Talk:Equations for a falling body

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what would be great- imho if the article confined itself to the math, but for each of the equations had two references: one to a page where the equation is derived and one to a page of the most relevant history of that derivation. 209.77.152.101 (talk) 14:25, 28 January 2016 (UTC)daniel[reply]

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Are there any equations already worked out to show the changes in height, velocity, and accelecteration with respect to the change in gravitional acceleration due to the proximity to a the massive body.

Possibly change equations?

I don't like how it uses Earth's acceleration due to gravity. Should the equations be changed to these?

Distance d travelled by an object falling for time t:	$\ d={\frac {Gmt^{2}}{2r^{2}}}$
Time t taken for an object to fall distance d:	$\ t=\ {\sqrt {\frac {2dr^{2}}{Gm}}}$
Instantaneous velocity v_i of a falling object after elapsed time t:	$\ v_{i}={\frac {Gmt}{r^{2}}}$
Instantaneous velocity v_i of a falling object that has travelled distance d:	$\ v_{i}={\sqrt {\frac {2Gmd}{r^{2}}}}\$
Average velocity v_a of an object that has been falling for time t (averaged over time):	$\ v_{a}={\frac {Gmt}{2r^{2}}}$
Average velocity v_a of a falling object that has travelled distance d (averaged over time):	$\ v_{a}={\frac {\sqrt {2Gmd}}{2r}}\$

with G being the gravitational constant and r being the distance from the center of the main body

Tell me if I made a mistake in these, I'm not used to working with Wikipedia's math coding. Yanah 20:40, 17 October 2006 (UTC)[reply]

Well, g is more convenient than GM/r² for Earth uses. I would favor the equations in terms of g, with the qualifier that for other planets and for significantly different radii than Earth's, g must be replaced with GM/r² where the M and r are chosen appropriately as other than Earth-surface normal. S B H arris 22:39, 17 October 2006 (UTC)[reply]

- I agree with you, these equations are way too dependant. I also think it would be better for first-time readers if we clarify m/s² as (m/s)². -HuBmaN!!!! 06:16, 19 November 2006 (UTC)[reply]

- - Huh? If you're talking about acceleration or gee, the units are m/s² and certainly not (m/s)² = m²/s². S B H arris 06:42, 19 November 2006 (UTC)[reply]

If you are trying to perform a calculation like landing an object on the Moon both g and r change with time. The integral formula may be better. Andrew Swallow 06:31, 20 September 2007 (UTC)[reply]

Might this get into the article?

People have survived falls from airplanes and from tall buildings.^[1]

--Ancheta Wis (talk) 19:50, 4 January 2008 (UTC)[reply]

References

^ http://www.nytimes.com/2008/01/04/nyregion/04fall.html?pagewanted=2&ei=5087&em&en=a6cd1e14f02e1178&ex=1199595600 New York Times January/4/2008: Man survives fall from 47-story building.

Galileo's priority?

Galileo was the first to demonstrate and then formulate these equations. Not according to the Wikipedia article Two New Sciences. At least not as far as one of them is concerned. Eroica (talk) 13:15, 8 July 2010 (UTC)[reply]

Equations for a falling body

Just thinking; the following are either ignored or missed while formulating the equation for falling mass.

An equation of instantaneous velocity for falling mass is v = g x t in which a magnitude of 9.8 m/s/s [acceleration] can't be preceded in the very first infinitesimal falling of an object as g also incepts from zero [besides v and t] and should increase gradually to its constant value of 9.8 m/s/s.

Similarly "v" is directly proportional to "t" but there is no chance at all for the numerical value of v to meet with numerical value of t in the very first development of mass falling as v is always ahead of t at 9.8 times t [9.8xt].

Since g = GM/R^2; where M = mass of earth and R = the on centre distance between gravitating earth and a resting mass [radius of earth]. So this means g decreases by increasing the on centre distance of masses and also it is an axiomatic truth that g decreases with altitude. Therefore should there be any range beyond which it starts varying in equation.68.147.41.231 (talk) 18:36, 3 November 2010 (UTC)khattak#1[reply]

The equation v = gt presumes that g is constant, and is not valid over regions where r changes enough that g varies (your second equation). S B H arris 19:42, 3 November 2010 (UTC)[reply]

Over a distance where the radius (R) changes significantly, it is preferable to integrate the variable force (g) over the change in radius to obtain the work done by gravity, then equate this to the change in energy. Over short ranges, the simple equations give the same answer to a good degree of accuracy. It is wrong to say that g starts from zero. It is equal to GM/R^2 before the body starts to fall. Dbfirs 08:27, 28 December 2010 (UTC)[reply]

But Newton didn't know that g= GM/R^2 = 9.8 m/s/s as G of the Newton idea [F=GMm/R^2, g= GM/R^2 = 9.8 m/s/s] was not discovered untill 71 years of his death. Newton had a problem of action at a distance and his gist based on no physical connection between point masses while all masses of the tortion balance (balls, spheres, wires, wooden rod, frame of torsion balance etc) used in Cavendish experiment are physically connected to each other one way or the other. Thus as a corollary the whole torsion balance is itself a raggle-taggle MASS of balls, spheres, wires, wooden rod, frame etc. Therefore such ramification of physical connection and local gravitational attraction such as wooden box, torsion balance, shed , telescope, observer were igored either erroneously or blatantly. So after understanding the finer nuances, the least precisely known G appears in the general relativity and universal law of gravitation may not be its true picture.68.147.41.231 (talk) 01:35, 4 January 2011 (UTC)Khattak#1-420[reply]

On the contrary, Newton claimed the "inverse square law" in his Philosophiae Naturalis Principia Mathematica , and Robert Hooke claimed that he had previously devised this law in the early 1660s (though he hadn't done the mathematics), and the thinking of both geniuses was based on Kepler's area law, based on the observations of Tycho Brahe. Of course, we are still looking for the "true picture", but the simple equations of motion under gravity will never be superseded for simple calculations in a gravitational field that is almost constant. Dbfirs 07:49, 4 January 2011 (UTC)[reply]

Equations fail miserably for helium balloons

I tried testing some of these equations by dropping helium balloons, I am disappoint.
The acceleration of a body in a dense medium is $\textstyle {\frac {GM}{r^{2}}}\left(1-{\frac {\rho _{medium}}{\rho _{object}}}\right)$ , an object will weigh ~~less~~ more in a vacuum chamber than in the atmosphere. The difference for the average person is about 4 ounces.

Feathers fall slowly primarily because of buoyancy, not wind resistance (viscosity).

Also, the section "Gravitational potential" seems out of place, anyone oppose deleting it? NOrbeck (talk) 23:12, 11 January 2011 (UTC)[reply]

The article does make clear that the equations only apply to objects that are much denser than air, but I agree that a mention of buoyancy would be much more appropriate than the paragraph on gravitational potential. I think you meant that objects weigh more in a vacuum chamber than in the atmosphere, and if you try testing feathers I think you will see that they fall more slowly because of both buoyancy and air resistance. I, too, have experimented with "dropping" helium balloons, but mine had a small weight attached (as is common when they are sold). I was not disappointed, because, until air resistance became significant, they did obey the equations, but with a lower acceleration, as you correctly claim. Dbfirs 07:25, 12 January 2011 (UTC)[reply]

Yes, thank you for the corrections. My point was the title "Equations for a falling body" includes bodies experiencing wind resistance and buoyancy.

"The equations only apply to objects that are much denser than air" -- actually the equations are exact for objects of any density, so long as they are in a vacuum.

"This means all results below will be quite inaccurate after only 5 seconds of fall" -- this is false for very dense objects. NOrbeck (talk) 19:25, 14 January 2011 (UTC)[reply]

Yes, I agree that the article could be made clearer and more precise. Would you like to do this or shall I? Dbfirs 09:01, 16 January 2011 (UTC)[reply]

How is the idea of Speed Gun

It would be an intersting if we can trace/ record an instanateneous velocity/ g of of falling object with the help of speed gun placed on the ground facing up right towards the falling object with the help of other electronic arrangement which can precisely record covered distance and time. I hope it wouldn't agaist the rule.68.147.41.231 (talk) 08:55, 16 January 2011 (UTC)khattak#1-420[reply]

Is Galileo's statement correct, theoretically?

Galileo was first to demonstrate that in the absence of air, all things would truly fall with the same acceleration and 300 years later demonstrated this by the crew of Apollo-15 on the lunar surface (which has gravity & also lacks air) by dropping a hammer and a feather.

Assume hammer and feather dropped simultaneously from same ANTIPODEAN altitude. As moon can be seen from two different gravitational fields ["g_f" of feather "g_h" of hammer] therefore cognizance shows that hammer and moon should strike each other first as g_h > g_f 68.147.43.159 (talk) 06:30, 4 November 2011 (UTC) Eclectic Eccentric Khattak#1[reply]

"This assumption is reasonable for objects falling to earth over the relatively short vertical distances of our everyday experience, but is very much untrue over larger distances, such as spacecraft trajectories.". So... you're right. — Preceding unsigned comment added by 94.36.161.252 (talk) 15:06, 28 August 2012 (UTC)[reply]

Adding mass instead of multiplying?

Sorry if I'm missing something here, but I was using this article to help check some gravity equations and ran across this paragraph:

"For astronomical bodies other than Earth, and for short distances of fall at other than "ground" level, g in the above equations may be replaced by G(M+m)/r² where G is the gravitational constant, M is the mass of the astronomical body, m is the mass of the falling body, and r is the radius from the falling object to the center of the body."

According to the page on the Gravitational constant, that should be G(Mm)/r^2. Is this an error/typo? If not why are the masses added here instead of multiplied as in other calculations for g? Also, isn't that first part irrelevant since it still applies to Earth? It just allows for the calculation of g rather than plugging it in as a constant. — Preceding unsigned comment added by 173.85.79.164 (talk) 01:09, 6 March 2015 (UTC)[reply]

     This is because g represents acceleration and A=F/M. This means that the mass of  the object doesn't matter for acceleration, and in fact, I don't know why it is in there at 
     all.76.93.160.48 (talk) 04:18, 9 March 2019 (UTC)[reply]

Problems with Acoustic Fall Time vs. Height

Why is the Acoustic Fall Time vs. Height figure here? The article provides no explanation for the figure. What is the relevance of "Acoustic"? And most important, why are there disparities between the data and the prediction? The disparities are clearly not due to air resistance or some other parameter that would cause error in a consistent direction. Haven't there been experiments with more accurate results?192.249.47.204 (talk) 19:38, 28 June 2017 (UTC)[reply]

==

What about starting velocity?

I would like for there to be an equation that takes into account both changes in G and also a starting velocity, both away and towards the central object. — Preceding unsigned comment added by 76.93.160.48 (talk) 04:20, 9 March 2019 (UTC)[reply]

[1] ttp://www.nytimes.com/2008/01/04/nyregion/04fall.html?pagewanted=2&ei=5087&em&en=a6cd1e14f02e1178&ex=1199595600 New York Times January/4/2008: Man survives fall from 47-story building.

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