Why does it cost more to circularize the mun clockwise?

[Lymark1]

If going clockwise around the mum would slow down/lower enough apoapsis for a free return trajectory back to Kerbin, shouldn't I able to use the same method to lower apo even more in order to circularize around the mum using less Delta-V? I'm planning a mum landing trip and I thought I could save some fuel for a 'clockwise circularize', but apparently it cost more! Why is that?

 

Edit: Thank you for answering guys!

Edited March 10, 2018 by Lymark1

[Plusck]

To restate the question: does a retrograde flyby of the Mun have to be faster than a prograde one?

Basically, there are only two things to be taken into consideration:

- to minimise the circularisation dv cost, you need to minimise the speed at which you enter the Mun's spherical SOI while having enough of an SOI-entry angle to miss hitting the surface of the Mun;
- for a "free return" flyby, the only rule is that you must exit the Mun's SOI at a speed and angle that sends you through or near to Kerbin's atmosphere.

My answer is therefore that for the vast majority of transfer trajectories from LKO, it's impossible to reconcile the above two factors: you enter and exit the Mun's SOI at the same speed, and a slow exit from the Mun's SOI is never going to send you down to a low Pe over Kerbin.

It's very easy to minimise your velocity on entering the Mun's SOI if the Mun comes up behind you as you are still climbing to Ap above Kerbin, but that won't give you a free return trajectory. Rather, you'll either exit on an escape trajectory (if prograde) or on a much more circular orbit around Kerbin (if retrograde).

For a free return trajectory you must be higher up from Kerbin, nearer the top of your orbit around Kerbin and therefore going slower with respect to the Mun's SOI when you meet it. Your relative speed is therefore higher, your entry into the SOI is faster and, necessarily, you have to be going faster on your clockwise fly-by past the Mun.

[HebaruSan]

7 hours ago, Lymark1 said:

If going clockwise around the mum would slow down/lower enough apoapsis for a free return trajectory back to Kerbin, shouldn't I able to use the same method to lower apo even more in order to circularize around the mum using less Delta-V?

No. The free return trajectory you speak of is an effect of using the Mun for a gravity assist, in other words, as @Plusck says, the Mun flings you out on a trajectory that is favorable for what you want to do away from the Mun. A gravity assist can only raise or lower your orbit around the assisting body's parent body, not with respect to the assisting body.

Quote

I'm planning a mum landing trip and I thought I could save some fuel for a 'clockwise circularize', but apparently it cost more! Why is that?

The cost to circularize directly depends on your orbital velocity at Mun periapsis. In my experience, a free return trajectory requires you to be moving faster to get the right exit vector from Mun.

Edited March 4, 2018 by HebaruSan

[Reactordrone]

Free return does cost more (only about 15m/s for the Mun). You first have to boost to a slightly higher apoapsis which costs more delta-v and the when you're at apoapsis you're travelling slightly slower relative to the Mun so you need to speed up more (retrograde relative to the Mun) in order to catch up to the Mun.

[Capt. Hunt]

in a nutshell, Free return saves delta-v on the return to Kerbin, however, entering that orbit costs more Delta-V because you're fighting the rotation of the mun.

[Zhetaan]

@Lymark1:

First, welcome to the forum.  Enjoy your stay.

Also, if you're planning a landing, then clockwise orbit (also known as a retrograde orbit) moves against the Mun's own rotation.  This means that in order to land, you expend more fuel to match the velocity of a surface that is rotating away from you.  To take off and make orbit again (assuming you want to go back to a retrograde orbit), you need to zero the velocity you have from moving with the surface and only then work to attain orbital velocity.  Because the Mun is tidally locked to Kerbin, its rotation speed is very slow, so it's not a tremendous waste of fuel--but it is still worth noting.

The reason the Apollo missions chose the free-return trajectory is because they:

1. Had the fuel to spare, and
2. Cared more for safety than for absolute fuel efficiency.  Apollo 13 specifically proved that reasoning correct; if that crew had needed to make a larger correction burn ... well, let's just say that they'd have an eternal monument.

[devikwolf]

5 hours ago, Zhetaan said:

The reason the Apollo missions chose the free-return trajectory is because they:

1. Had the fuel to spare

At least, until the final few missions in the program, when they used clever orbital maneuvers to run those puppies almost totally dry. Gotta admire the engineering genius that went towards squeezing every bit of performance out of that outstanding vehicle.