Best Direction For Landing

[The Flying Kerbal]

I wonder if you guys have any preference as to the direction of the orbit around a planet or moon when intending to land?

Is it better/easier/more efficient to be going in the same direction as the rotation of the body being landed on, or do you prefer the orbit to be in the opposite direction?

Thanks everyone.

[bewing]

It saves a bit of deltaV if you orbit and land prograde to the rotation. When you are choosing a landing site, it also gives you a little more time to look the terrain over if you are going the same direction.

 

[Blaarkies]

Same direction as planet rotation is cheaper

Reason: Same reason as why it is cheaper to Take-Off due east from Kerbin. Let's assume the celestial body you land on, needs you to move at 200m/s to stay in low orbit. The body itself rotates so that its surface moves at 40m/s (relative to the stars).
You can see this when landed, just click on the navball speed label to switch it from "Surface" mode to "Orbit" mode. Although the landed craft is standing dead still, its "Orbit" velocity will still be 40m/s, because it is actually moving in circles as the planet surface rotates.

Now, to land there at the Poles (they have none of these effects) it would cost at least 200m/s dv to cancel out all your orbit velocity.
To land in the direction due east, you need to reduce your orbit velocity by at least 160m/s(so that you reach that special 40m/s mark) to be at standstill relative to the surface
To land in the direction due west, you need to cancel your orbit velocity by at least 200m/s and then another 40m/s extra just to get up to speed with the surface rotation.

Total of 240m/s west Versus 160m/s east
* The KSP planets don't have ratio THIS big, but its still worth planning for in-game

Edited May 8, 2017 by Blaarkies

[The Flying Kerbal]

Two super answers guys!  When I read them, I suddenly realised how obvious it was and just how daft my question must have appeared to people who know what they're doing.

Thanks for taking the time to reply. 

[maceemiller]

8 hours ago, The Flying Kerbal said:

Two super answers guys!  When I read them, I suddenly realised how obvious it was and just how daft my question must have appeared to people who know what they're doing.

Thanks for taking the time to reply. 

I don't believe any question about KSP is daft....knowledge is power

[Alshain]

As many have said, same as rotation is cheaper however how much cheaper depends on the planet/moon.  The Mun for example is tidally locked so it only rotates as fast as it's orbit around Kerbin.  The savings are technically there, but negligible.  The faster the axial rotation, the more savings you will get.

Edited May 8, 2017 by Alshain

[Zeiss Ikon]

The other advantage of orbiting posigrade is that you'll have less heating during reentry or aerobraking, because your velocity relative to the atmsophere is by twice that magic figure (Kerbin equatorial rotation speed is about 300 m/s).  Enter atmosphere with the rotation, you get to subtract that 300 m/s from the speed your ablator or other heat management has to deal with; enter against rotation, and you have to add that figure.  With a Mk. 1 pod and standard heat shield, returning from Mun, it makes the difference between an easy reentry and one that gets close enough to burning through the ablator to leave you weak in the knees (with the same periapsis height).  Return retrograde from Minimus, and you'd better set a higher periapsis than you normally would.

Beyond that, you're generally burning more dV before you even consider landing, if you set up retrograde, vs. posigrade.  Talking about a Mun return again, making a retrograde reentry means you've burned more than you needed to by twice your posigrade Munar ejection velocity for the same periapsis.

[Blaarkies]

5 hours ago, Zeiss Ikon said:

Beyond that, you're generally burning more dV before you even consider landing, if you set up retrograde, vs. posigrade.  Talking about a Mun return again, making a retrograde reentry means you've burned more than you needed to by twice your posigrade Munar ejection velocity for the same periapsis.

"posigrade", first time i ever heard of that 

What exactly do you mean by "twice your posigrade Munar ejection velocity" ? 
Do you mean, we are in low Mun orbit, and want to get back to Kerbin. So we escape Mun SOI in the the Mun's orbit retrograde direction, but instead of burning "just enough" to fall back into Kerbin atmosphere, we now have to burn(coincidentally) almost double that amount to get into a retrograde, elliptical orbit around Kerbin?

[Zeiss Ikon]

4 hours ago, Blaarkies said:

Do you mean, we are in low Mun orbit, and want to get back to Kerbin. So we escape Mun SOI in the the Mun's orbit retrograde direction, but instead of burning "just enough" to fall back into Kerbin atmosphere, we now have to burn(coincidentally) almost double that amount to get into a retrograde, elliptical orbit around Kerbin?

Yep, that's about it.  The excess velocity (above the minimum for Munar escape) from Mun's SOI to get an in-atmosphere periapsis is going to be something close to Mun's orbital velocity -- your "ejection velocity" which is your residual orbital velocity (for a highly elliptical Kerbin orbit).  You have to reverse that orbit to reenter retrograde.  Now, that's not a bunch of extra dV on a Mun return, and less from Minimus, but it's still unnecessary, and the retrograde orbit it creates gives you, in effect, around 600 m/s additional velocity at atmospheric interface.  Do this from Minimus and it's about the harshest reentry you can make from within Kerbin's SOI.