Talk:Gravitational acceleration

Article

Wow it's nice to come back to an article you created a long time ago and see that others have added to it :) My IP address was 70.149.21.254. September 30, 2006 --Fastman99

Only problem is, the article it should be merged with (now at Acceleration due to gravity) was created a long, long time before yours was (3½ years earlier than your 11 month old article), has much more information, and it has about a zillion links compared to a handful here. Gene Nygaard 13:58, 10 October 2006 (UTC)[reply]

Ok, are you trying to flame me and the article? I was the one who suggested that maybe this article be merged with Acceleration due to gravity. But I think there are some differences between the two articles. The other article only talks about the unit g, which is a unit of acceleration. "g" can be used to describe any kind of acceleration. Like, the pilot experienced 3 g's. The article, Acceleration due to gravity, is confusing because it only talks about gravity on earth. What do you mean, "only problem is" by the way? I don't think you even stated the problem. I was happy that people contributed to the article, that's all. You could say, since this article is very similar to Acceleration due to gravity, I think they should be merged together. Is it a problem that this is a new, short article? That's how all articles start out. Small at first, and then get larger. --Fastman99 17:48, 12 October 2006 (UTC)[reply]

Wow someone vandalized it to r^3 on the bottom instead- fixed it. —Preceding unsigned comment added by 169.229.101.150 (talk) 20:20, 15 October 2007 (UTC)[reply]

Question

For an acceleration a body has to be in motion (speed/velocity) if not then technically its acceleration is zero

See the following link

http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/1DKin/U1L5b.html

At starting point

Initial velocity = 0, t=0 why g is not =0 Similarly when a body of mass reach the surface of earth then its motion stop and cannot accelerate further. Then how come we calculate the weight =

In acceleration problems acceleration is always given to be uniform. However, for a falling particle from a remote point this is not so. Gravitational acceleration ( g ) depends on the distance ( r ), the distance depends on the velocity which in turn depends on acceleration. For example what is the falling time of a small mass in free fall ( V_in = 0 ) from 150 000 000 km (earth orbit radius) to Sun ? ( G • M for Sun is approximatelly 1.327• 10²⁰ in SI units) Certainly $\mathbf {r} ={\frac {1}{2}}\cdot g\cdot t^{2}$ is not applicable in this case.(non uniform acceleration) This problem can be solved by numerical methods . But I'd like to know if there is an analytic solution for such distance dependent acceleration case (maybe with the help of NL differential equations) ? Nedim Ardoğa (talk) 06:27, 14 October 2009 (UTC)[reply]

Today, Norbeck kindly informed me that I can find the answer to my question on page Free fall. As far I can see on the History page, it was added just a month ago (17th of Dec.) Nedim Ardoğa (talk) 12:44, 25 January 2010 (UTC)[reply]

"For an acceleration a body has to be in motion (speed/velocity) if not then technically its acceleration is zero"

That statement is false. Acceleration integrated forward in time produces rate, which integrated again gives change in position. Acceleration can be calculated as the vector sum of all forces divided by the mass. If all forces balance (as for an apple sitting on a table), the acceleration of the apple relative to the table is zero.

Of course the concept changes if the reference frame is considered to be rotating. Then there is an acceleration toward the center of the rotation. Pick a reference frame suitable for the level of detail of your analysis.

PoqVaUSA (talk) 01:40, 29 December 2012 (UTC)[reply]

"In acceleration problems acceleration is always given to be uniform."

Aircraft simulations have varying aerodynamic, thrust and gravity forces applied to possibly varying masses all the time. The acceleration magnitude and direction vary throughout the simulation. Pick a reference frame: flat Earth, rotating spherical Earth, oblate Earth? The rotation of the Earth has produced an oblate (flattened) distribution to its mass. The equilibrium solution for the mean sea level has an additional component due to the centripetal acceleration of the rotation. The two factors together define 'down' as being perpendicular to the mean ocean surface, and the only places where that intersects the center of the Earth is at the poles and at the equator. For a spacecraft that does NOT rotate with the Earth, the rotation rate of the Earth is immaterial. Only things that rotate with the Earth have the 1 rotation per day rate, and that is a sidereal day relative to the stars (366.25 days/yr) vs. a solar day (365.25 days/yr).

For many problems, the local value of the acceleration of gravity can be considered to be constant. However, in the absence of other forces acting on a spacecraft, the gravity gradient over the length of the spacecraft will produce an equilibrium with the long axis of the spacecraft aligned with a radial from the Earth. The assumption of uniform gravitational field where the spacecraft is will not produce that experimentally verified result. Certainly interplanetary trajectories need to account for the change in gravitational field throughout the flight. The computational solution is to numerically integrate the acceleration vector for the vehicle as a function of time, with the acceleration accounting for a changing gravitational field produced by each large body that matters (e.g. Earth, Sun, Mars). Kepler gives a pretty good accounting for the midway trajectory by just accounting for the motion around the Sun, but his analysis does not account for operations in the vicinity of planets.

It seems like this wiki article should either avoid getting into details, or go into sufficient depth to be relevant. By introducing some complications and then glossing over the details, some of the statements that are made are just not supportable.

PoqVaUSA (talk) 01:40, 29 December 2012 (UTC)[reply]

Small and big masses

Its hard to distinguish between small and big masses in the article.

Shouldn't the sentence be "All bodies accelerate at same rate which can generate similar/ same relative acceleration towards the center of gravitating mass."

Similarly its a little confusing; should the same relative acceleration be considered in the equation of w =mg and universal law of gravitation between celestial bodies- big masses.

Furthermore, Einstein's gravity prevail over Newtonian but would it possible for two identical spherical masses to attract each other such that the space distorted by them don't coincide or just like two externally tangent circle- no chance for them to move in trajectories determined by the geometry of their spacetimes. 68.147.41.231 (talk) 02:32, 18 October 2010 (UTC)Khattak#1[reply]

IMO it is big mistake

"...The relative acceleration of two objects in the reference frame of either object or the center of mass is: : $\mathbf {\hat {g}} =-{G(M+m) \over r^{2}}\mathbf {\hat {r}}$ ..."
It is big mistake. Acceleration is independent of the inertial reference frame. In non inertial r. f. we have trouble with the interaction definition. In this article I first time, in my long life, I have seen " relative acceleration" term. I must repeat, the forces and the accelerations are independent of the inertial reference frame. It formula is truth only for two objects which are together as one body. The source is necessary in this case. Regards ZIpek83.30.76.49 (talk) 08:38, 30 July 2012 (UTC)[reply]

It's actually the reduced mass Mm/(M+m), not the sum of masses (M+m). Also added comment on effect of centropedal forces. Wendy.krieger (talk) 10:17, 25 October 2012 (UTC)[reply]

Higher derivatives of position w.r.t time and their falling equations

Although higher derivatives of the position are not discussed by the Newton while formulating “F = GMm/d^2” (where reduction in “d” is lopsided for the falling of small objects on earth ) but verily the subtle nuances of such spawning can easily be perceived through arcane reasoning if imagined two identical planets of mammoth masses “M” separated by conspicuous on-centre distance “d”. Since both masses "M" are under the influence of gravity "F" therefore reduction in “d” would be equal but abrupt on either side of point “c” lucidly if “c” is the mid point of “d" as both F = GM^2/d^2 and "g =GM/d^2" are time varying.

The simultaneous abruptness of motion starts equally on either side of "c" right at the inception of "g" of each M which sent the aforementioned masses “M” swiftly into different types of motion such as gravitational jerk, jounce, crackle, pop, lock, drop etc, wherefore, should the propinquity of higher derivatives of position as such be included in the topic if yes then what would be their falling equations and order of the derivatives? 74.200.19.65 (talk) 00:59, 21 October 2013 (UTC)Eclectic Eccentric Kamikaze[reply]

Altitude or Latitude ?

A part of the article states "At different points on Earth, objects fall with an acceleration between 9.78 and 9.82 m/s2 depending on altitude,". Should this not be "depending of latitude instead ? I'm quite sure of that g differs with latitude somehow. And depending on altitude should really be "from almost zero up to 9.82 m/s2, depending of altitude". With "altitude" I do mean also outside the atmosphere. Astronauts and Kosmonauts orbiting Earth becomes "weightless", hence gravity must be very close to zero (?) Boeing720 (talk) 05:36, 6 January 2014 (UTC)[reply]

G(45) to G(poles)

Just changed

\mathbf {g} =G_{45}-(G_{poles}-G_{equator})\cdot \cos \left(2\cdot lat\cdot {\frac {\pi }{180}}\right)

to

\mathbf {g} =G_{poles}-(G_{poles}-G_{equator})\cdot \cos \left(2\cdot lat\cdot {\frac {\pi }{180}}\right)

as it the old formula doesn't comply to the constraints of G(poles) and G(equator) (example: the gravitational accelaration will be lower at 25.2 degrees latitude than at the equator, which is wrong in my opinion)

unfortunately the formula is cited and I don't have the book. Maybe the reference underneath to G(45 degrees) should be removed as well. — Preceding unsigned comment added by Agrijseels (talk • contribs) 09:34, 18 February 2014 (UTC)[reply]

I don't have the source either, but the only way the formula can make sense is to add a factor of 1/2:

g=g_{45}-{\tfrac {1}{2}}(g_{poles}-g_{equator})\cos \left(2\,lat\,{\frac {\pi }{180}}\right)

It would be good if someone could check this against the source, or any other source. -- Dr Greg talk 20:17, 18 February 2014 (UTC)[reply]

Proposed merge of Gravitational acceleration into Gravity field

wide overlap, interrelated definitions fgnievinski (talk) 21:17, 2 January 2022 (UTC)[reply]

Oppose merge of Gravitational acceleration as I think that there is sufficient scope within this important topic for articles covering the topic from different points of view: one with a focus on fields (a challenging read for most readers ...), and another on the effects of acceleration (more accessible ...). Klbrain (talk) 13:05, 28 November 2022 (UTC)[reply]

Closing, given the uncontested objection and no support. Klbrain (talk) 23:29, 25 December 2022 (UTC)[reply]

The technology to travel 1g?

? Aidepikiw1186 (talk) 04:05, 21 December 2022 (UTC)[reply]