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When an apple falls from a tree, Earth exerts a gravitational pull on the apple that amounts to an acceleration of around $9.81\,m/s^2$. But as the apple falls, it gets closer to Earth and thereby the gravitational pull as well as the acceleration increase.

When the apple orbits Earth, the gravitational pull is (nearly) constant, but it moves perpendicular to the pull. Can we have both: follow the direction of the pull while having constant acceleration for a non-trivial time and path length?

Put another way: Is it possible to form a mass such that its gravitational pull on other objects is constant along the path these objects follow when moving in the direction of the pull? Can the path be straight or does it have to be straight?

(No, not a homework question. Just curious.)

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  • $\begingroup$ How do you define a "non-trivial time and path length" here? What do you mean by these terms? $\endgroup$ – sammy gerbil Jul 13 '18 at 17:41
  • $\begingroup$ Anything markedly longer than a dx or dt: 1mm, 1m, 1km or more. $\endgroup$ – Harald Jul 13 '18 at 18:26
  • $\begingroup$ Is it possible to form a mass such that its gravitational pull on other objects is constant along the path these objects follow when moving in the direction of the pull? Can the path be straight or does it have to be straight? Not clear what this means. Are objects allowed to move in a direction other than the direction of the force? If not, I think the only path possible is a straight line, because otherwise inertia will carry the object away from the line of force. $\endgroup$ – sammy gerbil Jul 13 '18 at 18:33
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The idea that the gravitational force on earth is constant is really not true. It is just an approximation. It works because if you drop a ball from even some height of order 100m, the displacement is very small compared to the radius of the earth.

When you are farther away, or even at the surface if you wanted the most exact value, you would compute it using Newton's Law of Universal Gravitation (as has been done to determine the value $9.81m/s^2$.

$$ \mathbf{F}_{grav} = \frac{Gm_1m_2}{r^2}\mathbf{\hat r} $$

This still assumes that we can treat the mass as if it is a point mass which is good for most spherically symmetric bodies.

I suppose in theory you could "construct" a mass that gives a truly constant field. You would have to make a mass that is analogous to the infinite plane of charge (see here). So if you had a very large and flat slab of mass (infinite to truly have a constant field) you could in theory construct a constant field. Clearly this is not a realistic scenario though.

So for all practicality, he gravitational field can be treated as constant when you are near a large object but farther away or for smaller or more complicated objects, the field must be manually computed by Newton's Law of Gravity or some other method.

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  • $\begingroup$ I don't think this really answers the question. They are asking about the possibility of creating a mass such that there is some line extending out from the mass to infinity on which the acceleration is of constant magnitude and direction. An infinite plane of mass wouldn't achieve this effect. $\endgroup$ – Ulthran Jul 13 '18 at 11:40
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    $\begingroup$ If you apply Gauss' law for gravity then you can show that a hypothetical infinite plane has a constant field same as infinite plane of charge. You can show that $g = 2\pi G \sigma$. Though obviously the mass scenario is not realistic to any real world problems. $\endgroup$ – fhorrobin Jul 13 '18 at 11:43
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    $\begingroup$ I was indeed after a theoretical answer, not something really practical to build. I did not know about the infinite plane example for the electric field (khanacademy.org/science/electrical-engineering/…), but it looks like the derivation should work much the same. $\endgroup$ – Harald Jul 13 '18 at 11:45
  • $\begingroup$ This link goes through Gauss' Law for gravity for many cases including infinite plane. pgccphy.net/ref/gravity.pdf $\endgroup$ – fhorrobin Jul 13 '18 at 11:46
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    $\begingroup$ Is the infinite plane the only solution? If so, what is the proof? $\endgroup$ – sammy gerbil Jul 13 '18 at 17:39

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