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**Quaoar****Member**- Registered: 2013-12-13
- Posts: 665

As a SF writer, I often try to figure a future where interplanetary travels are routine: many scientific station orbit around Mercury, Venus, Earth, Mars, the main belt asteroids and the jovian satellites, connected by big orbit-to-orbit spaceship, like GW Johnson's modular space-train

https://exrocketman.blogspot.com/2013/1 … -2013.html

This kind of spaceships are not supposed to perform atmospheric entry and landing, but only to travel from the low orbit of the planet to the low orbit on another, doing all maneuvers propulsively and being refurbished in orbit by shuttle-tankers.

For a Earth-Mars or Mars-Earth transfer it's perfect, but what happens when this kind of spaceship returns from Venus, using a Hohmann orbit?

As far as I can understand, she is forced to insert herself in a retrograde orbit around Earth, where it's very difficult to be refurbished (a retrograde LEO mission has almost 10.2 km/s of deltaV vs. 9.2 km/s of a classic prograde orbit insertion) and there is also a high risk of satellite debris impact.

So, how to solve this problem, without consuming tons of propellant to invert the orbit?

I figured to insert the spaceship in a very high retrograde circular orbit (200-300,000 km), then make a change of plain, to match the inclination of Moon orbit, perform a fly-by around the Moon and use her gravity to invert the orbit with a half 8 maneuver. At this point we can lower the perigeo at 400 km and circularize the orbit.

A better idea is highly appreciated.

*Last edited by Quaoar (2022-01-30 14:48:10)*

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**GW Johnson****Member**- From: McGregor, Texas USA
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Hi Quaoar:

I think you might be onto something with your elongated orbit. Normally, orbit plane changes are very expensive in terms of delta-vee, bt if your speed about the primary is low enough, that is a much smaller problem.

I have no trajectory code or computer software to investigate the kind of thing you are proposing. I have yet to compute the relative gravitation influences in a pair of 2-body problems to approximate a real 3-body problem. So I cannot yet answer your question.

But I have been thinking about the pair of 2-body problems that might lead to an orbit stability criterion, even if it is rough. As soon as I come up with something, I'll let you know here.

GW

GW Johnson

McGregor, Texas

"There is nothing as expensive as a dead crew, especially one dead from a bad management decision"

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**tahanson43206****Moderator**- Registered: 2018-04-27
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For Quaoar re "hard" science fiction ...

It is possible you might have missed it, but Terraformer recently opened a topic about "hard" science fiction.

I'm bringing this to your attention because (quite some time ago) (as I recall) we briefly discussed the possibility someone might be interested in translating your work for a market beyond the ones you already serve.

If you and Terraformer strike up a conversation, I (and hopefully others) will be happy to see what might come of it.

(th)

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**kbd512****Administrator**- Registered: 2015-01-02
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With very high-Isp propulsion, is there some reason why you can't approach ahead of or behind a planet, and then slow down or catch up to be "captured" into orbit?

Is that not exactly how we've sent a growing number of these probes to investigate the outer planets and asteroids or comets?

If those Mach-effect thrusters actually work, then it's only a matter of supplying power, rather than propellant. At that point, the solution is to put more power behind the propulsion system or to simply accept longer transfer times and to appropriately shield the ship's occupants from radiation effects.

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**RobertDyck****Moderator**- From: Winnipeg, Canada
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Could you please explain the detailed physics why orbital insertion would require entering a retrograde orbit? As far as I can see, you could directly enter a prograde orbit.

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**Quaoar****Member**- Registered: 2013-12-13
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RobertDyck wrote:

Could you please explain the detailed physics why orbital insertion would require entering a retrograde orbit? As far as I can see, you could directly enter a prograde orbit.

When I travel from a inner planet to an outer planet, I reach it at the aphelion of my hohmann transfer orbit, and the orbital speed of my ship is smaller than the orbital speed of the planet. So when I enter in the planet SOI, to minimize the deltaV of my insertion burn, I suppose to approach the planet with a retrograde hyperbola facing the dark side and firing my rocket with the nose of my ship in the direction of the insertion orbit...

But, wait a minute: can I also maneuver my ship to enter in the planet SOI with a prograde hyperbola facing the sunlight hemisphere of the planet? It's the same?

In this case the problem is solved.

*Last edited by Quaoar (2022-01-31 17:13:38)*

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**tahanson43206****Moderator**- Registered: 2018-04-27
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For Quaoar ... this question has been considered, and discussed, at some length by GW Johnson.

There is a topic called "Orbital Mechanics" where your question would seem to be a good fit.

Dr. Johnson may not have seen your question in the new topic.

In the case of Mars (if I recall correctly) the difference in velocity is on the order of 2 km per second. This is well worth considering if you do not have infinite fuel.

(th)

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**GW Johnson****Member**- From: McGregor, Texas USA
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Going from Earth to Mars, when you get to Mars via Hohmann transfer, you are moving parallel to Mars's motion, but slower by around 2 km/s. Mars is trying to run over you swiftly from behind. As the vehicle and Mars get closer, Mars gravity adds to the velocity difference by accelerating the vehicle toward the planet, so you have a lot more speed to deal with than just the 2 km/s. If you land direct, you can use hypersonic friction drag to dissipate the difference in velocities. But you have to strike near tangentially at the edge of the disk, to keep your angle below local horizontal very shallow (around 2 degrees). Otherwise, you have no chance for air friction to slow you, you just snack the planet at interplanetary speeds.

For a prograde orbit about Mars, it is "Mars's speed" plus the orbit speed that you have to deal with. If you look at a retrograde orbit about Mars, then it is "Mars's speed" less the orbit speed that you have to deal with. But the "Mars speed" I am talking about is the oncoming planet's 2 km/s plus the attraction of its gravity as it nears you. Off Hohmann, this is near 5.5 km/s. Low orbit speed is near 3.5 km/s.

So entering prograde orbit your delta-vee is nearer 5.5+3.5 = 9 km/s. Entering retrograde your delta-vee is nearer 5.5 - 3.5 = 2 km/s.

That same effect (on a smaller scale) is EXACTLY why Apollo orbited the moon retrograde.

You can estimate the gravitational speed-up effect as the difference "far" from Mars (the 2 km/s difference between Mars's speed and your speed about the sun) using Mars escape velocity: Vnear^2 = Vfar^2 + Vesc^2. Use escape at orbital altitude, not surface escape. They are not the same.

Returning to the Earth is different. Your transfer perihelion velocity is greater than Earth's orbital velocity about the sun, so you are effectively running into the Earth from behind. Correcting for the attraction of gravity, your Vnear at direct entry interface is around 12+ km/s. You need to strike tangential to the disk to have a low trajectory angle below horizontal, in order not to just smack the planet still at 12 km/s.

A prograde orbit is trying to move away from you, so the delta vee to enter it is your Vnear of 12 km/s, less the low orbit velocity of 8 km/s, for about 4 km/s. On the other hand a retrograde orbit is moving toward you as you approach, so that your Vnear of 12 km/s plus orbit speed of 8 km/s produces a 20 km/s delta vee, or thereabouts.

The lesson here is that moving out from the sun, you want to enter a retrograde orbit about your target planet. When moving in toward the sun, you want to ender a prograde orbit. It makes a big difference to the orbital transport ship's delta-vee, and there is no real way around that, that I understand. But I'm no expert.

GW

GW Johnson

McGregor, Texas

"There is nothing as expensive as a dead crew, especially one dead from a bad management decision"

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**Quaoar****Member**- Registered: 2013-12-13
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GW Johnson wrote:

Going from Earth to Mars, when you get to Mars via Hohmann transfer, you are moving parallel to Mars's motion, but slower by around 2 km/s. Mars is trying to run over you swiftly from behind. As the vehicle and Mars get closer, Mars gravity adds to the velocity difference by accelerating the vehicle toward the planet, so you have a lot more speed to deal with than just the 2 km/s. If you land direct, you can use hypersonic friction drag to dissipate the difference in velocities. But you have to strike near tangentially at the edge of the disk, to keep your angle below local horizontal very shallow (around 2 degrees). Otherwise, you have no chance for air friction to slow you, you just snack the planet at interplanetary speeds.

For a prograde orbit about Mars, it is "Mars's speed" plus the orbit speed that you have to deal with. If you look at a retrograde orbit about Mars, then it is "Mars's speed" less the orbit speed that you have to deal with. But the "Mars speed" I am talking about is the oncoming planet's 2 km/s plus the attraction of its gravity as it nears you. Off Hohmann, this is near 5.5 km/s. Low orbit speed is near 3.5 km/s.

So entering prograde orbit your delta-vee is nearer 5.5+3.5 = 9 km/s. Entering retrograde your delta-vee is nearer 5.5 - 3.5 = 2 km/s.That same effect (on a smaller scale) is EXACTLY why Apollo orbited the moon retrograde.

You can estimate the gravitational speed-up effect as the difference "far" from Mars (the 2 km/s difference between Mars's speed and your speed about the sun) using Mars escape velocity: Vnear^2 = Vfar^2 + Vesc^2. Use escape at orbital altitude, not surface escape. They are not the same.

Returning to the Earth is different. Your transfer perihelion velocity is greater than Earth's orbital velocity about the sun, so you are effectively running into the Earth from behind. Correcting for the attraction of gravity, your Vnear at direct entry interface is around 12+ km/s. You need to strike tangential to the disk to have a low trajectory angle below horizontal, in order not to just smack the planet still at 12 km/s.

A prograde orbit is trying to move away from you, so the delta vee to enter it is your Vnear of 12 km/s, less the low orbit velocity of 8 km/s, for about 4 km/s. On the other hand a retrograde orbit is moving toward you as you approach, so that your Vnear of 12 km/s plus orbit speed of 8 km/s produces a 20 km/s delta vee, or thereabouts.

The lesson here is that moving out from the sun, you want to enter a retrograde orbit about your target planet. When moving in toward the sun, you want to ender a prograde orbit. It makes a big difference to the orbital transport ship's delta-vee, and there is no real way around that, that I understand. But I'm no expert.

GW

Thanks GW is ever a pleasure to hear your lessons

So an orbit-to-orbit ship traveling from Venus to Earth in a Hohmann orbit, arrives in the Earth SOI with a hyperbolic speed of 2.5 km/s and reaches the perigee at 400 km of altitude (escape velocity 10.85 km/s) with a speed of SQR(2.5^2+10.85^2) = 11.13 km/s.

The orbital speed at 400 km of altitude is almost 7.67 km/s, so our ship will have a LEO insertion burn of 11.13-7.67 = 3.46 km/s of deltaV if her orbit is retrograde or 11.13+7.67 = 18.8 km/s (!!!) if her orbit is prograde.

18.8 km/s are prohibitive even for a solid core NTR, so our ship is forced to insert in a retrograde orbit as I thought (It's correct?).

It's very difficult to refurbish an orbit-to-orbit ship parked of a retrograde LEO, so a solution may be to insert our ship in a very high elongated retrograde orbit with the apogee at the L1 Earth-Sun Lagrange point (perigee velocity 10.82 km/s) and dock her on a deep space L1 habitat connected by shuttles to LEO: the perigee insertion burn will have 11.13-10.2 = 0.31 km/s of delta V plus almost 0.54 km/s of apogee L1 insertion burn. The whole maneuver has a deltaV of only 0.85 km/s, but the transfer orbit to Earth-Sun L1 adds 37.5 days to our Venus-Earth travel (plus other 37.5 days of shuttle travel from L1 to LEO).

However, the crew can leave the spaceship with a small entry vehicle and perform a direct EDL from the transfer orbit, while the mother ship continues her travel to L1 unmanned.

*Last edited by Quaoar (2022-02-04 10:38:46)*

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**GW Johnson****Member**- From: McGregor, Texas USA
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Hi Quaoar:

I think your notion of capturing into a highly-elliptic retrograde orbit about the Earth is the way to go. It reduces your orbit insertion delta-vee to do so, if you enter orbit at its perigee.

Then, way out at its apogee, where your orbital speed is actually quite low, you burn for a delta-vee at twice the orbit speed, to convert to prograde. Then you can burn once more at perigee to circularize prograde.

That's a lot of burns, and a significant amount of delta vee, but lots less than trying to enter prograde low orbit directly.

The limit here is two-fold: either (1) the highly-elliptic period is intolerably long, or (2) the moon or the sun can pull you out of orbit (orbital stability too weak).

I'm not sure how to figure that last, being the amateur orbital mechanic that I am. An uneducated guess says your apogee altitude ought to be short of where the moon's pull equals the Earth's pull, which is somewhere about 3/4 of the way to the moon or thereabouts.

Once your orbital transport is in low orbit prograde, you can launch a lot of surface-based transports up to refill and refit it, at min delta-vee required of the surface-based transports.

Whether this is better than launching retrograde from the surface depends mainly on how many you need. I'm showing a velocity at the Earth's equator due to its rotation of about 0.46 km/s. That says you add about 0.92 km/s (unfactored for gravity and drag losses) to the required launch dV (of surface orbit velocity 7.905 km/s), for a westward launch at the equator, a bit less further north or south. (The factor is typically 1.10). Factored, the mass ratio-effective dV's are then 8.700 km/s eastward, and 9.708 westward.

It's a tradeoff: burdening the orbital transport to reach low prograde orbit via the elongated elliptic capture orbit off the trajectory from Venus, versus burdening your surface-based supply fleet with a bit more delta-vee to launch westward. I don't know the "right" answer to that, or even if there is a "right" answer.

GW

*Last edited by GW Johnson (2022-02-04 14:12:26)*

GW Johnson

McGregor, Texas

"There is nothing as expensive as a dead crew, especially one dead from a bad management decision"

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**GW Johnson****Member**- From: McGregor, Texas USA
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After thinking about it, I might well be wrong thinking that one has to enter low orbit in a retrograde direction when outbound from the sun.

Prograde orbital direction at Mars (from Earth) or at Earth (from Venus) is parallel to planetary motion on the shadow side, but opposite it on the sunward side. It is the opposite orbit speed that gets subtracted from the vehicle's "near V" to determine the lower dV, while the parallel orbit speed gets added for the far higher dV.

The analog to the Apollo figure 8 orbital entry (retrograde on the side facing away from Earth) has the apohelion just past the center of Mars (or Earth) for the gravity-driven swing-around into the retrograde orbit direction, on the shadow side.

But, what if your transfer apohelion is a tad short of the center of Mars (or Earth) and you time things to arrive a bit before Mars (or Earth) can get there? The notion is to have the planet gravity pull you from Vfar to Vnear, except toward the planet on the sunward side, ending more-or-less tangent to the adjacent surface, probably around the backside (as referenced to planetary motion direction).

It would take a real orbital trajectory code capable of analyzing 3-body problems, to work this out. But my hunch is that there has to be a way to do this, so that journeys outbound from the sun can still enter low prograde orbits about planets, without incurring huge dV's to do so.

I have no such software. I only have the equations for the 2-body problem solution that is in all the old textbooks. That is what I implement in the spreadsheet stuff, with jigger factors for gravity and drag losses, and the Vnear vs Vfar energy correction for local 3rd-body gravity pull. What I do isn't navigation, it is just estimating reasonable design requirements for vehicle mass ratios.

Somebody with access to real 3-body software needs to answer this planetary orbit approach question for both Quaoar and me.

GW

GW Johnson

McGregor, Texas

"There is nothing as expensive as a dead crew, especially one dead from a bad management decision"

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**Quaoar****Member**- Registered: 2013-12-13
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GW Johnson wrote:

After thinking about it, I might well be wrong thinking that one has to enter low orbit in a retrograde direction when outbound from the sun.

Prograde orbital direction at Mars (from Earth) or at Earth (from Venus) is parallel to planetary motion on the shadow side, but opposite it on the sunward side. It is the opposite orbit speed that gets subtracted from the vehicle's "near V" to determine the lower dV, while the parallel orbit speed gets added for the far higher dV.

The analog to the Apollo figure 8 orbital entry (retrograde on the side facing away from Earth) has the apohelion just past the center of Mars (or Earth) for the gravity-driven swing-around into the retrograde orbit direction, on the shadow side.

But, what if your transfer apohelion is a tad short of the center of Mars (or Earth) and you time things to arrive a bit before Mars (or Earth) can get there? The notion is to have the planet gravity pull you from Vfar to Vnear, except toward the planet on the sunward side, ending more-or-less tangent to the adjacent surface, probably around the backside (as referenced to planetary motion direction).

It would take a real orbital trajectory code capable of analyzing 3-body problems, to work this out. But my hunch is that there has to be a way to do this, so that journeys outbound from the sun can still enter low prograde orbits about planets, without incurring huge dV's to do so.

I have no such software. I only have the equations for the 2-body problem solution that is in all the old textbooks. That is what I implement in the spreadsheet stuff, with jigger factors for gravity and drag losses, and the Vnear vs Vfar energy correction for local 3rd-body gravity pull. What I do isn't navigation, it is just estimating reasonable design requirements for vehicle mass ratios.

Somebody with access to real 3-body software needs to answer this planetary orbit approach question for both Quaoar and me.

GW

Thank's GW

Do you know if Mars Reconnaissance Orbiter and other probes insert themselves in prograde or retrograde orbits?

*Last edited by Quaoar (2022-02-05 10:51:14)*

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**tahanson43206****Moderator**- Registered: 2018-04-27
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For Quaoar,

Your question seems (to me at least) worthy of serious consideration by NASA customer service personnel. Please consider submitting your question to them, and (please) be sure to report their reply here.

For all ... we (forum) would certainly benefit if someone with current software for orbital calculations and skill in running it were to volunteer to help out.

See the Recruiting topic for procedures.

(th)

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**Quaoar****Member**- Registered: 2013-12-13
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kbd512 wrote:

With very high-Isp propulsion, is there some reason why you can't approach ahead of or behind a planet, and then slow down or catch up to be "captured" into orbit?

Is that not exactly how we've sent a growing number of these probes to investigate the outer planets and asteroids or comets?

If those Mach-effect thrusters actually work, then it's only a matter of supplying power, rather than propellant. At that point, the solution is to put more power behind the propulsion system or to simply accept longer transfer times and to appropriately shield the ship's occupants from radiation effects.

My novel is set in future where humanity is in decline and has lost interest in space exploration, so even if they live in the year 2222, they still use 1970 nuclear rocket technology with NERVA derived solid core NTR.

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**Quaoar****Member**- Registered: 2013-12-13
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GW Johnson wrote:

Somebody with access to real 3-body software needs to answer this planetary orbit approach question for both Quaoar and me.

GW

I've no 3-body software but I would try to solve the problem anyway using the patched conic approximation with impulsive maneuvers, given that the laws of mechanic are time and space symmetrical.

Case 1

Let's think to the inverse case: an ideal ship with a prograde and perfectly circular low Earth orbit at 400 km of altitude (orbital radius 6771 km). She fires her ideal rocket over the sunward quarter of the LEADING hemisphere, gaining instantaneously 3.46 km/s of deltaV, and escape in an hyperbolic orbit, flying over the sunward side of the Earth in opposite direction of the planetary motion vector.

When our ship crosses the edge of the Earth's sphere of influence (1,496,505 km of radius) she finds herself at the aphelion of a prograde elliptic Homann orbit direct to Venus.

It's quite easy to calculate the aphelion of this orbit: it is tangent to the asymptote of the escape hyperbola, so its radius is equal to the Sun-Earth distance at the insertion point minus the impact parameter (the distance of the asymptote from the Earth center) of the hyperbola. But I'm not able to do the same with the time, so let's simply call it T and say that our ship reaches the aphelion of the Earth-Venus transfer orbit T hours AFTER the Sun-Earth vector has passed this point.

Surely, the real trajectory of our ship will be different from my approximation, but we have sent many probes to Venus, so we know for sure that this kind of maneuver is possible.

Case 2

So let's imagine to use the other half of the same transfer orbit, traveling this time from Venus to Earth: this time our ship will reach the aphelion T hours BEFORE the Sun-Earth vector will pass this point: our ship will enter in the Earth SOI and find herself on a hyperbolic trajectory whose asymptote is tangent to the aphelion of the transfer orbit.

The arrival hyperbola is exactly specular to the departure one, so our ship will fly over the sunward quarter of the Earth's TRAILING hemisphere, fire her rocket, losing instantaneously 3.46 km/s of deltaV, and insert herself in a prograde LEO.

Even in this case, the real trajectory will be different from my approximation, but if Case 1 is possible (and it is), Case 2 must be possible the same.

*Last edited by Quaoar (2022-02-07 03:26:12)*

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**SpaceNut****Administrator**- From: New Hampshire
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Post made for 3-body software in orbital mechanics

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**Quaoar****Member**- Registered: 2013-12-13
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SpaceNut wrote:

Post made for 3-body software in orbital mechanics

I've found this study where is explained how to insert in a Mars prograde orbit coming from Earth, flying over the sunward hemisphere

chrome-extension://efaidnbmnnnibpcajpcglclefindmkaj/viewer.html?pdfurl=https%3A%2F%2Fhanspeterschaub.info%2FPapers%2FUnderGradStudents%2FConicReport.pdf&clen=1695636&chunk=true

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**GW Johnson****Member**- From: McGregor, Texas USA
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Quaoar:

I pretty much came to the same conclusion myself. Sunward capture at Earth is the way to enter a prograde orbit, when coming from Venus or inward. Same is true at Mars or outward, when coming from Earth.

GW

GW Johnson

McGregor, Texas

"There is nothing as expensive as a dead crew, especially one dead from a bad management decision"

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