Talk:Gravity assist

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Saturn example

For the record, the Saturn example was somewhat imprecise. It claimed the delta-v for the Hohmann transfer from Earth to Saturn is 15.7 km/s. However, this figure disregards two effects that alter the situation drastically:

The planets' own gravity. To get from a low-Earth orbit to a low-Saturn orbit actually requires about 18 km/s.
Aerobraking. Upon arriving at Saturn, it's only necessary to alter the spacecraft's orbit around Saturn from a hyperbola to an ellipse with a periapsis passing through Saturn's atmosphere; after that, subsequent passes through periapsis will circularize the orbit. Taking this into account leads to a delta-v more like 10 km/s.

I added "disregarding Earth's and Saturn's own gravity wells". (I forgot to mention aerobraking, but I figure that's not worth disrupting the flow of the sentence any more than I already have.) --Doradus 21:09, Jun 18, 2005 (UTC)

For clarity, it ought to be a comparion of the same mission flown with and without the slingshot. What happens at the destination should be the same, whether it's a flyby, braking into orbit with a rocket, or aerobraking. Since the paragraph uses Cassini as an example, what would the total delta-V be, going directly to the same orbit around Saturn? The cited 2 km/s figure obviously doesn't include Cassini's intial boost.

—wwoods 00:04, 19 Jun 2005 (UTC)

Question on powered slingshot

Under the section Powered slingshots, does the last sentence of the first paragraph make sense? "The extra energy comes from the propellant being "left behind" in the planet's gravity well." ~ Rollo44 15:02, 9 February 2007 (UTC)[reply]

No, it's nonsense. If that were true they could just dump some propellant to "leave it behind" - Rod57 (talk) 14:56, 29 November 2019 (UTC)[reply]

Error in the lead? No

to decelerate a spacecraft (useful when traveling to an inner planet) or accelerate a spacecraft (useful when traveling to an outer planet).

I have always thought it should be vice versa, the inner planets are traveling much faster than the outer. (Igny (talk) 16:26, 6 October 2008 (UTC))[reply]

All explained in, eg. Hohmann transfer. You slow down to drop your perihelion or shrink the orbit. You speed up to raise your aphelion or expand the orbit. - Rod57 (talk) 14:47, 29 November 2019 (UTC)[reply]

Data/source for animations?

What is the data and/or source of the data for the animations of the satellite launches? The animations are very pretty but they are not cited or defined in any way. BigBadAlgae (talk) 11:42, 29 March 2020 (UTC)BigBadAlgae[reply]

Regarding the animations, a tick-mark per frame (or every N frame) could visually emphasize the speed change, because the distance between ticks would change. --146.233.255.212 (talk) 17:42, 29 August 2022 (UTC)[reply]

Orphaned references in Gravity assist

I check pages listed in Category:Pages with incorrect ref formatting to try to fix reference errors. One of the things I do is look for content for orphaned references in wikilinked articles. I have found content for some of Gravity assist's orphans, the problem is that I found more than one version. I can't determine which (if any) is correct for this article, so I am asking for a sentient editor to look it over and copy the correct ref content into this article.

Reference named "PionOdys":

From Pioneer 10: Fimmel, R. O.; W. Swindell; E. Burgess (1974). SP-349/396 PIONEER ODYSSEY. NASA-Ames Research Center. SP-349/396. Retrieved January 9, 2011.
From Pioneer 11: Fimmel, R. O.; Swindell, W.; Burgess, E. Pioneer Odyssey. SP-349/396. Washington, D.C.: NASA-Ames Research Center. OCLC 3211441. Retrieved January 9, 2011.

I apologize if any of the above are effectively identical; I am just a simple computer program, so I can't determine whether minor differences are significant or not. AnomieBOT ⚡ 02:22, 18 April 2021 (UTC)[reply]

And Ulsses?

OK, momentum from the planet's orbital movement is transferred to the spacecraft. This means: if the planet stood still (sun reference frame) gravity assist would not work. However: how then could Jupiter be used to shoot Ulysses out of the ecliptic plane? Jupiter's orbit has no component perpendicular to the ecliptic plane. Or was is just a deviation with the absolute value of Ulysses's velocity unchanged? If so, why didn't they shoot Ulyssed off the plane right away? -- Wassermaus (talk) 07:08, 6 May 2021 (UTC)[reply]

Jupiter's orbit is slightly inclined to Earth's elliptical plane by about 1.3° degrees.
Shooting the spacecraft off the plane would have required a much bigger rocket than the Space Shuttle.
For the calculations involved, see here, p. 10.

Hawkeye7 (discuss) 09:26, 6 May 2021 (UTC)[reply]

Jupiter doesn't need any velocity component perpendicular for this to work. Ulysses and Jupiter had zero velocity perpendicular to the ecliptic before the encounter (well, nearly zero). After the encounter, Ulysses had a sizable northward velocity, and Jupiter has a very small southward velocity.

Vz_Jupiter = - Vz_Ulysses * (m_Ulysses/m_Jupiter) So momentum was conserved and exchanged between Ulysses and Jupiter. Fcrary (talk) 17:30, 6 May 2021 (UTC)[reply]

Too much irrelevant detail

Too much irrelevant detail in the summary of Voyager 1 & 2 (and maybe other missions). Only the details of the trajectory and gravity assist are needed in this article ? - Rod57 (talk) 19:52, 26 September 2021 (UTC)[reply]

Incorrect Derivation

I removed the section "Derivation" as it was incorrect.

The elastic collision equations used in the section were for a one-dimensional space. A flyby in a one-dimensional space will result in a big "splotch", that is a crash with no bouncing back :)

You should not expect an elastic collision between a spacecraft and a planet/star if they really come in contact. The beauty of gravity assist is that is the spacecraft seems to "bounce" off the planet/star without ever coming in contact.

Nor should you expect to be able to plot the trajectory of a spacecraft with just equations for the conservation of linear momentum and kinetic energy without solving differential equations or using parabolic/hyperbolic trajectories (which are derived from solving differential equations).

The derivation for a simplified two-dimensional space gravity assist (where all relevant vectors lie in the same two-dimensional plane) is not difficult. Approximate the planet's path during the flyby to be a straight line, with a velocity u2. Then change the inertial frame so that the planet is stationary. A flyby where the spacecraft is not captured by the planet providing the gravity assist has a parabolic or hyperbolic trajectory (both of which are symmetric) in this inertial frame. The component of velocity of the spacecraft along the axis of the planet's travel prior to the flyby increases from u11 to u11 + u2 as you change the inertial frame.

Therefore, the *magnitudes* of the two components of velocity after the flyby by symmetry will be: v11 = u11 + u2 and v21 = u21 in the changed inertial frame (at the same distance from the vertex), and v11 = u11 + 2*u2 and v21 = u21 in the original inertial frame (where the planet was traveling with a velocity of magnitude u2).

The deleted section mistakenly ignored v21 = u21 (the component of velocity orthogonal to the planet's velocity) etc.

Also note that the kinetic energy of the spacecraft of mass m after the flyby will be (1/2)*m*((u11 + 2*u2)^2 + u21^2) compared to (1/2)*m*(u11^2 + u21^2) prior to the flyby.

I will add the above to the article when I have the time. If someone else wants to add it, please do so.

JS (talk) 21:24, 4 October 2022 (UTC)[reply]

I think this section should be restored. The one dimensional case is an intuitive illustration of the basic principle. That's useful even it it doesn't capture the true, two-dimensional nature of a flyby. We should, however, add text to the effect that the true process is two dimensional, and this is just an illustration. Fcrary (talk) 21:39, 6 October 2022 (UTC)[reply]

I think the basic idea of a gravity flyby is how a planet/star may provide momentum and kinetic energy to a spacecraft without ever having come in contact with it.

Giving an example where the spacecraft actually crashes into the planet and then bounces off is rather misleading in my opinion. Another aspect of a gravity assist is the variable acceleration that the spacecraft experiences as the distance from the planet/star changes. This requires calculus to analyze and is missed in the elastic collision example.

However, if no one is ready to spend the time to add a two-dimensional section, and you feel the deleted section is better than no section at all, then please restore it.

The simplified two-dimensional case is not hard to explain (in the real world it will really be three-dimensional), and does capture the mechanism of momentum and kinetic energy transfer to a spacecraft without contact. I have already provided the "derivation" in my post above, but it needs to be more "formal".

If later I can add the two-dimensional section I will add it and remove the one-dimensional section. It may take a bit of time as I have other things going on, and I have worked this out myself so I do not have any references handy.

Best,

JS (talk) 04:37, 7 October 2022 (UTC)[reply]

Is this super-villainy subplot really warranted in the article?

> Another potential application of gravity assist is alteration of the Earth's orbital distance from the Sun to reduce increasing global temperatures.

Thus paragraph sounds like a mad scientist running amok. Firstly, a civilization capable of such a grand feat would have already mastered controlled thermonuclear fusion and tapping of Casimir-vacuum for clean energy and thus need not worry about AGW. On the other hand, significantly displacing Earth and Moon within the Solar System might make other planets deviate as a side-effect and maybe end with a collision of (AGW-free) Earth and a perturbed asteroid, exterminating all life... 78.131.76.69 (talk) 20:15, 15 December 2022 (UTC)[reply]

'space itself is dragged at the speed of light in the same direction as the black hole's spin'???

Not sure this is correct: 'General relativity predicts that a spinning black hole is surrounded by a region of space, called the ergosphere, within which standing still (with respect to the black hole's spin) is impossible, because .'

I think that would only be the case in the limit where the black hole is spinning so fast it can't grow anymore, and in all other cases, a spinning black hole would cause the space around it to spiral inwards rather than moving in exactly circular motion. 125.253.110.41 (talk) 06:45, 21 August 2023 (UTC)[reply]