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why falling on a planet from a moon is much cheaper than doing it from high orbit?


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I just thought about it, and i'm now very curious. I am decently expert in orbital mechanics, but not enough to solve this.

Say you are orbiting Mun. It's orbit is at 12000 km. from low mun orbit, you can use 280 m/s to lower your kerbin periapsis enough to aerobrake.

Yet, if you were just orbiting at 12000 km in a circular orbit around kerbin, it would take over 1000 m/s to lower your periapsis that much.

and when you are orbiting mun, you are also orbiting kerbin at that height. why being around Mun makes lowering your periapsis so cheap?

 

first possible explanation: oberth effect. being around mun gets you some oberth effect, so that going back is cheaper. however, that explanation does not hold up to close scrutiny, for several reasons. First, mun is smaller than kerbin. if it takes 850 m/s to get from kerbin to mun, the reverse trip being cheaper for oberth effect would require mun being bigger than kerbin. second, you get that cheap return even from high orbit, where oberth effect is small. third, you get it even in very small moons with too low gravity to have a significant oberth effect, like minmus and - most notably - gilly. Gilly is especially the last nail in the coffin for the oberth effect theory, and for any theory involving different orbital speed. Orbiting around gilly you move at 15 m/s. It can't have any significant effect on your orbital speed around Eve.

Though gilly also corroborates the idea that oberth effect may play a role in it, because returning to eve from gilly requires 500 m/s, which is about half as much as it takes from eve to gilly. A much smaller saving than for Mun. So, perhaps oberth does play some role there. But it certainly cannot explain everything.

Second possible explanation: an artifact of the game gravity. in reality, gravity from all bodies is affecting you all the time. in ksp, only one body is considered at any time. this is a decent approximation when you are in a low orbit, but it certainly generates artifacts when you are close to the boundary between spheres of influence. It's possible. I don't know enough of orbits to figure it out. Yet I can look up real deltaV budgets from wikipedia, and here I find for the earth-moon system: 4 km/s for LEO-Low Lunar Orbit, and only 900 m/s from LLO to LEO. So, this effect happens for real too.

At this point I have discarded the few explanations I could come up with, and I have no idea what's the actual reason. I would be very curious to find out, if anyone knows.

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Its pretty simple, the gravity is weaker thus less  deltav it takes to change your orbit. Also you're orbit distance isn't as long as Kerbin so your orbital velocity is less and that means the less you need to slow down.

But what do you mean by aerobrake because you need an atmosphere for that.

Edited by SpaceFace545
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If you drive up Mt. Everest in a car, it would take more gas than coasting back down.

Likewise, the dV cost to Mun from Kerbin has nothing to do with the dV cost of the return trip.

Mun orbits Kerbin at 1000 m/s. You orbit Mun at 600 m/s. The burn home is 300 m/s. The rest is Oberth and the fact that you don't need to completely stop dead to fall back to Kerbin.

And rounding errors because I didn't check my numbers because I'm on my phone right now.

Edited by Superfluous J
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Once you exit Muns SOI, you are in a highly eccentric orbit. Compare that to the "High altitude circular orbit" you mentioned earlier, and you have your answer. 

Post edit:

This depends of course on which direction you exit from the moon... Exiting in it's prograde direction would put you in a high circular orbit about Kerbin I believe, making your returning situation worse. 

Edited by James M
Afterthought?
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The reason it takes less delta-v to return to Kerbin from a Mun orbit than it does from an equivalent altitude from Kerbin orbit is you're in a way using a powered gravity assist. The Mun itself has an orbital velocity of 543 m/s with respect to Kerbin. And let's say your spacecraft is orbiting around Mun at 280 m/s.

That means that from the perspective of someone on Kerbin, your spacecraft is going 543 + 280 = 823 m/s when your spacecraft is orbiting in the same direction of the Mun. But on the other side of the orbit, your spacecraft is going in the opposite direction, so an observer on Kerbin would see your velocity as 543 - 280 = 263 m/s. It's only when your spacecraft is going neither with nor against Mun's orbit that your spacecraft looks like it's going 543 m/s around Kerbin (the equivalent of another spacecraft orbiting Kerbin at the same altitude as Mun).

So if you want to go slower with respect to Kerbin, would you rather be the spacecraft orbiting Kerbin at 543 m/s, or the one going effectively 263 m/s during a certain part of its orbit around Mun?

If you burn to escape Mun while going the opposite direction of its orbit, you end up with a lower orbital velocity with respect to Kerbin.

zB4yiZ1.png

This is the video that helped me understand gravity assists the best. Although it doesn't explain your specific question, the concepts are related: https://www.youtube.com/watch?v=utpzQfDdUJs and pay attention to the 5:00 - 6:00 part the most. The key thing to understand here is that you take into account BOTH the velocity of your spacecraft AND the Mun, all compared to the parent body.

 

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2 hours ago, king of nowhere said:

Yet, if you were just orbiting at 12000 km in a circular orbit around kerbin, it would take over 1000 m/s to lower your periapsis that much.

 

26 minutes ago, Xavven said:

zB4yiZ1.png

 

 

543 is much less than 1000. One of these things is wrong and I'm thinking it's the OP?

This is what I get for involving myself in threads when I can't play the game to check my facts.

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1 hour ago, Superfluous J said:

If you drive up Mt. Everest in a car, it would take more gas than coasting back down.

Please elaborate! This is quite confusing....how???!?! How does this work? If you coast down, aren’t you using a minimal/almost no amount of fuel??

Edited by Lewie
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Just now, Superfluous J said:

My sarcasm detector is in the shop so I can't tell if you mean this or not. I'm going to assume not.

Ok....I’m not crazy after all! Whoo, that’s a relief.

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6 minutes ago, Superfluous J said:

 

543 is much less than 1000. One of these things is wrong and I'm thinking it's the OP?

This is what I get for involving myself in threads when I can't play the game to check my facts.

OP is wrong.  The orbital velocity of a circular orbit at Mun's altitude (12 km) is 543 m/s.  As 543 m/s of delta-v would put you at a dead stop, meaning your periapsis would be literally the dead center of Kerbin, it couldn't take more than 543 m/s to lower your periapsis to aerabraking altitude. I'll fire up KSP right now to check the actual.

Ok, here we go. Screenshots say it all:

z9HuDGK.png

AOPXkSe.png

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If you’re orbiting the Mun while the Mun is orbiting Kerbin, there will be times (on the far side for a prograde orbit) when you’re moving in the same direction round the Mun as the Mun is round Kerbin making your relative velocity to Kerbin higher than the Mun’s; conversely on the other side (the near side for a prograde orbit) your relative velocity to Kerbin is lower than the Mun’s.

It’s this difference in speed, worth a few hundred m/s in Kerbin’s SOI and even more around Jool’s moons or for interplanetary transfers, that makes the biggest difference in delta-V between returning from orbit of a moon and just returning from the same orbit as a moon (but not inside said moon’s SOI)- the moon you’re orbiting is giving you some of the ‘extra’ delta-V for free, which you paid for when you arrived and braked into orbit in the first place. Think of it like a stored gravity assist, ready to be released when you make your transfer burn.

There’s a little bit of Oberth effect going on too, but for a Mun orbit the effect will be pretty small.

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8 hours ago, Xavven said:

OP is wrong.  The orbital velocity of a circular orbit at Mun's altitude (12 km) is 543 m/s.  As 543 m/s of delta-v would put you at a dead stop, meaning your periapsis would be literally the dead center of Kerbin, it couldn't take more than 543 m/s to lower your periapsis to aerabraking altitude. I'll fire up KSP right now to check the actual.

Ok, here we go. Screenshots say it all:

z9HuDGK.png

AOPXkSe.png

This is the only answer that makes sense.

I was basing my assumption on kerbostationary orbit, actually; I am pretty sure, the time i had a contract to reach it, it took more deltaV to raise periapsis than to raise apoapsis, because of oberth effect. So i assumed something similar would be in effect here, where going down would be just as expensive as going up, plus oberth effect.

but i must be remembering wrong, because i just made the attempt now, and it takes more deltaV to raise apoapsis than to circularize. And it takes less deltaV to lower orbit than to raise it, when i assumed it would be equal. this is what I get from going by memory.

 

The question then shifts, though; why is raising periapsis cheaper than raising apoapsis, despite oberth effect helping you in the second case? all this time, i've been told that prograde/retrograde manuevers are more effective at high speed. except, here i am seeing evidence to the contrary. there is still something missing from the puzzle

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16 hours ago, king of nowhere said:

I just thought about it, and i'm now very curious. I am decently expert in orbital mechanics, but not enough to solve this.

Say you are orbiting Mun. It's orbit is at 12000 km. from low mun orbit, you can use 280 m/s to lower your kerbin periapsis enough to aerobrake.

Yet, if you were just orbiting at 12000 km in a circular orbit around kerbin, it would take over 1000 m/s to lower your periapsis that much.

and when you are orbiting mun, you are also orbiting kerbin at that height. why being around Mun makes lowering your periapsis so cheap?

I can't really make sense of what you're saying.  aerobrake on the mun?

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This impinges on real world orbital mechanics but specifically involved perspective changes necessitated by the game's handling of SOIs. So Gameplay Questions seems to be the best sub for it. And so, moved. 

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1 hour ago, Corona688 said:

I can't really make sense of what you're saying.  aerobrake on the mun?

i said lower your kerbin periapsis enough to aerobrake. as in, leave mun's SoI and get back to kerbin with a 40-50 km periapsis.

10 minutes ago, Vanamonde said:

This impinges on real world orbital mechanics but specifically involved perspective changes necessitated by the game's handling of SOIs. So Gameplay Questions seems to be the best sub for it. And so, moved. 

actually, i was interested in the real world orbital mechanics. that's why i posted this specifically in the general forum. there are no questions about gameplay here

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26 minutes ago, king of nowhere said:

i said lower your kerbin periapsis enough to aerobrake. as in, leave mun's SoI and get back to kerbin with a 40-50 km periapsis.

actually, i was interested in the real world orbital mechanics. that's why i posted this specifically in the general forum. there are no questions about gameplay here

OH!  Now I understand what you're asking finally.

Depending what time you burn to leave the Mun's orbit, you either get a speed boost or a braking effect relative to kerbin.  This is because your velocity relative to Kerbin is high in one half of your orbit and low in the other.  Imagine watching, from Earth, a satellite orbiting the moon.  Sometimes it will be move to the right in the sky, sometimes it will move to the left.

Pick the right time and your orbital speed around the mun will partly cancel, getting you a lower periapsis for free.  Pick the wrong time and that velocity adds to the mun's own orbital velocity instead.

You can pick the right time by making a manuever node and dragging it around to change the time of the burn, watching your projected periapsis around Kerbin.

Edited by Corona688
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1 minute ago, king of nowhere said:

actually, i was interested in the real world orbital mechanics. that's why i posted this specifically in the general forum. there are no questions about gameplay here

If I understand your question correctly, the confusion arises from the fact that you're comparing figures from one SOI frame of reference to those from another, which is an artificial situation resulting from the game's method of simulation. 

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1 hour ago, Corona688 said:

OH!  Now I understand what you're asking finally.

Depending what time you burn to leave the Mun's orbit, you either get a speed boost or a braking effect relative to kerbin.  This is because your velocity relative to Kerbin is high in one half of your orbit and low in the other.  Imagine watching, from Earth, a satellite orbiting the moon.  Sometimes it will be move to the right in the sky, sometimes it will move to the left.

that's the wrong answer, because it works when orbiting gilly too. when orbiting gilly you have 15 m/s orbital speed, which is negligible, but you still can return to eve cheaply.  xavven actually found the answer to the original question

1 hour ago, Vanamonde said:

If I understand your question correctly, the confusion arises from the fact that you're comparing figures from one SOI frame of reference to those from another, which is an artificial situation resulting from the game's method of simulation. 

having different frames of reference adds to the confusion, but that applies to any discussion on orbital mechanics.

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8 hours ago, king of nowhere said:

The question then shifts, though; why is raising periapsis cheaper than raising apoapsis, despite oberth effect helping you in the second case? all this time, i've been told that prograde/retrograde manuevers are more effective at high speed. except, here i am seeing evidence to the contrary. there is still something missing from the puzzle

You already increased the apoapsis and have a higher Specific_orbital_energy  at the time you burn to raise de periapsis

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The way I think of it is like this:

* Don't forget "the reversability of orbits"
* It takes ~280m/s dV to go from an eccentric orbit of Kerbin to be captured by Mun and stay within its orbit
* Thus, it takes ~280m/s to go from an equivalent orbit of Mun to a similar eccentric Kerbin orbit.

So its not really the Oberth effect, its more that its significantly important WHEN (and of course, in what direction) you burn. The 280m/s burn to be captured by Mun just happens to be at the exact point which will mean you remain permanently in Mun's SOI. Similarly, that 280m/s to return to Kerbin needs to be done at the appropriate time and direction, otherwise it will cost much more if you just randomly left Mun's orbit. Since you'll be in Kerbin's SOI, you'll have to do another burn of some kind to return to Kerbin's atmosphere (and maybe to prevent yourself leaving Kerbin SOI etc)

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21 hours ago, king of nowhere said:

first possible explanation: oberth effect.

Ding ding ding.  Your orbital velocity around the Mun is significant.  It's over 500 m/s relative to the Mun.  That's large, compared to the amount of dV you need to do, so Oberth effect really matters.

In other words,

21 hours ago, king of nowhere said:

that explanation does not hold up to close scrutiny, for several reasons. First, mun is smaller than kerbin

Doesn't matter that it's smaller than Kerbin.  You get a local Oberth boost due to being close to the Mun.

 

The other thing you have going for you, if you eject from low Mun orbit to go back to Kerbin, is that you're ejecting when your orbital velocity around the Mun is directly opposite the orbital velocity of the Mun around Kerbin.  Therefore that gets you much of the way home, so to speak-- your actual orbital velocity around Kerbin at the moment you do your burn is equal to the Mun's orbital velocity around Kerbin, minus your orbital velocity around the Mun.

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I'm still suspecting that the confusion is arising from frames of reference, but I'll re-move the thread to Spaceflight since that was your intention. 

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On 1/14/2021 at 10:11 PM, paul_c said:

The way I think of it is like this:

* Don't forget "the reversability of orbits"
* It takes ~280m/s dV to go from an eccentric orbit of Kerbin to be captured by Mun and stay within its orbit
* Thus, it takes ~280m/s to go from an equivalent orbit of Mun to a similar eccentric Kerbin orbit.

So its not really the Oberth effect, its more that its significantly important WHEN (and of course, in what direction) you burn. The 280m/s burn to be captured by Mun just happens to be at the exact point which will mean you remain permanently in Mun's SOI. Similarly, that 280m/s to return to Kerbin needs to be done at the appropriate time and direction, otherwise it will cost much more if you just randomly left Mun's orbit. Since you'll be in Kerbin's SOI, you'll have to do another burn of some kind to return to Kerbin's atmosphere (and maybe to prevent yourself leaving Kerbin SOI etc)

This, simply enough, To get on an reentry trajectory you only have to lower your Pe into the atmosphere, you do not have to circulate at the end. 
Yes as you orbit the moon you will slower relative to earth than the moon and require less dV for an return burn, yes you need to escape moon gravity but burn is large enough for this anyway. 

 

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