Delta-V, Earth Moon L2 and Asteriods

[RuBisCO]

The earth-moon L2 is a good staging point (assuming lunar mining and fuel production first) because it removes most of the delta V needed to get out of the earth system, for getting to a NEA the difference is HUGE, the nearest NEA would require just ~200 m/s from L2 (according to the nice cartoon above), but 3300 m/s from LEO (The delta-V to reach earth escape velocity plus a little more).

Now here is my question to you, would it be a good rough bet then to subtract these delta-v figures here by 2.7 km/s (rule of thumb from mars and venus) to get a good rough estimate (say within 20% of actual) of what the delta-v from L2 would be to enter a matching orbit with any particular NEA on that list.

Say for example it says the asteroid Didymos (1996 GT) would require 5.098 km/s to reach from LEO (I assume LEO) so it should require just ~2.3 km/s to reach from EML2, right?

The list does say some inconsistent things though, it say 6.0 km/s is needed to "rendezvous" with the moon, but this is how much delta-v is needed to go from LEO and land on the moon (5.7 according the this cartoon, 5.93 according the Wikipedia but it is only 3.05 - 3.25 km/s to reach trans-lunar injection orbit, which would get you to "rendezvous" you would just fly around it and back, fly past it or crash into it without further delta-v spent once reaching moon space. So I'm a little confused if the table means it takes that much delta-V to just pass close by an asteroid or to match orbits with the asteroid. I'm assuming the later

And now on the return trip if I was to leave an NEA it should require the same Delt-V to return if conditions were right. So no aerobrake but instead swing around the earth entering into a lunar transfer orbit, spend a little delta-V slowing down and back into L2. Of course more delta-v could be taken off by using the moons gravity to provide capture into HEO as well. In short I can use L2 to avoid having to "go down" most of the gravity well of the earth, just fly through it without having to spend energy to stay, most of my escape velocity and asteroid to earth transfer orbit velocity would be sucked up by passing the earth and flying all the way back up to lunar space to enter EML2 with minimal delta-v spent.

So to get back from Didymos would require a delta-V ~2.3 km/s, via my rough estimate. Total from EML2 to Didymos and back would be 4.6 km/s.

Is there anything wrong with this logic?

[metaphor]

If you want to look at delta-v and actual launch windows to get to different asteroids, this website is a very good resource.

And now on the return trip if I was to leave an NEA it should require the same Delt-V to return if conditions were right.

That depends. Most of these near-Earth asteroids have an orbital period very close to that of Earth's. This means it might not take much delta-v to get to/from them, but the transfer windows are very far apart (since the synodic period is very long). So if you don't want to stay several years at the asteroid, the transfers are going to be asymmetric in terms of delta-v. For example, the asteroid Apophis passes very close to Earth in 2029, so you can either launch before that close approach, requiring a large amount of delta-v to rendezvous but a tiny amount of delta-v to come back, or launch right at close approach, requiring a small amount of delta-v to rendezvous but a larger amount of delta-v to come back. If you wanted both a small amount of delta-v to rendezvous and a tiny amount of delta-v to come back, you would have to wait on the asteroid for 7 years until the next Earth closest approach in 2036.

So I'm a little confused if the table means it takes that much delta-V to just pass close by an asteroid or to match orbits with the asteroid.

It's almost certainly the delta-v to match orbits with it. The delta-v to pass by a body from low Earth orbit is 3.2-4 km/s for pretty much anything in the inner solar system because of the Oberth effect.

If you were to leave from L2 to go to an asteroid, the most efficient way to do that for most asteroids would be to use ~0.3 km/s to drop down to a low Earth periapsis, then burn 0-0.8 km/s at periapsis. So basically, subtract 2.9 km/s from the delta-v needed to reach the asteroid from LEO. If you use the trajectory browser, subtract ~10.5 km/s from the re-entry speed to figure out how much delta-v would be needed to capture in L2 when coming back from an asteroid.

I'm not sure what that site means by rendezvous with the Moon or Mars. The delta-v is close to the amount needed to land on the Moon, or to get into a low Mars orbit.

Edited February 10, 2014 by metaphor

[RuBisCO]

If you want to look at delta-v and actual launch windows to get to different asteroids, this website is a very good resource.

Thanks for the link.

That depends. Most of these near-Earth asteroids have an orbital period very close to that of Earth's. This means it might not take much delta-v to get to/from them, but the transfer windows are very far apart (since the synodic period is very long). So if you don't want to stay several years at the asteroid, the transfers are going to be asymmetric in terms of delta-v. For example, the asteroid Apophis passes very close to Earth in 2029, so you can either launch before that close approach, requiring a large amount of delta-v to rendezvous but a tiny amount of delta-v to come back, or launch right at close approach, requiring a small amount of delta-v to rendezvous but a larger amount of delta-v to come back. If you wanted both a small amount of delta-v to rendezvous and a tiny amount of delta-v to come back, you would have to wait on the asteroid for 7 years until the next Earth closest approach in 2036.

I'm aware of that, for my example of Didymos it has an orbital period of 2.1 years so launch windows would be available roughly every 2 years, as the earth will pass it by every ~2 years. 2006 RH120, the lowest delta-v asteroid on the list has an orbital period of just 1.05 years, so the earth would pas it by roughly ever 20 years, so a launch window for near minimum delta V would be every 20 years!

Of course for the Didymos the delta-V to get to it would be variable depending on how far it is (between Aphelion and Perihelion and inclination) it is from the earth every time earth passes it by and a transfer could take place. The 5.091 is probably the lowest delta-v possible. If it was at aphelion then a space craft launched from earth would need to put in alot of detla-v just to get up to it, then match orbit, if it was at Perihelion very little delta-v would be needed to get to it as it would pass frightening close to earth, but then delta-V would need to be spent to match orbit. I would suspect matching orbit at aphelion would take much less delta-v then matching at perihelion and this would equalize things out somewhat.

Your app was very helpful, my assumption about 2 years windows was right as well as variable delta-V, the best window for the next 30 years is 2022, with a 144 day stay time and 2.1 years flight time total, total delta-V would be 6.33. Using your assumptions though injection from L2 would be 1.78 km/s and return would be 3.72 km/s for a total of 5.5 km/s... that does not seem right.

[metaphor]

Using your assumptions though injection from L2 would be 1.78 km/s and return would be 3.72 km/s for a total of 5.5 km/s... that does not seem right.

That's because I'm assuming propulsive capture instead of aerocapture. That app doesn't have any constraints on re-entry speed, and 14.22 km/s is actually really high (like coming back from Jupiter on a Hohmann transfer). If you do aerocapture then go to L2 you would only need about 0.3-0.4 km/s of delta-v since you burn the rest of your speed using the atmosphere.

So without aerocapture that mission from/to L2 would be 1.78 + 1.65 + 3.72 = 7.15 km/s of delta-v. With aerocapture it would be 1.78 + 1.65 + 0.3 = 3.73 km/s.

Edited February 11, 2014 by metaphor

[RuBisCO]

How is the delta-V greater from L2 and back then for LEO and back? Would not that 1.65 km/s be in the re-entry velocity?

The re-entry velocity seems alright to me, falling from the edge of earth space to earth would impart a maxium velocity of 11.2 km/s, since instead of hitting the earth I'm flying right past it and back up the gravity well, thus most of that velocity is sucked off as potential energy, all I need now is to add the delta-v needed to go from escape velocity to L2, which according to wiki is 0.14 km/s, so there for it would be re-entry velocity - 11.2 + 0.14 which is pretty close to your 10.5 estimate.

Likewise going form LEO to escape is 3.22 km/s going form L2 to escape is only 0.14 so there for I save 3.08 km/s leaving from L2. On my return I would save all of escape velocity except the 0.14 to get to L2.

So by this rule of thumb:

injection to Didymos (I assume somehow this includes delta-v to leave earth and delta-v to match orbit with didymos)

LEO = 4.68

L2 = 4.68 - 3.22 + 0.14 = 1.6

Return from Didymos

Earth = 1.65

L2 = 14.22 - 11.2 + 0.14 = 3.16

Total from L2 and back would be 4.76 km/s, if we assume none of 1.65 return velocity is in the re-entry velocity it would be 6.41 km/s

There is also the option of using the moon to suck velocity when coming it, via a 2-body capture. That could take off at most 2.2 km/s, so my return delta-V could be as low as 0.96 km/s.

There is also the problem of L2 point not being optimally placed for flying off to an asteriod, alternatively I could fly into a really high earth orbit beyond the moon, I will assume this is equal to L4/5, so that would add 0.33 (L2->L4/5) and 0.43 (L4/5 to C3) or 0.76 km/s total. There is of course your idea of flying around the moon and back towards earth and around the earth as well.

[metaphor]

There's several parts to that delta-v. According to that chart, the opportunity for Didymos in 2022 says 4.68 km/s injection delta-v, 1.65 km/s post-injection delta-v, and 14.22 km/s re-entry. That means this breakdown of delta-v: (click the dots on the chart or click the 'view' button to see more details)

4.65 km/s burn from LEO towards the asteroid

1.05 km/s burn at the asteroid to get captured and enter orbit/land

0.60 km/s burn at the asteroid to leave it and go towards Earth

14.22 km/s re-entry speed means an extra 6.4 km/s over the 7.8 km/s orbital velocity of LEO (so it would take 6.4 km/s to brake into LEO)

If you leave from L2 instead:

0.3-0.4 km/s to get from L2 down towards a low Earth periapsis

1.45 km/s burn at Earth periapsis towards the asteroid (since you're already going 3.2 km/s faster than LEO, you only need 4.65-3.2=1.45 km/s more)

1.05 km/s burn at the asteroid to get captured and enter orbit/land

0.60 km/s burn at the asteroid to leave it and go towards Earth

3.22 km/s burn at periapsis to slow down and get captured by Earth (since you need to slow down to 11.0 km/s to get to L2, 14.22 - 11.0 = 3.22 km/s)

0.3-0.4 km/s to get captured at L2

If you leave straight from L2 towards the asteroid, or come straight from the asteroid to L2, without going down to a low Earth periapsis first, you can't take advantage of the Oberth effect from being close to Earth. A 14.22 km/s re-entry speed means you're traveling about 9 km/s with respect to the Earth when you enter its SOI, and the L2 point is far enough out of Earth's gravity well that you will effectively have to cancel out the entire 9 km/s if you want to stop right at L2.

Another example to show the Oberth effect: If you're in LEO and burn 4.65 km/s, your velocity will be 7.8+4.65=12.45 km/s. Since the Earth's escape velocity from LEO is 11 km/s, when you exit Earth's SOI you will be going sqrt(12.45^2 - 11^2 ) = 5.8 km/s. L2 is very close to being outside Earth's SOI, so if you leave straight from L2 you will need to burn about 5.8 km/s (plus or minus 1 km/s from the Moon's orbital velocity depending on where the Moon is in its orbit). If instead of doing that, you first spend 0.35 km/s to get from L2 down to a low Earth periapsis, you only need an extra 1.45 km/s burn at periapsis to go 12.45 km/s and reach the asteroid.

When coming back from an asteroid, you can also use Earth's atmosphere to burn off that 3.22 km/s and aerocapture into an elliptical orbit that would take you back up to L2, and use 0.3-0.4 km/s of delta-v to get captured at L2. That would be extremely hard though since you would need a precision of something like a meter in periapsis altitude to get the right apoapsis after aerobraking.

Edited February 12, 2014 by metaphor