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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
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.
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: | |
Time t taken for an object to fall distance d: | |
Instantaneous velocity vi of a falling object after elapsed time t: | |
Instantaneous velocity vi of a falling object that has travelled distance d: | |
Average velocity va of an object that has been falling for time t (averaged over time): | |
Average velocity va of a falling object that has travelled distance d (averaged over time): |
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)
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)
People have survived falls from airplanes and from tall buildings. [1]
-- Ancheta Wis ( talk) 19:50, 4 January 2008 (UTC)
References
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)
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
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
I tried testing some of these equations by dropping helium balloons, I am disappoint.
The acceleration of a body in a dense medium is , 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)
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
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 ["gf" of feather "gh" of hammer] therefore cognizance shows that hammer and moon should strike each other first as gh > gf 68.147.43.159 ( talk) 06:30, 4 November 2011 (UTC) Eclectic Eccentric Khattak#1
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)
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)
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)
==
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)
![]() | This ![]() It is of interest to the following WikiProjects: | ||||||||||
|
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
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.
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: | |
Time t taken for an object to fall distance d: | |
Instantaneous velocity vi of a falling object after elapsed time t: | |
Instantaneous velocity vi of a falling object that has travelled distance d: | |
Average velocity va of an object that has been falling for time t (averaged over time): | |
Average velocity va of a falling object that has travelled distance d (averaged over time): |
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)
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)
People have survived falls from airplanes and from tall buildings. [1]
-- Ancheta Wis ( talk) 19:50, 4 January 2008 (UTC)
References
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)
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
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
I tried testing some of these equations by dropping helium balloons, I am disappoint.
The acceleration of a body in a dense medium is , 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)
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
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 ["gf" of feather "gh" of hammer] therefore cognizance shows that hammer and moon should strike each other first as gh > gf 68.147.43.159 ( talk) 06:30, 4 November 2011 (UTC) Eclectic Eccentric Khattak#1
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)
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)
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)
==
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)