be mentioned when it comes to landing on gilly or minmus but a significant threat when you are coming down on Tylo.

So your best bet is to set your peri-/apoapsis as low as the tallest mountain (or even lower when you are bold enough) and wait for your landing site to be 90° ahead in your Orbit, place your deorbit burn and try to bleed of your excess Velocity by burning retrograde PLUS into a Radial+ direction "skywards" at the same time to rule out the gravity that tries to kill you.

Your answer is close, but unfortunatly I cant help you with the math

[Streetwind]

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.

(...)

Or am I looking at it entirely wrong?

You're looking at it wrong. I know it's counterintuitive - I asked myself the same question - but landing from a low orbit always costs less dV than landing from a high orbit. And you don't need to do any math to prove it.

Just consider the following statements:

- You need more dV to ascent from the ground into a high orbit than you need to ascent into a low one. Therefore, you accelerated more, and the high orbit is a higher energy state that requires you to decelerate more to return from.

- To get from a high orbit into a low orbit, you have to expend dV. If landing from low orbit costs X dV, then landing from a higher orbit can be expressed as first expending Y dV to go into the lower orbit, and then expending X dV to land. Therefore your total dV to land from high orbit is (X + Y) > X.

Edited March 26, 2014 by Streetwind

[daniu]

You're looking at it wrong. I know it's counterintuitive - I asked myself the same question - but landing from a low orbit always costs less dV than landing from a high orbit. And you don't need to do any math to prove it.

Oh, I don't find it counterintuitive... by now I'm really mostly looking for the math.

Not sure how I'm looking at it wrong.

- To get from a high orbit into a low orbit, you have to expend dV.

Yeah, but because you burn retrograde; but that only takes away speed that you had to accelerate to in the first place.

But since a higher altitude has less orbital velocity, I should have to burn retrograde less to cancel it out. This part does feel a bit counterintuitive. It's probably wrong because I have to think surface velocity.

[lockpicker]

In my opinion, both aproaches should require about the same dv.. Suppose you are on a highly elliptical orbit.. Your speed would be much bigger on your periapsis rather than your apoapsis.. However since in space there is no friction, your energy (that is kinetic and potential energy summed) remains the same at all times.. So when you are high from the planets surface, you might need to slow yourself less but you have less kinetic energy, but your potential energy is increased and vice versa when you are low.. So in general you require the same amount of energy to bring yourself to a complete stop at the surface of the planet, which should translate to the same amount of dv..

[Streetwind]

But since a higher altitude has less orbital velocity, I should have to burn retrograde less to cancel it out. This part does feel a bit counterintuitive. It's probably wrong because I have to think surface velocity.

You burn less to cancel out your horizontal speed in a high orbit, yes. And then you have a long way down accelerated by gravity, which generates more vertical speed for you to cancel out than you saved in the horizontal.

To figure out what that vertical speed is, you need to sum up the ever-increasing acceleration as you approach the surface... in other words, you want an integral over gravity. The resulting function describes your increasing speed over time. With that, you can figure out how long it takes you to fall from a given height, and what your final velocity will be as you impact the surface.

This value, finally, is what you must add to the cost of zeroing out your horizontal velocity in order to get the total dV cost... a value which of course assumes you can perform a perfect, near-instant suicide burn. Because in addition to the cost comes the fact that if your burn to decelerate vertical speed is not instant, you lose extra dV to gravity drag in addition that what you would normally spend. This is because your engine expends a part of its thrust just to maintain the status quo ("hovering"), which does in no way contribute to decelerate you. And if you come in vertically from a very high orbit, you need a very long burn to cancel out your vertical velocity, which will be very inefficient.

The burn to zero out horizontal velocity does not suffer from this, because there is no gravity trying to pull you forward. 100% of the engine thrust goes towards decelerating you. Therefore you want to burn as much horizontally as you can, and as little vertically as you can. (This is the reason we do a gravity turn - and yes, you can do a "reverse gravity turn" to land. It is in fact the best possible maneuver, but so hard to pull off that in practice nobody bothers. And on airless bodies it basically translates into "burn horizontally one meter above the surface, then flip and instantly touch down" anyway.)

Okay, enough edits for now, I promise

Edited March 26, 2014 by Streetwind

[FREEFALL1984]

I would of thought that (in a perfect scenario) without any unnecessary burns or direction changes the amount of DV used landing on a body with no atmosphere would always be the same regardless of approach since you always need to kill the same amount of total kinetic energy, and therefor you should always chose the most simple approach in order to eliminate any possibility of errors... in fact the only varying factors are orbital direction and height of landing sight.

[Streetwind]

You don't need to think in terms of kinetic energy, you need to think in terms of specific orbital energy. And that's a whole lot more than just kinetic (and no, I don't claim to understand that myself.)

Also, the aforementioned gravity drag applies even on airless bodies. There's no such thing as a drag-less landing/takeoff.

Edited March 26, 2014 by Streetwind

[daniu]

In my opinion, both aproaches should require about the same dv..

I would of thought that (in a perfect scenario) without any unnecessary burns or direction changes the amount of DV used landing on a body with no atmosphere would always be the same

Hmm, interesting...

You burn less to cancel out your horizontal speed in a high orbit, yes. And then you have a long way down accelerated by gravity, which generates more vertical speed for you to cancel out than you saved in the horizontal.

Yes, that is exactly what I'm saying

To figure out what that vertical speed is, you need to sum up the ever-increasing acceleration as you approach the surface... in other words, you want an integral over gravity.

That sounds as if it could be right. You'd need to integrate gravity over distance because that's what is known I supposed... but that math is so long ago for me

a value which of course assumes you can perform a perfect, near-instant suicide burn. Because in addition to the cost comes the fact that if your burn to decelerate vertical speed is not instant, you lose extra dV to gravity in addition that what you would normally spend. This is because your engine expends a part of its thrust just to maintain the status quo ("hovering"), which does in no way contribute to decelerate you. And if you come in vertically from a very high orbit, you need a very long burn to cancel out your vertical velocity, which will be very inefficient.

I don't think that is correct. The acceleration due to gravity will add up to a velocity (in m/s) you'll have to cancel; whether you burn it at once or over time shouldn't make any difference.

In other words, whether you wait until you have gained 500 m/s and then thrust it away should burn the same amount of fuel as if you'd been throttling the whole way decreasing it gradually (if you don't have an atmosphere which would help slowing things down with drag).

EDIT: hmmm just read the Gravity drag article, seems like at least the Vector part still applies without atmo... but I'm considering a vertical landing so not in this case. Then again, burning all the time removes fuel so loses mass, which may actually end up saving fuel.

Edited March 26, 2014 by daniu

[ScottyDoesKnow]

I don't think that is correct. The acceleration due to gravity will add up to a velocity (in m/s) you'll have to cancel; whether you burn it at once or over time shouldn't make any difference.

In other words, whether you wait until you have gained 500 m/s and then thrust it away should burn the same amount of fuel as if you'd been throttling the whole way decreasing it gradually (if you don't have an atmosphere which would help slowing things down with drag).

EDIT: hmmm just read the Gravity drag article, seems like at least the Vector part still applies without atmo... but I'm considering a vertical landing so not in this case. Then again, burning all the time removes fuel so loses mass, which may actually end up saving fuel.

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).

[Tidus Klein]

I like to park my self in between 6,000 and 8,000 any lower and your suicidal.....well more suicidal then I am

[Reynaard]

In my opinion, both aproaches should require about the same dv.. Suppose you are on a highly elliptical orbit.. Your speed would be much bigger on your periapsis rather than your apoapsis.. However since in space there is no friction, your energy (that is kinetic and potential energy summed) remains the same at all times.. So when you are high from the planets surface, you might need to slow yourself less but you have less kinetic energy, but your potential energy is increased and vice versa when you are low.. So in general you require the same amount of energy to bring yourself to a complete stop at the surface of the planet, which should translate to the same amount of dv..

You're forgetting the Oberth effect. IMO it's best to land from an elliptical orbit, because then you're going the fastes when you do your deorbit burn at periapsis, which means it's doing the most work (work equals force times speed).

[Streetwind]

I don't think that is correct. The acceleration due to gravity will add up to a velocity (in m/s) you'll have to cancel; whether you burn it at once or over time shouldn't make any difference.

In other words, whether you wait until you have gained 500 m/s and then thrust it away should burn the same amount of fuel as if you'd been throttling the whole way decreasing it gradually

You can easily verify that gravity drag exists and (very drastically applies) by doing the following:

1.) In a 50km orbit over the Mun, zero out horizontal velocity and then decouple your fully fueled lander for the vertical descent

2.) Fall straight down and make a short, hard burn at the last possible moment in order to land

3.) Note how much fuel you have spent for the descent

4.) Reload your quicksave and repeat step 1.)

5.) Activate your engine right away and keep it on low power over the entire descent, maintaining a constant 10 m/s vertical velocity (safe for landing) via manual throttling

8.) After landing, note how much fuel you have spent for the descent

If what you claim is true, both descents will consume the same amount of fuel, because you're dropping from the same height and it doesn't matter how long your engine is running.

However, what you will find in reality is that your first attempt will get you safely landed, while your second attempt will likely run out of fuel before you even get near the surface. The reason is that because you're floating down slowly, you give gravity far, far more time to accelerate you. Every second, the Mun applies acceleration. The longer you spend descending (regardless of height), the more you get accelerated downwards, and thus the more fuel you need to spend on cancelling it out. That is gravity drag.

To minimize gravity drag, you want to spend the shortest possible time being accelerated by gravity. And this is achieved by spending the least possible time decelerating, because any deceleration causes your total descent to take longer. An instant, perfect suicide burn on the last millimeter would give the theoretically optimal result.

Edited March 26, 2014 by Streetwind

[FREEFALL1984]

Also, the aforementioned gravity drag applies even on airless bodies. There's no such thing as a drag-less landing/takeoff.

Gravity drag is defined as the dV required to hold an object in a gravitational field, it is applied to climbing rockets and added to the total dV required to reach orbital velocity on top of atmospheric drag.

Based on that then gravity drag is not relevant to a falling body since no energy is required to hold the item against gravity.

[Streetwind]

I invite you, too, to perform the above experiment and then report back on your findings

[FREEFALL1984]

I invite you, too, to perform the above experiment and then report back on your findings

Well in the event of that approach then you're right, you will burn FAR more fuel, but I thought the OP was relating moving the periapsis to the surface and performing a last minute horizontal suicide burn to kill horizontal velocity which would use about the same fuel as a vertical burn but with less margin of error, you're talking about having a long constant burn against gravity and indeed holding the craft against gravity which would be insanely inefficient

[FREEFALL1984]

You can easily verify that gravity drag exists and (very drastically applies) by doing the following

By taking off in a lander and hovering stationary with a zero vertical speed and a zero orbital speed.

[Streetwind]

but I thought the OP was relating moving the periapsis to the surface and performing a last minute horizontal suicide burn to kill horizontal velocity

No, that's not what daniu was saying at all. He said: "The acceleration due to gravity will add up to a velocity (in m/s) you'll have to cancel; whether you burn it at once or over time shouldn't make any difference." And that is incorrect because anytime you do not have an instant burn, you're prolonging the duration of gravity acting on your vessel. Also, we've been talking about "killing horizontal speed", i.e. rendering it 0, not lowering your periapsis a bit.

If you do it as you describe, lowering the periapsis to just graze the surface, then you're approximating a reverse gravity turn maneuver in which you spend just about all your dV burning horizontally and just the absolute minimum required in the vertical. It's an incredibly efficient maneuver because you're removing almost all gravity drag from the equation. It also tends to plaster you in a thin film across a cliff face during the last few dozen kilometers, unless your celestial body is a perfect sphere With enough TWR, you could brake and land early with an inefficient vertical descent part, or otherwise a deflection burn will let you clear the cliff but screw up your projected landing site, requiring further corrections that cost more dV. That's why it's difficult to pull off and most people choose nominally less efficient landing approaches instead.

Scott Manley actually has a hilarious video of this where he skims close across the lunar surface for minutes, frantically flailing to decelerate and dodge cliffs with tiny engines that have nearly no thrust.

Edited March 26, 2014 by Streetwind

[Kasuha]

I did not really read all the conversation in your thread but what you forgot in your calculation is the actual landing.

Even if we disregard "fighting gravity", we can calculate something. And actually, on Mun these results are very close to reality.

So, assuming you start with circular parking orbit at altitude A, the "perfectly optimal" landing (assuming surface is made of pillows) is Hohmann transfer to circular orbit at altitude 0, then killing all your horizontal velocity.

And that number is the higher the higher your parking orbit is. Because killing the horizontal velocity is the same regardless where you start (you just kill the orbital velocity at zero altitude) but the higher the difference, the more you spend on Hohmann transfer.

The "fuel-optimal" Apollo-style mission approach does not work with circular orbits at all. You need to come on a trajectory which is no more than safe distance from your target's surface. You brake at periapsis so your trajectory just closes within the SOI. No circularizing, your ship stays on highly elliptic trajectory. Then (still at periapsis) you decouple your lander and land from there.

When returning you launch into circular orbit which touches the ship's trajectory at its periapsis and then burn at that periapsis to have rendezvous there the next orbit. The lander must have enough dv to be able to match the ship's elliptic parking orbit. But it's still less fuel spent than bringing the whole ship (with lander) down and then up again.

Edited March 26, 2014 by Kasuha

[Red Iron Crown]

It seems to me that the Oberth effect should apply to this problem. When landing, the craft is going from a higher specific orbital energy to a much lower one, so as much of the burn as possible should be made at low altitude/high speed where each unit of dV spent results in the largest energy change.

[FREEFALL1984]

purely mathematically the Hohmann transfer to 0 alt then killing all velocity does make for the most efficient landing, (such as a horizontal landing on minmus ice and using wheel breaks and RCS downforce thrusters to stop) but then as everyone has mentioned, that's a very difficult thing to achieve. So often the best course of action is the easiest, even if it is a little more costly on fuel.

In fact I'm gonna a horizontal landing on minmus as soon as I can, because it sounds really fun

[maccollo]

Gravity drag is defined as the dV required to hold an object in a gravitational field, it is applied to climbing rockets and added to the total dV required to reach orbital velocity on top of atmospheric drag.

Based on that then gravity drag is not relevant to a falling body since no energy is required to hold the item against gravity.

Newtonian physics work exactly the same in reverse. Gravity drag applies to descent just as it does to ascent.

I don't think that is correct. The acceleration due to gravity will add up to a velocity (in m/s) you'll have to cancel; whether you burn it at once or over time shouldn't make any difference.

That's incorrect.

It works like this.

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.

However if your starting velocity was 0 m/s then the velocity you will end up with after falling the same distance is 215 m/s.

Another way to to look at it is just to reason that gravity will accelerate you at a certain rate each second. In a vertical descent, the longer your descent takes the more time gravity has to accumulate velocity.

[FREEFALL1984]

Newtonian physics work exactly the same in reverse. Gravity drag applies to descent just as it does to ascent.

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.

[juanml82]

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?

[FREEFALL1984]

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?

I use that technique on all small orbital bodies with low gravity and no atmo, if not for a more efficient landing then for a safe and more easily targeted one