For example, can a space ship overshoot Mars in a reverse gravitational slingshot mixed with a little air breaking, steering to keep the spacecraft from orbit and then land on Mars with a near 0 ground speed during entry? No sharp turns it would just be going slow enough to stall at the edge of the planet's gravity and start falling back to the planet. The picture is not a perfect model on how this could work. The space after each pass would give the heat shield time to cool.


Could a Dynastat enter near 0 MPH ground speed into the atmosphere of Mars?

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    $\begingroup$ Isn't this skip re-entry from a highly elliptical (or perhaps initially hyperbolic) orbit? $\endgroup$ – Anthony X Feb 11 '17 at 18:12
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    $\begingroup$ @Muze if that drawing is from someone else, or a website, you should show a link were it comes from. This is to give credit to the artist, and to allow people to read further about the image. $\endgroup$ – uhoh Feb 11 '17 at 18:52
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    $\begingroup$ Illustration: Andrews Space. From: newscientist.com/article/… $\endgroup$ – user10509 Feb 13 '17 at 8:47
  • $\begingroup$ You STILL haven't done your homework, as proven by your "reverse slingshot" picture. $\endgroup$ – SF. Feb 18 '17 at 7:57
  • $\begingroup$ @Muze: it's more like relativistic speed meteorite with half of it exploding upon touching the atmosphere. You still think in straight lines. $\endgroup$ – SF. Feb 18 '17 at 21:15

You can use skip reentry to gradually reduce your speed when arriving from an interplanetary journey. NASA also studies skip reentry for the Shuttle, as a way to reduce the heat load during reentry. So you could use this technique to get below orbital speed before the final reentry starts.

The lower your speed gets, the more difficult it will be to keep skipping. When you're in orbit, you're falling towards the planet (because its gravity attract you), you just keep missing the planet because your forward speed balances out the speed at which you drop.

When you reduce speed, you predominantly reduce the forward component of your speed vector. The downward force remains the same. So as your forward speed reduces, you'll need more and more lift to stay at the high altitude you need for the skip maneuver. There comes a point where the lift you need for the next skip is more than the lift your spacecraft can provide, and then you're committed to going down.

No sharp turns it would just be going slow enough to stall at the edge of the planet's gravity and start falling back to the planet.

That will not work out the way you think. You can put the spacecraft in a very elliptical orbit and make it 'stall' (sort of, you can get the speed at apogee to a very low value). But when you're past apogee and the spacecraft starts moving towards the planet, the gravitational pull will start to accelerate the spacecraft until you're at perigee with your original speed.

So any deceleration will have to be done by aerobraking (i.e. the skip reentry described above).

I suspect you can use a gravitational slingshot to reduce the spacecraft's speed only when the spacecraft is not in orbit. Once you're in orbit, you can't transfer any more momentum from the spacecraft to the planet.

In a gravitational slingshot, the spacecraft accelerates (under the gravitational pull of the planet) until you're at the closest approach point. Then the spacecraft will be decelerated while it moves away from the planet.

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    $\begingroup$ I wonder though, if a gravity assist against the same planet can be used to reduce capture burn. Just like you can perform a slingshot against Earth after launching from Earth (about a year earlier) - could you, e.g. use Mercury assist to get your orbit much closer to Mercury orbit, and then (half a year later) perform a standard powered capture into Mercury orbit, spending way less fuel than you would if doing Hohmann transfer from Earth directly. $\endgroup$ – SF. Feb 8 '17 at 9:40
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    $\begingroup$ Good question, go ahead and ask it :) $\endgroup$ – Hobbes Feb 8 '17 at 9:45
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    $\begingroup$ @Muze:That looks more as airbraking than a gravitational slingshot. First, how would you achieve these sharp turns? Then - passing the planet in direction as in the first pass, would be an accelerating assist, likely quite counterproductive. $\endgroup$ – SF. Feb 8 '17 at 18:09
  • $\begingroup$ @Muze edge of what?? $\endgroup$ – SF. Feb 8 '17 at 18:18
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    $\begingroup$ @muze: If it was airbraking, these would be ellipses of decreasing size, with one of the focus points common. Quadratically decreasing size, too. (O))-)---)-------)---------------). For gravity alone, without skirting the atmosphere - you'd be just in a stable elliptical orbit, period. $\endgroup$ – SF. Feb 8 '17 at 21:32

Sometimes, and not as well as you are after. The problem here is that you don't understand the gravitational slingshot maneuver.

In the reference frame of the planet there's no net velocity change. The spacecraft leaves with the same speed as it approaches, just on a different bearing.

It is only in the reference frame of a larger body that you see a speed gain (or loss--you can do the same thing in reverse in order to slow a spacecraft, say to get it close to the sun.)

If this were the whole story it would be completely useless for an approach. However, the same thing applies to moons and planets as applies to planets and the sun. You can do gravitational slingshots off a moon to shed velocity--that how planets capture asteroids and turn them into baby moons. You won't get below orbital velocity this way, though. It's a capture maneuver (not that it can be done in most places due to the lack of an adequate moon), not a landing maneuver.

It can also work the other way, ask the Kerbal (Kerbal Space Program. Play that and you'll get a better feel for orbital maneuvering than any amount of looking at formulas will give you) I've chased all the way to Eve. I didn't see whether it was the Mun or Minmus that caused the ejection.

  • $\begingroup$ You CAN get below orbital velocity that way - and drop like a rock to the planet from the moon's altitude. Rather pointless, but possible. $\endgroup$ – SF. Feb 11 '17 at 7:08
  • $\begingroup$ @SF You can put yourself on an impact trajectory but isn't your total velocity still at least orbital? (If you could actually slow below orbital it would make the aerobrake easier and thus not be pointless.) $\endgroup$ – Loren Pechtel Feb 11 '17 at 15:45
  • $\begingroup$ There are no moons good for this in Solar System , but imagine a moon in low orbit - would need to be massive but firm, e.g metallic, to survive below Roche Limit. Encounter "head on" (retrograde), flyby on the planet side, ejection radial out. With approach slow enough and moon massive enough you could be below orbital speed. If not, repeat. $\endgroup$ – SF. Feb 11 '17 at 18:26
  • $\begingroup$ @SF I don't think "firm" will protect a massive moon below the Roche limit. $\endgroup$ – Loren Pechtel Feb 11 '17 at 21:42
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    $\begingroup$ Primarily, Saturn has a couple moons within its rings, and these don't seem to be getting torn apart at all. $\endgroup$ – SF. Feb 17 '17 at 14:56

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