Landing on airless bodies

[Mighty1]

Jeb told me once that if you are in a high orbit you are slower, if you are lower then you are faster. So if I bring my surface velocity down to 0 m/s and fall twards an airless body I will pick up speed. My dilema is are the delta-v requirements for landing the same or diferent for:

A) Deorbit from 100 km or

Deorbit from 30 km

In case of A your orbit is slower but you have a larger fall (pick up more speed) but in case of B your orbit is faster but you have a shorter fall eg. less speed to pick up.

Thank you for sharing your thoughts.

CONCLUSION: I did some tests and found out that lower is better. Most of you guys were right. Thank you for your participation.

Edited April 17, 2014 by Mighty1
Changed "deorbit" to "landing", Thread ANSWERD

[Kobymaru]

What do you actually want to do? To you want to lithobrake a derelict craft against a body, or do you want to safely land on it?

If you want to crash it: brake at the highest point. You will need the least dV, and the speed you pick up will be vertical and not horizontal (you do not have to work against it). If you slow down to 0 m/s orbital speed at apoapsis, your craft will just drop onto the body like a stone.

If you want to land, you should use a different approach: get as low as you possibly can (for Mun ~10 km, for Minmus ~7 km) in a *circular* orbit. There you will need to kill the least dV before landing.

Otherwise, I don't understand your question.

[DeMatt]

The delta-V for DEORBITING (that is, stopping your orbit) will be less at the 100km orbit than it will at the 30km orbit.

The delta-V for LANDING will be greater at the 100km orbit than it will be at the 30km orbit.

You have greater kinetic energy in the lower orbit, but lesser gravitational potential energy. If this confuses you, it might be worth reading over Specific Orbital Energy. Think of it this way: how much delta-V would you need to get from the surface to the orbit, that is, do the maneuver in reverse?

[BudgetHedgehog]

My instinct tells me that deorbiting from 30km is more efficient, but I don't know why..

[Kobymaru]

My instinct tells me that deorbiting from 30km is more efficient, but I don't know why..

I think that's because 30 km is lower than 100 km. But your instincts are wrong. To deorbit, you need to kill all forward velocity and that is much lower in a higher orbit.

[Mighty1]

That is my problem. In a high orbit you have to bring your speed down from 200 m/s but in a low orbit it could be 600 m/s. As DeMatt stated before the question here is potential gravitational energy that will speed you up on your way down. I got to test this when i get back home.

Edited April 16, 2014 by Mighty1

[FenrirWolf]

I think that's because 30 km is lower than 100 km. But your instincts are wrong. To deorbit, you need to kill all forward velocity and that is much lower in a higher orbit.

You don't necessarily have to kill all your velocity to deorbit. Once you cease to have a periapsis you just let the ground take care of the rest. Or your engines if you're planning on surviving the encounter.

[BudgetHedgehog]

You don't necessarily have to kill all your velocity to deorbit. Once you cease to have a periapsis you just let the ground take care of the rest. Or your engines if you're planning on surviving the encounter.

Yes, that's it. I think it takes less dV to bring your periapsis into the ground from 30km than from 100km. Then, it's a just a case of a suicide burn when you get close.

[FenrirWolf]

Actually it ought to take less dV to eliminate your periapsis from a high orbit as was mentioned. But then you'll be coming in on a steep incline picking up a lot of speed as you keep falling through the gravitational gradient, and will need a rather large suicide burn to make up for that.

Now a lower orbit will require you to spend more propellant on the initial burn to reduce periapsis to 0, but the benefit is that you don't have to fall as far before the landing.

[MarkyJ279]

I only have A-level physics qualification so if someone of Manley levels of knowledge could check my working I'd appreciate it but as best I understand the maths. I'm using real world numbers since I don't have KSP open at the moment. I'm assuming a normal scale earth with no atmosphere. I used an online physics calculator (http://www.calctool.org/CALC/phys/astronomy/earth_orbit) to get the figures for orbital velocity.

Impact: if you just want to hit the surface vertically that's easy. All you need to do is counter your horizontal velocity which is lower in a higher orbit. You'll hit the surface faster than if you'd droped from a lower orbit but you'd still hit vertically and if all you want is an impact then higher is cheaper (in delta-V). No complicated maths needed, it's just whichever orbital velocity is lower.

For a 1000km orbit the speed it 7.35km/s.

For a 10000km orbit it's 4.93km/s.

The 10000km (higher) orbit needs less delta-V

Landing: This is slightly more difficult. For a suicide burn landing we need to counter orbital velocity and then counter velocity built up from falling under gravity.

I'm going to use the SUVAT equations. Specifically: V^2 = U^2+2as Where V is final vertical velocity, U is initial vertical velocity, a is acceleration due to gravity (9.81m/s/s) and s is distance (orbital height). From a stable orbit we can assume U to be 0m/s and so can be safely removed and the formula rearranged to give: 2as = V^2

For 1000km, we can calculate: 2*9.81*1000000m = V^2 = 19620000m/s. Rooting that and converting to km/s gives 4.43km/s. Add that to the horizontal orbital velocity above and total delta-V to land from 1000km is 11.78km/s.

However for 10000km, we get: 2*9.81*10000000m = V^2 = 196200000m/s. Rooting that and converting to km/s gives 14.0km/s. Add that to the horizontal orbital velocity above and total delta-V to land from 10000km is 18.93km/s.

If anyone want to call out my maths as wrong I'll defer to them but at the moment according to my working it takes less delta-V to impact from the higher orbit and less delta-V to land from the the lower one. Exact figures may vary depnding on body and height but the trend should stay the same.

[Claw]

Marky, your methodology is on the right track but you have to account for the fact that the gravitational value (9.81) is not constant.

---

This question comes up from time to time.

Deorbiting from a lower orbit is always cheaper dV wise than deorbiting from a higher orbit. That's all assuming you execute everything perfectly.

DeMatt said it right. Killing orbital velocity will take less dV at the higher altitude. However, the dV to stop the descent will be more. When you look at the differences, the extra dV needed to stop the higher descent is more than the extra dV to kill the higher orbital velocity.

That being said, killing your orbital velocity and dropping straight down is not the most efficient deorbit method. But from your two options, the lower orbit takes less dV.

None of this discussion takes TWR into account...

[MarkyJ279]

Marky, your methodology is on the right track but you have to account for the fact that the gravitational value (9.81) is not constant.

I knew I would overlook something. Still nice to get confirmation though

[cantab]

The more efficient way to land is a small deorbit burn to bring periapsis close to or below the surface, then a large burn as low as you dare to kill horizontal velocity (a "suicide burn").

[wilt57]

Think of it this way: how much delta-V would you need to get from the surface to the orbit, that is, do the maneuver in reverse?

This. It always takes more delta-v to get to a higher orbit, so it'll always take more delta-v to come back down from that orbit. Also, due to Oberth's you will get more delta-v per unit of reaction mass burned in the lower orbit. In many cases this won't be very much, maybe even unaccountable, but the math is still there.

[J.Random]

Also, due to Oberth's you will get more delta-v per unit of reaction mass burned in the lower orbit.

You get the same dV from the same amount of fuel. You're just using it more effectively.

[Redshift OTF]

I'm pretty sure Mr Manley said that deorbiting from a lower orbit was the most efficient way of doing this. Basically the closer you are to the ground when you kill your horizontal speed the less time gravity has to counteract your thrust when you are killing your vertical speed. The only disadvantage of this is you will need a high TWR to kill the horizontal speed before you end in a mountain or smeared over a plain.

[ScottyDoesKnow]

I'm pretty sure Mr Manley said that deorbiting from a lower orbit was the most efficient way of doing this. Basically the closer you are to the ground when you kill your horizontal speed the less time gravity has to counteract your thrust when you are killing your vertical speed. The only disadvantage of this is you will need a high TWR to kill the horizontal speed before you end in a mountain or smeared over a plain.

I would think that theoretically, it would be more efficient (starting at a high orbit) to drop periapsis to nothing and suicide burn than to transfer to a lower orbit and then do the same. You get the most benefit from the Oberth effect that way, it's just more dangerous.

[Mighty1]

I tested it. DeMatt was right. The test was done on the Mun with a lander with 2352 delta-v. First I killed surface horizontal velocity to 0 and went straight down. Landing from 20 km orbit left me with 1578 delta-v left and I landed on 6500 ASL. Doing the same from 100 km orbit left me with 1391 delta-v while landing at 3500 ASL. Now I have to try landing on the same spot.

[Skorpychan]

In my experience, it's best to go at periapsis, and burn off that excess horizontal velocity. That's easier to get rid of than vertical, and leaves you falling slower.

A long descent is stressful, and you'll have to time the suicide burn just right. A shorter one is easier, and your craft will be lighter after killing more horizontal velocity.

Oberth effect aside because I don't really understand it, it's the same dV wherever. It's just a tradeoff between horizontal and vertical.

Of course, with an atmosphere, that goes out the window. You're coming in either low and fast from an apoapsis burn to aerobrake and chute, or dropping vertically to land precisely.

[nobbers12345]

I'm pretty sure burning a small amount to land somewhere, then suicide burning lower is more efficient, because you wont have to deal with gravity accelerating you as much.

[MockKnizzle]

A landing is an ascent in reverse. An ascent to a higher orbit costs more dV than an ascent to a lower orbit. Thus, a descent from a higher orbit costs more dV than a descent from a low orbit.

The critical parameter to look at is orbital energy, not orbital speed. While a 250km orbit has a lower orbital speed than a 75km orbit, it has a higher orbital energy. In order to land, you need to get rid of that extra energy.

Edited April 17, 2014 by MockKnizzle

[Nemecle]

Being in a higher orbit doesn't mean you are going slower, in fact you are going faster (seems logical) but your angular speed is smaller : I think the lower you are, the easier it is, because of the reverse-ascent thingy that MockKnizzle is speaking about

[Dunbaratu]

There is one case where you need to de-orbit to land from a higher orbit and that's if your thrust capacity isn't very good. For example, if you currently have 500 m/s speed, and it's going to take you 2 minutes of constant burn to get rid of all of it, but in those two minutes you hit the surface because you started very low, then that's not very good. You need to be able to kill your horizontal speed in the time it takes to fall the vertical distance to the ground as you are no longer going fast enough to orbit but still not going slow enough to land.

If you assume infinite ability to thrust, then lowest is most efficient, yes. But if you have a weak wimpy engine as is often the case for a thing like a Mun lander, then starting your burn as low as possible might not be safe.

[Dunbaratu]

Being in a higher orbit doesn't mean you are going slower

False. Yes it does, and not just in angular terms, but in linear terms as well.

In a circular orbit, the speed is sqrt( GM / r ). The bigger the 'r', the smaller the result.

[Nemecle]

False. Yes it does, and not just in angular terms, but in linear terms as well.

In a circular orbit, the speed is sqrt( GM / r ). The bigger the 'r', the smaller the result.

I looked again, and it seems you are right : my bad

but I still think de-orbitate from a lower orbit is better, especially because you will have to cancel your vertical velocity which will come when you will "fall" from your altitude