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Ok, so let me clarify: a planet that overcame all structural problems with being planetary size, I.E. collapsing into a ball.

  1. What would be the gravitational strength on a flat planet with the diameter and mass of, say, Mars?

(I believe Mars has .38 surface gravity of Earth). This theoretical planet has sufficient dense materials to have the same mass of Mars while only being disk-shaped.

The outer edge of the planet is flat (think of it like a massive coin). My understanding would be that the gravitational strength on the edge of this coin would be .38 the strength of Earth.

  1. but what would the gravitational pull if you were standing on the center?

  2. and ditto for different places on the flat surface?

I've see the Vsauce video, btw.

  1. Also, would rotating the planet help take stress off of any necessary structure support built into the planet to keep it from collapsing into a ball shape?

I'm talking rotating like a record player, not like a coin.

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    $\begingroup$ The field wouldn't be homogeneous, so one can't just characterize it with a single number. $\endgroup$ – CuriousOne Jul 25 '16 at 18:57
  • $\begingroup$ What vsauce video? I presume you mean this one : youtube.com/watch?v=VNqNnUJVcVs Doesn't that video answer your questions? $\endgroup$ – sammy gerbil Jul 26 '16 at 10:49
  • $\begingroup$ Related : Gravity on flat object (physics.stackexchange.com/q/126796) The potential and the intensity of the gravitational field in the axis of a circular plate (physics.stackexchange.com/q/62637) $\endgroup$ – sammy gerbil Jul 26 '16 at 11:27
  • $\begingroup$ My understanding would be that the gravitational strength on the edge of this coin would be .38 the strength of Earth. Have you calculated this value? If yes, how? If no, where did you get it from? $\endgroup$ – sammy gerbil Jul 26 '16 at 11:30
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Neglecting the effects of rotation on the 'apparent' strength of gravity, the gravitational field strength is similar to the electric field strength around a uniformly charged circular disk (or cylinder). This is because gravitational and electrostatic forces both obey $1\over r^2$ laws.

The field strength along the axis of the disk is quite easy to derive (eg The potential and the intensity of the gravitational field in the axis of a circular plate) and is :
$g(x)=2\frac{GM}{R^2}(\frac{x}{\sqrt{a^2+x^2}}-1)$
where $R, M$ are the radius and mass of the disk.

For a disk with the same radius and mass as Mars, the field strength at the centre of the disk would be $g(0)=2g_m$, ie twice that on the surface of Mars, because $g_m=\frac{GM}{R^2}$. (This is the answer to your Qn 2.)

By contrast, deriving the field strength off-axis but still in the plane of the disk (your Qn 3) is a very difficult task involving elliptic functions (see Electric field due to charged disc, on the plane of the disc). The variation of potential in this plane is depicted in the graph in the following question; the field strength is the slope of this graph.

Surprisingly, although the potential at the rim is finite, according to the following sources it would appear that the field strength at the rim of an infinitesimally thin disk is infinite!

Is the electric field at the edge of a uniformly charged disk infinite?
citing
Applied Mathematics Letters, vol 24, issue 11, November 2011, Pages 1919–1923
The electrostatic potential of a uniformly charged disk as the source of novel mathematical identities
by Orion Ciftjaa & Irena Hysib
http://www.sciencedirect.com/science/article/pii/S0893965911002564

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First you should understand why gravity does not change as you move around on Earth's surface. The easiest way to explain this is using symmetry. The Earth is roughly a sphere. On a sphere there are no special points so gravity must act the same everywhere.

A flat planet can actually be made to have constant gravity. If you had a flat planet that extended infinitely off into space (a geometric plane) then you can make the same symmetry argument. Because the planet extends infinitely there is no special point, there is no center of the plane. If you claim something is the center I can take two steps to the side and claim my point is actually the center. There is no test you can use to prove your point is more special than mine. Thus, the force of gravity will be the same at our two points and everywhere on the plane.

Finally I come to your coin shaped planet. Near the edge of the coin the force of gravity will be down and towards the center of the coin. Near the center of the coin the force of gravity will only pull downward. This is easy to see from the symmetry of the coin. When you are at the center of the coin gravity cannot pull you towards one side or the other because the coin looks the same in all directions from the center. It is not possible to say that the force of gravity near the edge is the same as on the surface of Mars if you do not know the thickness of the coin. To calculate the force anywhere on the coin you will need to know the thickness of the coin.

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