# Talk:Gravity drag

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## Example section

I hope I have not started an edit war by re-deleting the example. Personally I found it not only unnecessary (because the text of the article explains the concept quite well), but also confusing because its relation to the topic of gravity drag was never explained. --Doradus 03:36, Dec 6, 2004 (UTC)

It says
When applying delta-v against gravity to increase specific orbital energy, it is advantageous to spend the delta-v as early as possible, rather than spending some, being decelerated by gravity, then spending some more.
This is demonstrated with the simple concrete example.--Patrick 09:55, Dec 15, 2004 (UTC)
Sorry chief, you're the victim of your own clear explanation. Looking at the example again, I just can't imagine someone not understanding the concept, then reading the example and saying "oh! NOW I get it!". --Doradus 21:42, Dec 15, 2004 (UTC)
More importantly, it claimed to be an example of gravity drag, yet the words "gravity drag" never appeared in the example, which makes it a poor example by inspection, but is also symptomatic of the example's lack of clarity and precision. A person who doesn't "get" gravity drag from your perfectly adequate explanation is going to need some serious hand-holding. Crafting a good example will take some work, but if you think it's important, we can give it a try. --Doradus 04:35, Dec 16, 2004 (UTC)

## Relativity

Are we sure "gravity drag" is a real term from relativity? I think we need a better source for this than a single geocities page. Also, the relativistic effect is unrelated to the astrodynamical one (which is purely classical), so this article should be restructured to reflect this. --Doradus 21:40, Dec 15, 2004 (UTC)

seems to me they're the same -- help me out if i don't understand it fully -- i'll clarify to show that it's not strictly relativistic ... just describes the phenomenon. [1] Ungtss 22:40, 15 Dec 2004 (UTC)
Firstly, why are you linking to a shadow of Wikipedia, when you could just go to Gravity? :-)
Secondly, I think you're confusing 'gravity drag' with frame-dragging. the text on the link says that the term is the author's invention: "I call this effect gravity drag". He's not a reliable source. He's certainly wrong about the tidal locking of moons to their primaries, which was understood without any need for relativity. I think he's also wrong about Mercury, which is precessing but not losing energy. Double stars rotating fast enough to emit gravitational radiation could be said to be experiencing drag, but I don't recall anyone doing so. In sum, this section should be cut.
—wwoods 00:53, 16 Dec 2004 (UTC)
well alright then, i'll yield to the more educated mind -- cut away:). Ungtss 02:18, 16 Dec 2004 (UTC)

## Bogosity?

"It might be expected that the gravity losses would up to 9.8 m/s per second of thrusting. However, by thrusting at an angle wherever possible, the overall losses are reduced by Pythagorus."

Um no. If you thrust at, say, 4g diagonally at the right angle so that you get 1g vertical component, do the maths and you get 3.87g horizontal component. So it's more efficient to thrust diagonally than to have two engines, one vertical and one horizontal. It really is due to pythagorus, and it really does reduce the gravity losses.

Ok, I see what you mean. You're just reminding us that forces are a vector quantity. That's a good point that was missing from the article. I don't think an unexplained reference to Pythagoras is sufficient, though... I'll see if I can think of anything better. --Doradus 20:52, Jun 17, 2005 (UTC)

### Centrifugal Force

A may be pedantic, but centrifugal force is ficticious. Second, the reason we want horizontal speed is never explained. I am guessing that it is due to the fact that at a sufficiently high velocity you enter a state of constant freefall, but this is just a guess.

It's not really ficticious in a rotating frame of reference, such as an object rotating around the center of the Earth. Or if you think it is ficticious, many other everyday forces in physics are also ficticious.WolfKeeper 02:24, 15 July 2006 (UTC)

This article says: "horizontal speed... provides a centrifugal force". Speed providing force is a very poor thing to say without a large amount of explanation about the reference frame in use and real vs. ficticious forces. Also, it is stated as a given when it also depends upon the location of the object! There are better ways to describe this effect without adding to the confusion.

"Fictitious force" doesn't mean "bogus term", it's merely indication that the apparent force comes from the frame of reference being non-inertial, but it still serves a perfectly valid tool for calculations. The article on Coriolis force is doing pretty fine; one could similarly protest the imaginary number. Gravity drag is fictitious, because it's "tainted" by centrifugal force component due to the "horizontal" speed being in fact circular motion above (round) planet's surface. It absolutely doesn't make it "bogus" though, merely classifying it as a specific class of forces. Sharpfang (talk) 08:13, 9 February 2017 (UTC)

## Vector Considerations

The comment about directing the thrust downwards to increase acceleration would presuppose either a really weak surface or that the rocket is already airborne. Since gravity drag seems to be concerned with getting up higher, this example seems silly and confusing (gravity generally tends to make it easy to accelerate downwards.) 128.151.161.49 Iain Marcuson

I removed that.--Patrick 06:39, 21 June 2007 (UTC)
I've restored it, hopefully in a way that highlights that the downward thrust is absurd. I also removed the remark about GEO versus escape velocity, because that has little to do with gravity drag. It is a general statement about Hohmann transfer orbits. --Doradus 23:52, 15 November 2007 (UTC)

## Numerical examples

It would really help me as a reader if the article included a few real-world examples with actual numbers for specific spacecraft launch trajectories. Is data available for the gravity drag penalty incurred by the Space Shuttle on a launch to the ISS? Would other vehicles (e.g. Soyuz with Progress or Ariane with ATV or H-IIB with HTV) pay a lower penalty because (without humans aboard) they can accelerate more rapidly? Also, this would help make clear what units (is it meters per second squared ?) would be appropriate for measures of gravity drag. (P.S.: I don't mean the article needs to show how these values are derived -- just a table of the results would be fine. ;-) (sdsds - talk) 05:18, 16 July 2007 (UTC)

## New equation

In the new section Formal expression the formula

${\displaystyle {\frac {dp_{r}}{dt}}=-\nabla (p_{v}\cdot g)-{\frac {1}{m}}\nabla (p_{v}\cdot A)\,.}$

is quoted from the paper General Theory of Optimal Trajectory for Rocket Flight in a Resisting Medium, with the explanation:

"The first term on the right hand side is the part of the force which may be called "gravity drag".

I am struggling to see the relevance of this equation to "gravity drag" because:

1. The paper makes no mention of the terms "gravity drag" or "gravity losses" in relation to this equation (or anywhere else that I can see).
2. In the equation quoted, the vector A is identified as "aerodynamic force" i.e. lift (see Figure 1 in the paper). As there is no lift in space, this term will be 0 for a spacecraft in orbit.
3. The vectors pr and pv appear from context to be momentum vectors, but their connectrion to physical characteristics of the spacecraft is not made clear.

Can anyone provide a clearer explanation of the relevance of this equation ? Gandalf61 (talk) 08:48, 12 September 2008 (UTC)

1. It does not surprise me that the paper does not name the term since it does not require lengthy discussion in the text as some other terms do, e.g. the aerodynamic lift and drag. Its appearance in the equation is sufficient for defining the theory. However, I agree that this reference does not support my identification of that particular term (rather than something else in the theory) as the "gravity drag".
2. The vector A is the whole aerodynamic force including both aerodynamic lift and aerodynamic drag. Unfortunately, the reference seems to omit the effect of the Earth's rotation on the atmosphere. So in the equations (3) for the lift and drag, one should replace V by V-Vatm with ${\displaystyle \mathbf {V} _{atm}=\mathbf {\omega } _{\oplus }\times \mathbf {r} \,.}$ Even in outer space where the atmosphere is negligible, this theory should apply. Simply replace the density of the atmosphere by zero and many of the equations should be simplified.
3. At the end of section 2 of the reference, it says "The solution is obtained by integrating the systems of Eqs. (7) and (9), subject to specified end-conditions, and selecting the control T and A in their bounded spaces such that at each instant the Hamiltonian defined by (8) is an absolute maximum.". Thus the relationship of the "momentum" vectors to the coordinates is quite complicated. Actually, since this is a optimal control problem rather than a purely physical phenomenon, the Hamiltonian refers to the utility of the trajectory rather than the action. Thus the "momenta" are really more analogous to prices, that is, the marginal utility of an increase in the corresponding coordinate at that time.
I hope I can give you more information later after I have studied this more. JRSpriggs (talk) 06:36, 13 September 2008 (UTC)

I too am concerned that the Vinh 1973 reference is not directly discussing the "gravity drag" concept. The Vinh paper is about vehicles that are powered by a rocket engines but that rely on aerodynamic lift for flight. That puts their flights outside the discipline of astrodynamics, which is the scope for the article as specified in its lead section. (sdsds - talk) 07:40, 13 September 2008 (UTC)

Agreed. To properly explain the relevance of this equation I think the article needs to include a couple of extra things:
• A full explanation of the meaning of its variables, especially A, pr and pv.
• A version of the equation that applies to outer space - ideally, to avoid OR, I guess this should come from a reference that deals with the dynamics of space travel rather than atmospheric flight.
Also, I am now confused by two other issues:
1. The term ${\displaystyle -\nabla (\mathbf {p} _{v}\cdot \mathbf {g} )}$ that is identified with "gravity drag" is a vector quantity, whereas "gravity drag" appears to be a scalar quantity (like delta-v). Should "gravity drag" actually be the magnitude of ${\displaystyle -\nabla (\mathbf {p} _{v}\cdot \mathbf {g} )}$ ?
2. I think from the definition of pv in the paper that it has dimensions L. In which case the dimensions of ${\displaystyle \mathbf {p} _{v}\cdot \mathbf {g} }$ are L2T-2 and the dimensions of ${\displaystyle -\nabla (\mathbf {p} _{v}\cdot \mathbf {g} )}$ are LT-2. But surely "gravity drag", like delta-v, has dimensions LT-1 ? The dimensions don't seem to match, do they ?
Hoping JRSpriggs or someone else can shed some more light on this. Gandalf61 (talk) 08:12, 13 September 2008 (UTC)

### Formal expression

According to the General Theory of Optimal Trajectory for Rocket Flight in a Resisting Medium (see equation 30 at the external link below),

${\displaystyle {\frac {d\mathbf {p} _{r}}{dt}}=-\nabla (\mathbf {p} _{v}\cdot \mathbf {g} )-{\frac {1}{m}}\nabla (\mathbf {p} _{v}\cdot \mathbf {A} )\,}$

where ${\displaystyle \mathbf {p} _{r}\,}$ is the amount of fuel that would be saved in the portion of the flight so far per unit change in the current position (kilograms per meter), ${\displaystyle \mathbf {p} _{v}\,}$ is the amount of fuel that would be saved in the portion of the flight so far per unit change in the current velocity (kilogram seconds per meter), ${\displaystyle p_{m}\,}$ is the amount of fuel that would be saved in the portion of the flight so far per unit change in the current mass (1), t is the current time (units seconds), ${\displaystyle \mathbf {g} \,}$ is the acceleration of gravity at the current location (units meters per second squared), and ${\displaystyle \mathbf {A} \,}$ is the current aerodynamic force (lift plus drag) on the rocket (units kilogram meters per second squared).

The first term on the right hand side is the part of the "force" which may be called "gravity drag".

I moved the above section here from the article because Gandalf61 has raised questions which I cannot answer presently. Thus it appears that it may be incorrect in some significant respects. JRSpriggs (talk) 20:15, 13 September 2008 (UTC)

There's several things here. There's the instantaneous gravity drag, and there's the overall effect on the delta-v that the vehicle needs which is the integral over time for a particular launch. Both are not uncommonly referred to as gravity drag. The thing is that the gravity drag is an acceleration due to gravity which is independent of the weight of the vehicle, hence you can divide through the expression by the mass to get an acceleration, which when you integrate up over a mission, you get a velocity (the velocity vector delta-v (physics) due to gravity drag). There's also the extra delta-v that the vehicle needs to expend to oppose the gravity drag. That's very much a scalar, it's a measure of fuel burnt.- (User) Wolfkeeper (Talk) 22:36, 13 September 2008 (UTC)
I'm sure it all seems hopelessly complex, ill thought-out, poorly referenced, obscure etc. etc. to people who haven't thought about these issues much, and at first blush it seems to violate most of the rules of physics. But it's more that rockets have that awkward exhaust where plenty of momentum and energy end up, and so, for example, you have bizarre behaviours like the Oberth effect where the same rocket almost magically gains extra speed in some situations. But really it's the exhaust making the books balance. I guess this is why it's called 'rocket science' sometimes. Gravity drag is another one of those oddities. And yeah, gravity drag is negative when the rocket points down too, check out the dot product(!)- (User) Wolfkeeper (Talk) 22:36, 13 September 2008 (UTC)
Let me try to make sure I understand this clearly. Are you saying:
1. One definition of "gravity drag" is a velocity vector, and is the integral of the acceleration due to gravity (i.e. weight/mass) over a mission (or perhaps over a phase of a mission).
2. The other definition of "gravity drag" is a scalar which is the additional delta-v that the rocket needs to expend to oppose gravity.
Have I got that right ? How are the vector and scalar definitions connected - is the scalar "gravity drag" equal to the magnitude of the vector "gravity drag" ? Gandalf61 (talk) 08:22, 14 September 2008 (UTC)
Let's clarify this. Gravity drag is a force, and thus, a vector.. It's the difference between weight and centrifugal force (making it a fictitious force, applicable in non-inertial frame of reference of an accelerating spacecraft in flight over (round) planet. Its frame of reference is bound to the planet's surface at craft's nadir and moves as the craft moves, rotating with planet's curvature.
It bears many similarities to atmospheric drag. It's relation to velocity is quadratic (although the respective "immobile" is in circular orbit). It's relation to local gravitational acceleration is linear similarly to air drag's relation to air density (think "density of gravity"). It degrades to weight (as understood in aviation) at low speeds; substituting it for weight in aviation equations allows graceful adaptation of them for calculations concerning spaceplanes - gravity drag counteracts lift, while atmospheric drag counteracts thrust, comprising the four fundamental forces acting on an airplane.
The expression causing worst confusion is "gravitational losses". It can be understood momentarily, as a scalar proportional to magnitude of gravitational drag, magnitude of gravitational acceleration as experienced by craft in given conditions (m/s^2); or it can be integrated over ascent time through adapting F=ma, v=at; the result is gravitational loss to delta-V in m/s. - or even as energy spent to produce thrust needed to produce given delta-V, in Joules, or momentary power output needed to counteract it, in Watts. The term "gravitational losses" is poorly defined. Well-defined gravitational drag is a helpful expression that allows you to find "gravitational losses" in whatever terms you desire, being the neutral, fundamental quantity behind all of the many forms of "gravitational losses".
this answer presents a simple overview of mathematics behind gravity drag, although obviously I can't submit this - "original research". Sharpfang (talk) 09:00, 9 February 2017 (UTC)
The vectors pr and pv in the Vinh paper are not momentum vectors. The pv vector is the "primer vector". this paper gives an overview of primer vector theory. The primer vector points in the direction of thrust (forwards). It has a magnitude which is typically close to 1, and its interpretation is that if its magnitude exceeds 1 your rocket should be firing at full thrust, while if it is below 1 you should insert a coast phase, and is an example of Bang-bang control. In equation 37 on page 199 Vinh has the equations for rocket flight in free space which eliminates the aerodynamic forces.
${\displaystyle {\frac {d{\vec {p}}_{r}}{dt}}=-\nabla ({\vec {p}}_{v}\cdot {\vec {g}}),{\frac {d{\vec {p}}_{v}}{dt}}=-{\vec {p}}_{r},{\frac {dp_{m}}{dt}}={\frac {1}{m^{2}}}\left\Vert p_{v}\right\|T}$
${\displaystyle {\frac {d^{2}{\vec {p}}_{v}}{dt^{2}}}=\nabla ({\vec {g}}\cdot {\vec {p}}_{v})}$
While that formulation is very elegant and you can read off "thrust vector dotted into gravity vector" so it becomes obvious that it is zero when thrust is horizontal to the gravity field, I'm not sure if you can usefully include that since it requires a background in primer vector theory to really be able to "read" it. Lamontcg (talk) 01:20, 31 October 2017 (UTC)

## Original research

This article has no citations at all, and thus no proof is given that anyone in the real world uses the word drag to refer to the gravitational resistance on a space vehicle. Does it need to be moved to a new title? I don't think so; it seems to actually fit the intended meaning of WP:Original research: someone's presentation of a subject, and probably should be deleted. The topic is covered in Flight dynamics (spacecraft)#Powered flight. After powered flight (i.e. achieving orbit or greater), it's generally just considered gravity, not drag. It has nothing at all to do with Drag (physics). JustinTime55 (talk) 16:49, 21 February 2014 (UTC)

Gravity drag is a term in common use in the space community though. A search for phrase "gravity drag" through KSP forum produces 373 hits. ("gravitational drag", another variant of the same, yields additional 30.) space.stackexchange.com yields 30+4. I'm yet to see it in any scientific papers (doesn't mean it's absent there) but it definitely deserves an article, to serve as aid for people who encounter it for the first time - even if actual, peer-reviewed materials are hard to come by. Sharpfang (talk) 08:02, 9 February 2017 (UTC)