Orbits Q&A

[MP99]

On a completely different note, I was looking at the proposed Exploration Platform at EML-2 the other day, and got to thinking about the powered swing-by maneuver that really helps L-2's economics (compared to L-1). It occurred to me that during the powered swing-by, you have to pass quite close to the Moon, so that you could take the lander, undock, perform an LOI (to a phasing orbit; I assume that the swingby is not necessarily positioned perfectly for the desired landing site) and land, all the while Orion goes on and does the swing-by, then rendezvous and docks with the platform.

If that works, I suspect it would only allow for an equatorial landing site.

cheers, Martin

[truth is life]

Do Mars missions that land do a direct insertion?

Usually, yes. The Vikings and possibly some of the Russian probes were exceptions out of necessity, but nowadays direct insertion to the Martian surface is the rule.

Thus, do the Mars orbiters need to spend fuel to break into orbit?

Yes and no. They need to spend propellant to insert into initial Martian orbit; no one has ever aerocaptured (without subsequently and immediately landing), to my knowledge. However, there have been several probes, most prominently Mars Global Surveyor, which used atmospheric drag (and thus much less propellant) to circularize their orbit.

If that is the case, a Mars Synchronous Orbit would require more delta-v than a lander, but less than Low Mars Orbit orbiter?

It would depend on the details. In general I would expect that inserting into MSO would require a pretty substantial amount of delta-V, though. There would be no real scientific benefit from it, so I'm not sure why you would want to, at least in the near future.

Any idea of how much extra delta-v after a TMI for a MSO orbit?

Unfortunately, no.

@MP99: Well, if you would read the rest of the post that is exactly what I am asking, isn't it? To state it more technically, the question is whether the departure vector from LEO can be selected so that coplanar (ie., minimum delta-V) insertion into an arbitrarily inclined (ie., allowing you to land wherever you please) low lunar orbit is possible while still allowing for the powered L-2 swing-by maneuver.

[kevin-rf]

Doesn't Deimos' orbit pass near or through MSO? I though it did, but I could be wrong.

[truth is life]

Doesn't Deimos' orbit pass near or through MSO? I though it did, but I could be wrong.

Almost, but not quite. It's got an orbital period of 30.3 hours, about 6 hours too long to be in MSO. Semi-major axis is apparently about 3000 km too large as well.

[baldusi]

So, my next question is: doing the standard Hoffman transfer, would Mar's gravity capture an Orbiter on some "free" orbit, or would it simply deviate it? In other words, does the transfer speed relative to Mars is more or less than Mars escape velocity?

[ugordan]

It's more than Mars escape velocity. Without some fancy 3-body action (and nothing of significant mass is available at Mars) you cannot get a "free" capture by Mars and need to shed some energy.

You can minimize the incoming hyperbolic excess velocity if you can match Mars' heliocentric orbit parameters as closely as possible with your incoming spacecraft, but as you can imagine this is not possible with a Hohmann transfer from Earth. Such matching can be done by solar powered ion craft like Dawn, though.

[sdsds]

The intuition behind the "patched conics approximation" technique is that as you approach Mars its gravitational influence begins to dominate the trajectory and the influence of the Sun can be largely ignored. The spacecraft is effectively approaching Mars "from infinity" and its Mars-relative trajectory will thus be parabolic at a minimum, and most likely hyperbolic.

It's possible that the gravitational influences of the moons of Mars could be used in some tricky way, but that would depend on them being in exactly the right positions for it, which might constrain your launch window a bit!

[Hop_David]

So, my next question is: doing the standard Hoffman transfer, would Mar's gravity capture an Orbiter on some "free" orbit, or would it simply deviate it? In other words, does the transfer speed relative to Mars is more or less than Mars escape velocity?

With regard to Mars, the Earth to Mars Hohmann path is approximately a hyperbola.

Speed of a hyperbola is sqrt(Vesc^2 + Vinf^2). This can be easily remembered with this device:


In this case Vinf is about 2.7 km/s.

Vesc gets smaller and smaller as you get further from Mars

To get the burn to circularize an orbit, you would subtract circular orbit velocity from hyperbola velocity.

Since escape velocity is sqrt(2) circular orbit velocity, escape velocity can be visualized as the hypotenuse of an isosceles right triangle with circle orbit speed as the the legs.



The red portion is circle orbit speed subtracted from hyperbola speed. This is the burn needed.

Assuming a Vinf of about 2.65 km/s, it'd take a 1.9 km/s burn to go from a hyperbolic orbit to a circular orbit at Mars synchronous altitude.

I used Hohmann.xls to get that figure:

[baldusi]

Wow! That's the best answer ever!!! Thank you so much!!! I'll print it for posterity. BTW, those graph are golden!
So, forgetting the issue of plane change for a while, it's 100m/s more than a GTO transfer with a 1800m/s deficit. Doesn't seems like that much outside of what a stock commercial satellite does, right? Of course the plane change could be a nasty issue, and the navigational issues are way beyond what a stock communications satellites does. But at least from a delta-v perspective doesn't seems that difficult. An Atlas V 431 can do 3.87tonnes to TMI. Thus that's a normal commsat with a 1.800m/s deficit.
Now that I come to think of it, is such a problem the plane change? Since you are coming from outside the gravitational sphere of influence, you could chose any plane you want by selecting where you do your burn. I mean, assuming no rotational tilt with respect to your transfer plane, if you approach from the north you could get to polar, if you approach from the east you could get to equatorial, right?

[Jason1701]

Would the delta-v to get from the lunar surface to a Lagrange point be equal to that required to make the return trip?
Thanks.

[Hop_David]

Now that I come to think of it, is such a problem the plane change? Since you are coming from outside the gravitational sphere of influence, you could chose any plane you want by selecting where you do your burn. I mean, assuming no rotational tilt with respect to your transfer plane, if you approach from the north you could get to polar, if you approach from the east you could get to equatorial, right?

My models assume coplanar orbits so plane change expense is ignored. If an equatorial orbit is your goal, plane change expense would be hefty since Mars tilts 25 degrees to the ecliptic plane.

A way to mitigate would be doing multiple burns to park at Mars synchronous orbit. The 1.9 figure assumes a single burn to enter Mars synchronous orbit.

1st burn: .7 km/s, just enough to exit Hohmann to a Mars capture orbit with a 300 km periapsis and 570,000 km apoapsis (apoapsis right at the edge of Mars' sphere of influence). At a 570,000 km apoapsis, the ship is only moving .031 km/s

2nd burn: .05 km/s Apoapsis plane change/raising periapsis. At this high altitude a .05 km/s burn suffices for a 25 degree plane change  and raising periapsis to Mars synchronous altitude. The apoapsis should be in the equatorial plane during plane change burn, this would be a matter of timing.

3rd burn: .6 km/s to circularize at Mars synchronous altitude.

These 3 burns total less than 1.4 km/s. You may notice the total is less than the 1.9 km/s single burn figure I gave earlier.

Although this triple burn saves delta V, it'd be more time consuming.

[baldusi]

I'm surprised that once you're past TMI, this would be less expensive even than a GTO->GSO transfer!
But the thing I'm not quite getting about the orbital mechanics is this: I understand that I'm coming from outside the sphere of influence, and let's say that the minimum approach to the planet takes me over a line that passes through the center of the planet and is normal to my transfer plane. At that time the gravity pull of the planet would be normal to my plane, and thous I would get no "help" of the planet's gravity to counter some of my velocity, but would result (at that instant) in a small plane change, right?
So, any burn that I do at that point would be spent all on my current speed. In other words, I would have to burn in my transfer plane, to get captured by the planet.
I don't know if I made myself clear. My first intuition told me that if you are approaching from the infinity your plane matters not as long as your intended plane ascending/descending node coincides with your approach vector (if you are pointing straight to the center of the planet).

[Hop_David]

I'm surprised that once you're past TMI, this would be less expensive even than a GTO->GSO transfer!
But the thing I'm not quite getting about the orbital mechanics is this: I understand that I'm coming from outside the sphere of influence, and let's say that the minimum approach to the planet takes me over a line that passes through the center of the planet and is normal to my transfer plane. At that time the gravity pull of the planet would be normal to my plane, and thous I would get no "help" of the planet's gravity to counter some of my velocity, but would result (at that instant) in a small plane change, right?
So, any burn that I do at that point would be spent all on my current speed. In other words, I would have to burn in my transfer plane, to get captured by the planet.
I don't know if I made myself clear. My first intuition told me that if you are approaching from the infinity your plane matters not as long as your intended plane ascending/descending node coincides with your approach vector (if you are pointing straight to the center of the planet).

No matter what direction the ship is coming from, the path will be a hyperbola with the hyperbola's focus at Mars' center. So the hyperbola's plane will be passing through Mars' center.

To get the most bang out of your capture burn, the burn should be anti-parallel to the velocity vector at hyperbola's periapsis. So no plane change during the 1st burn.

This periapsis burn should be at an ascending or descending node of the target orbit, as you say.

I would postpone the plane change until apoapsis. Plane changes are cheaper at a distant apoapsis.

[mgfitter]

So is there an easy way (excel spreadsheet like the one above?) to calculate the approximate dv for just a plane change?

-MG.

[Hop_David]

I'm surprised that once you're past TMI, this would be less expensive even than a GTO->GSO transfer!

I was also surprised first time I did the numbers. Phobos and Deimos are two of the lowest delta V destinations in the solar system.