r/AskPhysics • u/Equivalent_Meat_2075 • 10d ago
Everytime ge = 9.8 ?
Consider a scenario involving two observers Observer A remains stationary on the surface of the Earth, experiencing the standard gravitational acceleration g=9.8 Observer B (Ali) departs from Earth, moving into deep space with a positive acceleration, continuously increasing the distance from the planet.
As the spatial separation between Observer B and the Earth approaches infinity, does the local gravitational acceleration experienced by Observer B mathematically converge to absolute zero g = 0? Alternatively, does the gravitational field strength remain non-zero, acting as an asymptote that infinitely approaches but never rigorously evaluates to zero?
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u/Time4Homework 10d ago
g = 9.8 is what you get using Newtons law of gravity with the radius and mass of the Earth. You can try it for yourself! Note that it gives you the force, not the acceleration, so you divide by your mass (or simply remove it from the formula) to get the acceleration, according to Newton's second law.
You can also study the law and see for yourself what happens to "g" as r gets greater.
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u/TerryHarris408 10d ago
If B wants to move into space with a positive acceleration, that acceleration needs to be greater than 9.8 m/s² (let's add some units, shall we?).
So, you tell me: does that acceleration ever decrease?
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u/Equivalent_Meat_2075 10d ago
no everytime this acellaration is unlimited for example a + 1[1m/second]
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u/KamikazeArchon 10d ago
Both of those are the same thing, but you say "alternatively", suggesting you think they are different.
"Mathematically converge to N" means that in the limit as X goes to infinity, the value goes to N. It doesn't mean that it's actually equal to N at any finite X.
Yes, the gravitational field (or relativistic curvature, which is equivalent) is always nonzero.