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[[User:Jayanta Sen|JS]] ([[User talk:Jayanta Sen|talk]]) 21:24, 4 October 2022 (UTC)
[[User:Jayanta Sen|JS]] ([[User talk:Jayanta Sen|talk]]) 21:24, 4 October 2022 (UTC)

: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. [[User:Fcrary|Fcrary]] ([[User talk:Fcrary|talk]]) 21:39, 6 October 2022 (UTC)

Revision as of 21:39, 6 October 2022

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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:

  1. The planets' own gravity. To get from a low-Earth orbit to a low-Saturn orbit actually requires about 18 km/s.
  2. 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]

When

In the text I found this:

"That discovery was made by Michael Minovitch in 1961.[5] The gravity assist was first used in 1959 and relied on research performed"

How it's possible it has be used before discovery? Maybe I miss something.... Sorry for my bad english.

Indaco1 (talk) 03:22, 16 April 2008 (UTC)[reply]

Minovitch developed practical computer codes to plan gravity assists, and may have discovered the planetary alignments later used for Voyager. - Rod57 (talk) 14:41, 29 November 2019 (UTC)[reply]

Historical origins

I think the article should mention that in popular culture the slingshot is considered to be Arthur C. Clarke's invention, because he was the first to use the term in a popular medium (2001) and had some influence in several missions that used it.

anonymous, 00:26:03 UTC, 2008/05/19

Michael Minovitch was not the first person to realize that a spacecraft could use the gravitational assist. This link gives some examples that far predate 1960 http://www.daviddarling.info/encyclopedia/G/gravityassist.html

Quorsav (talk) 20:00, 13 February 2009 (UTC)[reply]

Minovitch might have been the first to develop practical algorithms and computer codes for gravity assist planning. Heavily used for the Voyager mission planning. - Rod57 (talk) 14:37, 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]

  1. Jupiter's orbit is slightly inclined to Earth's elliptical plane by about 1.3° degrees.
  2. Shooting the spacecraft off the plane would have required a much bigger rocket than the Space Shuttle.
  3. 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]