Retrograde reentry

[NFunky]

Hi all,

So I recently ran into an interesting issue that I've never really payed attention to before.  First off, let me state that I am using RSS, but this topic should apply to stock as well, if perhaps to a lesser degree.

I was returning a three kerbal capsule from the moon, and accidentally overdid the lunar escape burn.  This meant that, while I had a good return trajectory, I'd be coming in retrograde relative to Earth's rotation.  I didn't think this really mattered, until my capsule exploded on reentry.  This was a capsule I'd returned from the moon multiple times before, and I hadn't modified my install or save game at all, so I was baffled until I realized the possible cause.  I did some tests and it seems that reentering retrograde produces much more intense heating (approximately 100-200 degrees more).

So here's my question.  Is this realistic?  I understand that the atmosphere is rotating along with the planet, but I wouldn't have thought it would cause any real differnce in airspeed during reentry.  Was I wrong and atmospheric rotation is a much bigger issue than I suspected?

[Aegolius13]

I don't play RSS, but in regular KSP (and real life) it can be a pretty significant issue.  As you say, the atmosphere is rotating along with the earth.  So when you reenter from a prograde orbit, to get your speed relative to the air (which determines heating), you can effectively subtract the earth's rotational speed from your orbital velocity.  But if you reenter in a retrograde orbit, you're ADDING the earth's rotational speed instead of subtracting it.  You can get an idea of the difference by comparing your "orbit" vs. "surface" velocity readouts for both orbit types.

This is also related to why rockets usually launch west-to-east.  When you launch that way, you get "free" velocity from the earth's rotation so your rocket does not need to spend as much delta-v.  The same thing happens in reverse -  you need to disperse energy into heat when you reenter, and you don't have to deal with as much heat if the earth's rotation is helping you out.  

[Aelfhe1m]

A point on the Earth's surface near the equator has a rotational speed of approximately 460 m/s. So if you re-enter prograde (parallel to the equator) with an orbital velocity v your velocity relative to the surface will be (v - 460) m/s but if you re-enter retrograde from the same orbital velocity your surface relative velocity is (v + 460) m/s or 960 m/s faster. The atmosphere rotates at the same speed as the surface. From LEO your initial orbital speed is about 7.5 km/s so the difference between prograde and retrograde will be around 13%. A lunar return will have a higher orbital velocity at atmospheric interface but since the difference in surface velocities is fixed the percentage change between prograde and retrograde will be less (around 8-10%) but still significant.

[bewing]

On top of this, heating tends to go as velocity cubed, so a small incremental increase in velocity has big implications.

 

[Aelfhe1m]

25 minutes ago, bewing said:

On top of this, heating tends to go as velocity cubed, so a small incremental increase in velocity has big implications.

Very good point as increasing a value by 10% increases any other value that's proportional to its cube by 33%!