Entering SOI's Clockwise or Counter-clockwise

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Your experiment is carefully designed to give you the answer you want. One doesn't normally land by zeroing out the orbital velocity and then speeding up (by rockets or by lithobraking) to match sidereal rotation speed.

Try again, but zeroing out surface speed. And try landing on the same spot. And run the landings multiple times, since the difference isn't huge on Mun. It would be easier to notice on Eeloo.

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So where in the solar system would it actually be noticed then?

It's very critical when trying to rendezvous with planets, moons or other ships. If you're orbiting the wrong direction you're gonna have a bad time of it.

And to reiterate Kurja's point: Just because the effect is small for places you can land, does not mean there is no difference.

Edit: Yeah, looks like Eeloo has the highest sidereal velocity of all the atmosphere-less bodies. If you want an extreme example, check out

.

=Smidge=

Edited by Smidge204
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In the experiment as performed, there is in fact no difference. You zero out orbital speed, which is the same from either direction. Then you have to speed up to zero out your surface speed. This is precisely the situation sketched here.

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It's very critical when trying to rendezvous with planets, moons or other ships. If you're orbiting the wrong direction you're gonna have a bad time of it.

And to reiterate Kurja's point: Just because the effect is small for places you can land, does not mean there is no difference.

Edit: Yeah, looks like Eeloo has the highest sidereal velocity of all the atmosphere-less bodies. If you want an extreme example, check out

.

=Smidge=

Yeah, with rendezvous, there's no question about it otherwise you're going to hit each other at the same speed.

Really, Eeloo?

In the video Scott Manley even says that an object like that can't exist in reality as a natural object.

Edited by 700NitroXpress
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As a practical question, do you actually land like that normally, Nitro? If you do, the fuel waste just from gravity losses far out-paces the difference from surface velocity and makes this whole conversation moot.

Also, again as a practical matter, you don't get to complain about where surface velocity matters when you've been using a great deal of your time saying it mathematically doesn't matter.

In the video Scott Manley even says that an object like that can't exist in reality as a natural object.

Now I'm beginning to get suspicious that you're being deliberately obtuse. The fact that such an object can't exist IRL is entirely irrelevant. The mod-planet doesn't create what we're talking about here. Rather it amplifies it to a level that can be easily discerned by people possibly being deliberately obtuse.

Edited by luchelibre
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Yeah, with rendezvous, there's no question about it otherwise you're going to hit each other at the same speed.

Really, Eeloo?

In the video Scott Manley even says that an object like that can't exist in reality as a natural object.

Your point being? In ksp Kerbin, Duna, Eve and Jool have atmospheres, Moho and Dres have slow rotational velocities (as do all the atmosphereless moons in the ksp universe). That leaves Eeloo.

No-one said that it would be a huge difference in dv, just that it's there.

edit - btw, what's your take on ascending off of planets? If surface speed does not matter, it should be the same if you launch east or west from ksc, right?

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Nono, you launch straight up to 100km then burn sideways to circularize... just like landing with zeroing out your orbital velocity and falling, but in reverse!

=Smidge=

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That's inaccurate. To reverse the process, you'd launch straight up, and west enough to cancel out sidereal rotation. Then you burn sideways.

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While there is a difference, I really don't think that the delta-V difference needed to match the surface rotation is significant in most cases. On planets with atmosphere, the air-resistance is going to eliminate whatever difference in surface speed that resulted from prograde/retrograde orbits. P.S., I'm pretty sure atmosphere rotates with the planet, otherwise reentering Kerbin with parachute will give you a surface velocity of 174 m/s (the sidereal rotation velocity of Kerbin) if the atmosphere don't rotate with the planet. In short, prograde/retrograde rotation shouldn't matter if your goal is to land on atmospheric planet (it might matter if, say, thermal damage is modeled, the difference of about 340 m/s on Kerbin might be the difference between your spacecraft entering atmosphere successfully or blowing up).

So, let's look at planets without atmosphere and has a fairly high sidereal velocity, Dres, and Eeloo.

Dres has a sidereal velocity of about 25 m/s, which translate to a delta-V difference of 50 m/s between prograde/retrograde. Using this Delta-V mapfor reference, it takes about 2500 m/s in delta-V to go from Kerbin orbit to Dres surface (or an additional 4000 delta-V from kerbin surface), so you're talking about 2% difference in delta-V needed to land on Dres compared to the delta-V needed to get there from Kerbin orbit.

Eeloo has sidereal velocity of 67.8 m/s, which translate to a delta-V difference of about 134 m/s between prograde/retrograde. It takes about 5000 m/s in delta-V to go from low Kerbin orbit to Eeloo surface. So about 2.5% difference in delta-V.

Basically, in worst case scenario, you're adding 2.5% of delta-V needed if you insert into retrograde orbit versus prograde. So, unless you're challenging yourself on reaching planet with the absolute minimum amount of delta-V/fuel tonnage, 2.5% delta-V is not going to matter in most cases (you're probably going to design your spacecraft with more tolerance, in delta-V, than that).

Edited by UberFuber
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Your point being? In ksp Kerbin, Duna, Eve and Jool have atmospheres, Moho and Dres have slow rotational velocities (as do all the atmosphereless moons in the ksp universe). That leaves Eeloo.

No-one said that it would be a huge difference in dv, just that it's there.

edit - btw, what's your take on ascending off of planets? If surface speed does not matter, it should be the same if you launch east or west from ksc, right?

I just didn't think of Eeloo, that's all. In the video, that planet can't exist because the rotational speed is greater than what the planet's gravity can hold together. The thing would fly apart as Scott Manley points out, he says it would be an artificial construct and not a natural body.

Actually, if you could tell me the exact amount of time it would take to descend to the surface of Kerban from an orbital altitude of 75,000 meters. Factor this value as if Kerban has no atmosphere just for the sake of negating the effects of aerobraking and wind resistance. Then I can show the point.

For your last question, that answer is no. And I never said it was the same for the ascent. When you are on something, you're moving at the same speed it is moving. When taking off from Kerban, you're already moving at 174.6 m/s, you're not starting from 0. If Kerban was stationary than you would take off at the same speed.

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We've established that in your favourite method of descent, it is true that sidereal rotation speed doesn't matter. But that's only because in your favourite method of descent you specifically pay extra to make sure the sidereal rotation speed doesn't matter. Everyone else is landing more efficiently.

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Nitro what you are saying is only true if you take that inefficient decent. The proper way to land would be to match speeds with the surface and touch down. Once again let's consider the surface moving in the positive direction, prograde in the positive direction and retrograde in the negative (opposite of prograde).

Surface speed of 10m/s and orbit of 15m/s, it would take a change of 5m/s to match the surface speed if you are going in the prograde direction. If you are going retrograde (negative) then you are moving -15m/s relative to the surface and must make a change of 25m/s (15 to get from -15 to 0, another 10 to go from 0 to 10) to match speed and land.

Conclusion is that it does matter which way you orbit a body if you want to land on it. It is more efficient to orbit the same direction as the rotation of the body. Once again, basic physics.

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I just didn't think of Eeloo, that's all. In the video, that planet can't exist because the rotational speed is greater than what the planet's gravity can hold together. The thing would fly apart as Scott Manley points out, he says it would be an artificial construct and not a natural body.

You're missing the point entirely.

Actually, if you could tell me the exact amount of time it would take to descend to the surface of Kerban from an orbital altitude of 75,000 meters. Factor this value as if Kerban has no atmosphere just for the sake of negating the effects of aerobraking and wind resistance. Then I can show the point.

Assuming you're not moving vertically at 75000m, t = 123.65 seconds.

Not sure why that matters at all, because the vertical part has absolutely nothing to do with the horizontal part being discussed - especially if there is no atmosphere.

=Smidge=

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Smidge, we might have to give up.

Also just a heads up to everyone in this thread, it's spelt KERBIN not KERBAN get it right!

Edited by Vanamonde
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While wearing my player hat:

700NitroXpress, I think the mistake you are making is that you are assuming that eliminating orbital velocity will automatically mean that one is moving at zero speed with respect to the surface. But a world's rotational speed is indepedent of the velocity required to orbit at a certain height. Let's take a simple example and start with an orbital speed of zero, in which the ship immediately begins to fall straight down. As it does so, the world will be seen to rotate beneath it, and when the ship reaches the surface, it will find the ground moving past it in the direction of the planet's rotation, and the ship will need to expend delta-V to match that speed in order to come to rest with a speed of zero with respect to the planet's surface. Matching surface speed and eliminating orbital velocity are two separate issues, though in some conditions the difference in magnitude is minimal.

I believe that when you think this through, you are unconsciously using a rotational frame of reference when you consider the orbit, and then switching to a stationary frame of reference (with respect to the world) when you consider the landing, without carrying over the velocity from one frame of reference to the other.

And under my moderator hat:

Thank you folks for keeping this discussion civil despite the heat of disagreement, and please continue to do so.

Edited by Vanamonde
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I don't know this forum's policy on reaction images but I FEEL LIKE I'M TAKING CRAZY PILLS.

If you're going 100 mph and landing on something going 50 mph you need to slow down 50 mph. If you're going 100 mph and land on something going -50 mph then you have to slow down 150 mph! That's it. Done. Finitio. Matching velocities with something depends exactly on the velocity of the thing you're matching. Of course it does. How could you think otherwise?

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Gilly is fun to play around with landing on because the gravity is so low that everything seems to happen in slow motion. Also, this little moon has a very fast rotation rate. Anyway, I wanted to see what would happen if I tried to land by only zeroing out my orbital velocity and ignoring the the fact that the planet is rotating (rather rapidly) beneath me.

Here in the first picture you see that my orbital velocity is 0 as I hover this little Kethane probe over the surface of Gilly.

However, the planet surface is actually whizzing past me at 3.8 m/s, as you can see when I switch the navball to to surface mode in this next picture.

If I try to set the craft down without matching the surface velocity of Gilly, I take a nasty tumble. (Not that any tumble on Gilly is ever very nasty, but you get the point.)

So, it is definitely possible to crash even when your orbital velocity is zero. That's why it matters which way the planet is rotating with respect to our orbit around it.

Edited by jwilhite
Fixed images
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I just didn't think of Eeloo, that's all. In the video, that planet can't exist because the rotational speed is greater than what the planet's gravity can hold together. The thing would fly apart as Scott Manley points out, he says it would be an artificial construct and not a natural body.

Actually, if you could tell me the exact amount of time it would take to descend to the surface of Kerban from an orbital altitude of 75,000 meters. Factor this value as if Kerban has no atmosphere just for the sake of negating the effects of aerobraking and wind resistance. Then I can show the point.

For your last question, that answer is no. And I never said it was the same for the ascent. When you are on something, you're moving at the same speed it is moving. When taking off from Kerban, you're already moving at 174.6 m/s, you're not starting from 0. If Kerban was stationary than you would take off at the same speed.

I suppose these points were pretty much covered by others already but since you're quoting me I'll respond anyway.

1) I can't calculate off the top of my head the time it takes to descend, fortunately someone else already posted with the numbers. Please do elaborate on how this matters at all? As far as I can tell the horizontal surface speed does not care if your descent takes a minute or a day.

2) It's exactly the same with the ascent, only in reverse. At launch, you have some initial orbital velocity but zero surface velocity!!! Thus to reach an orbiting velocity, you need less dv if you accelerate in the direction where you - the surface! - is already moving. Exactly the same with landing, you need to reach that same position as at launch, when you had some orbital velocity but zero surface velocity. Go in the other direction and it'll cost you more dv.

edit - Oh, and it's not like all other bodies in the KSP system could exist in reality, think of the sun. It's way too small. And then there are these green little beings on their home planet where the only built structures are at their space center....

Edited by kurja
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Conclusion is that it does matter which way you orbit a body if you want to land on it. It is more efficient to orbit the same direction as the rotation of the body. Once again, basic physics.

While it is always more efficient, I'd argue that there are good reasons for making the dV trade-off to do a retrograde insertion when still learning the mechanics of the game. Early in the Apollo missions, NASA used a retrograde insertion specifically for the free-return trajectory. And with the tidal-locked rotation of the Moon, this was a pretty cheap trade. Much like when dealing with the Mun. I used a retrograde insertion path in my early Mun missions for the same reasons. Last thing I wanted was to mis-judge the fuel requirements (or have energy run out), and get a Kerbal stuck in Mun orbit. I'm still spending a fair amount of time around Kerbin and the moons, where the trade-off is worth it. Efficiency certainly becomes paramount once you start trying to visit other planets, and I don't intend to do retrograde insertions there.

However, in KSP, where we may not care if Jeb makes it back or notÃ¢â‚¬Â¦ meh, why not? I can see someone only doing the efficient thing, just because there are always more Kerbals willing to be hurled into space.

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While it is always more efficient, I'd argue that there are good reasons for making the dV trade-off to do a retrograde insertion when still learning the mechanics of the game. Early in the Apollo missions, NASA used a retrograde insertion specifically for the free-return trajectory. And with the tidal-locked rotation of the Moon, this was a pretty cheap trade. Much like when dealing with the Mun. I used a retrograde insertion path in my early Mun missions for the same reasons. Last thing I wanted was to mis-judge the fuel requirements (or have energy run out), and get a Kerbal stuck in Mun orbit. I'm still spending a fair amount of time around Kerbin and the moons, where the trade-off is worth it. Efficiency certainly becomes paramount once you start trying to visit other planets, and I don't intend to do retrograde insertions there.

However, in KSP, where we may not care if Jeb makes it back or notÃ¢â‚¬Â¦ meh, why not? I can see someone only doing the efficient thing, just because there are always more Kerbals willing to be hurled into space.

Free return trajectory isn't really orbiting though, but yes retrograde flight paths have uses. If you are planning on landing though you are going to want, for the most efficient landing possible, an orbit parallel to the equator of the body and in the same direction of the rotation. If you aren't planning on landing then it's almost irrelevant, unless what is orbiting is going to be a refueling station and you have mining stations on the surface of said planet.

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Free return trajectory isn't really orbiting though, but yes retrograde flight paths have uses. If you are planning on landing though you are going to want, for the most efficient landing possible, an orbit parallel to the equator of the body and in the same direction of the rotation. If you aren't planning on landing then it's almost irrelevant, unless what is orbiting is going to be a refueling station and you have mining stations on the surface of said planet.

I think you missed my point, which is that the efficiency loss is worth the trade off in some cases. To be more specific about the trade-off as it applies to Apollo, Apollo 11 used a retrograde Lunar orbit, to leverage the free return trajectory they used in earlier missions which would reach the far side of the moon. The extra dV needed to do this was worth the trade-off of reduced risk to the pilots. Later Apollo missions did use different free-return trajectories that never entered the Moon's SOI, but that was because they became more comfortable with the issues, and wanted to tackle more interesting landing sites which required more inventive solutionsÃ¢â‚¬Â¦ but even then, safety of the pilots was a higher priority than how much dV they spent getting there.

Folks that are starting out may in fact be interested in this approach to get practice as a stepping stone, seeing as you aren't actually throwing away too much efficiency to do it with the Mun. It certainly is a bit easier and less risky, assuming you don't want to kill Jeb your first time out of Kerbin orbit.

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You're missing the point entirely.

Assuming you're not moving vertically at 75000m, t = 123.65 seconds.

Not sure why that matters at all, because the vertical part has absolutely nothing to do with the horizontal part being discussed - especially if there is no atmosphere.

=Smidge=

Right, so objects released from the same point will arrive at the same time. The ship stopping at 75000 meters will hit the ground at the exact same time that a ship traveling in an arc with the planets rotation will hit the ground when it's descent begins at 75000 meters. Both ships will land on the planet at the same time. That's because the force of gravity on the two objects is the same and constant. Even if you take it the other way around, a ship descending at an angle of 20 degrees both to and against the planets rotation will arrive on the ground at the same time. This is because the planet's rotational speed has 0% impact on the object that's falling to the ground.

Physics for my rocket chair

Total Mass = 15.21 tons 13798.3 kg

Starting Height = 75000 meters

The force of gravity, g = 9.8 m/s2

Gravity accelerates you at 9.8 meters per second per second. After one second, you're falling 9.8 m/s. After two seconds, you're falling 19.6 m/s, and so on.

Time to splat: sqrt ( 2 * height / 9.8 )

It's the square root because you fall faster the longer you fall.

The more interesting question is why it's times two: If you accelerate for 1 second, your average speed over that time is increased by only 9.8 / 2 m/s.

Velocity at splat time: sqrt( 2 * g * height )

This is why falling from a higher height hurts more.

Energy at splat time: 1/2 * mass * velocity2 = mass * g * height

If object A is descending at a 20 degree prograde arc to the planets rotation it's final speed when it hit's the surface will be 1212.44 m/s and it's descent time will be 123.72 seconds

If object B is descending at a 20 degree retrograde arc to the planets rotation, it's final speed when it hit's the surface will be 1212.44 m/s and it's descent time will 123.72 seconds

This means that both chairs will hit the ground at the exact same point in time, traveling at the same speed. This means that you would have to slow down 1212.44 m/s before you hit the ground. Therefore, the deceleration amount for both objects is the same, which means you use the same amount of fuel.

If object A is rotating around the planet at 2000 m/s counter clockwise, and Object B is rotating around the planet clockwise, they are both orbiting at the same speed. The planet is rotating at 174 m/s

Those speeds are constant because the planet isn't accelerating and the force of gravity is equal on both objects.

If Both object A and B decelerate 1500 m/s they are going to head towards the surface at the same speed and arrive at the same time. Thus it is possible for a person standing on the surface to see both objects come down towards him at exactly the same speed and land at the exact same time and they will both appear to have fallen from the exact same height. Thus, both pods can land in the exact same spot right next to each other at the exact same time. This is because both pods are moving at the same speed. This is because orbital speed and surface speed are independent of each other. That means that one has 0% effect on the other.

In conclusion, objects that deorbit from the same height and orbiting at the same speed will hit the ground at the same time. This means that their Velocities to the ground in both cases are equal. That means that they both have to decelerate the same amount in order to land. That gives us the wonderful conclusion that a craft will consume the same amount of fuel to decelerate the same velocity to ground in both cases because again V = GT

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Right, so objects released from the same point will arrive at the same time. The ship stopping at 75000 meters will hit the ground at the exact same time that a ship traveling in an arc with the planets rotation will hit the ground when it's descent begins at 75000 meters. Both ships will land on the planet at the same time. That's because the force of gravity on the two objects is the same and constant. Even if you take it the other way around, a ship descending at an angle of 20 degrees both to and against the planets rotation will arrive on the ground at the same time. This is because the planet's rotational speed has 0% impact on the object that's falling to the ground.

Physics for my rocket chair

Total Mass = 15.21 tons 13798.3 kg

Starting Height = 75000 meters

The force of gravity, g = 9.8 m/s2

Gravity accelerates you at 9.8 meters per second per second. After one second, you're falling 9.8 m/s. After two seconds, you're falling 19.6 m/s, and so on.

Time to splat: sqrt ( 2 * height / 9.8 )

It's the square root because you fall faster the longer you fall.

The more interesting question is why it's times two: If you accelerate for 1 second, your average speed over that time is increased by only 9.8 / 2 m/s.

Velocity at splat time: sqrt( 2 * g * height )

This is why falling from a higher height hurts more.

Energy at splat time: 1/2 * mass * velocity2 = mass * g * height

If object A is descending at a 20 degree prograde arc to the planets rotation it's final speed when it hit's the surface will be 1212.44 m/s and it's descent time will be 123.72 seconds

If object B is descending at a 20 degree retrograde arc to the planets rotation, it's final speed when it hit's the surface will be 1212.44 m/s and it's descent time will 123.72 seconds

This means that both chairs will hit the ground at the exact same point in time, traveling at the same speed. This means that you would have to slow down 1212.44 m/s before you hit the ground. Therefore, the deceleration amount for both objects is the same, which means you use the same amount of fuel.

If object A is rotating around the planet at 2000 m/s counter clockwise, and Object B is rotating around the planet clockwise, they are both orbiting at the same speed. The planet is rotating at 174 m/s

Those speeds are constant because the planet isn't accelerating and the force of gravity is equal on both objects.

If Both object A and B decelerate 1500 m/s they are going to head towards the surface at the same speed and arrive at the same time. Thus it is possible for a person standing on the surface to see both objects come down towards him at exactly the same speed and land at the exact same time and they will both appear to have fallen from the exact same height. Thus, both pods can land in the exact same spot right next to each other at the exact same time. This is because both pods are moving at the same speed. This is because orbital speed and surface speed are independent of each other. That means that one has 0% effect on the other.

In conclusion, objects that deorbit from the same height and orbiting at the same speed will hit the ground at the same time. This means that their Velocities to the ground in both cases are equal. That means that they both have to decelerate the same amount in order to land. That gives us the wonderful conclusion that a craft will consume the same amount of fuel to decelerate the same velocity to ground in both cases because again V = GT

No. Just.... No. Draw a couple vectors, will you. Your chairs or whatever are falling in different directions, yes they are falling at the same speed in a particular frame of reference but on impact their horizontal surface speeds will not be identical because the surface itself is moving. Seriously. Draw a picture. Those often help.

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The ship stopping at 75000 meters will hit the ground at the exact same time that a ship traveling in an arc with the planets rotation will hit the ground when it's descent begins at 75000 meters.

No, because at that altitude the curvature of the planet is no longer negligible. The craft in a ballistic trajectory will take longer to hit the planet because the planet is curving away from it, and thus it has farther to fall.

Path A-B is notably shorter than path A-C.

Note that if the horizontal velocity of the red dot is high enough, the trajectory path will miss the planet entirely - which we refer to as an "orbit." When this happens the time it takes to land on the surface becomes infinite, as it will never land at all.

The more interesting question is why it's times two

It's not an interesting question if you understand calculus, and that acceleration is the time derivative of velocity, which itself is the time derivative of position... so when you integrate acceleration over time a "2" pops out all on its own.

In conclusion, objects that deorbit from the same height and orbiting at the same speed will hit the ground at the same time. This means that their Velocities to the ground in both cases are equal. That means that they both have to decelerate the same amount in order to land.

In conclusion, objects that deorbit from the same height and orbiting at the same speed will hit the ground at the same time. But if one is in a ballistic trajectory and the other does not, then they don't have the same speed and won't hit the ground at the same time.

However, the velocity of the one deorbiting prograde will have a much lower velocity relative to the ground than the one orbiting retrograde, unless additional fuel is used to compensate. Again, this is without atmosphere effects.

Edit: Since I'm drawing pretty pictures, how about one more?

The velocity of the ground is shown as yellow arrows, and the horizontal velocity of the object is shown as black arrows. Magnitude isn't that important but direction is. Notice how the arrows on the left are pointing in the same general direction, while the arrows on the right are pointing opposite? That's the difference. To land safely you need to turn the black arrow on the right completely around to point in the same direction as the yellow arrow, otherwise you'll be traveling sideways when you hit the ground - that's called a crash.

Changing the direction of that arrow takes energy - energy which, if you don't have an atmosphere to suck it up, requires fuel.

=Smidge=

Edited by Smidge204
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Back to the OP, will a retrograde planetary orbit cause you to need more delta-v for encounters with said planets moons? I think it should but I'm not certain.

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