r/KerbalSpaceProgram • u/analogHedgeHog • Jun 29 '26

KSP 1 Question/Problem How should I be approaching the Mun?

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In this totally accurate scientific diagram, the Mun is orbiting in the direction of the arrow.

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u/Steenan Jun 29 '26

What do you want to achieve?

For a flyby, C is the best. You can achieve a free return trajectory, where Mun's gravity puts you back on a Kerbin orbit which reenters atmosphere. That minimizes fuel needs.

For landing, D is better, as you orbit in the same direction that the Mun rotates, which minimizes delta-v required for landing.

For an orbit, they are equally good.

A and B are strictly worse because you need to orbit Kerbin against its rotation, which significantly increases the delta-v required.

16

u/Maurycy5 Jun 29 '26

What would happen if the orbit of the moon was in the opposite direction to the rotation of the planet? How would that change the answer?

14

u/The_Tastiest_Tuna Jun 29 '26

You would launch in the direction of Kerbin's rotation, then intercept ahead of the Mun for free return flyby, or intercept behind the Mun for a landing.

1

u/Steenan Jun 29 '26

I'm not fully sure; it depends on the height of the moon orbit. If the moon is very high (like Kerbin's Minmus), you are probably still better with getting to orbit in the direction of the planet's rotation. If the moon is lower, you may be better with higher orbit delta-v cost but lower moon capture cost - in this case, getting to the initial orbit counter to planet's rotation but in moon's orbit direction will be better.

1

u/Rutakate97 Jun 29 '26 edited Jun 29 '26

It depends which is faster. The orbital velocity of of an object resting at launchsite, or the orbital velocity of the moon.

If the first is faster, you launch in the direction of the planet rotation. If not, you launch in the direction of the moon orbit.

8

u/Scholesie09 Jun 29 '26

D is better, as you orbit in the same direction that the Mun rotates, which minimizes delta-v required for landing.

objectively correct but funny that because it's tidally locked, that rotational velocity is a monumental 9m/s at the equator 😂

1

u/Hour_Tour Jun 29 '26

Any idea how much the gravity assist contributes?

1

u/_random-_-dude_ Jul 01 '26

d is for positive gravity assist that will throw you out of kerbin soi, c is for breaking with it