Gravity on surface of sphere

  • #1

Main Question or Discussion Point

Howdy. I'm not a physics expert, but I am checking for solutions to this problem.

For an individual standing on the surface of a sphere (assume a perfectly spherical earth),
gravity of all atoms "pull" as per their fixed relation to that individual and their distance.
While the only perceivable force is "down", there is a pull nearly sideways and and all angles downward. The question is, will this force (negated for its even-on-all-sides pull) continue
at a gradient from weak to strong until it reaches its maximum gradient at straight down, or
will it have "spikes" of stronger pull at definite angles downward?
 

Answers and Replies

  • #2
berkeman
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Howdy. I'm not a physics expert, but I am checking for solutions to this problem.

For an individual standing on the surface of a sphere (assume a perfectly spherical earth),
gravity of all atoms "pull" as per their fixed relation to that individual and their distance.
While the only perceivable force is "down", there is a pull nearly sideways and and all angles downward. The question is, will this force (negated for its even-on-all-sides pull) continue
at a gradient from weak to strong until it reaches its maximum gradient at straight down, or
will it have "spikes" of stronger pull at definite angles downward?
What are your thoughts? Why would there be spikes?
 
  • #3
One thread mentioned that gravity is null in the center of the sphere (earth) but 109 percent
at 55 percent of the distance towards the surface (from the center). At a certain angle
from a fixed point on the surface there might be a greater concentration of mass (like a giant doughnut) where distance and the essential pull of gravity is actually larger than at a further point on the sphere (closer to straight down). I imagine drawing a series of straight lines (chords) through the sphere to make a cone. While each line individually cuts through fewer points on the line (atoms) than a straight line to the bottom, the wider circumference allows for
a greater number of these lines, and so more points exerting gravitational pull. A longer cone (to the perpendicular surface) has longer lines, but fewer of them (assuming a point distance between each line origination). There must be a solution where the number of lines and their combined distance allows for the greatest number of points (assuming some fixed length for each point). Assuming the above is correct, what is the angle of the cone? I'm curious about what I observe in nature, and am looking for reasons why things appear as they do.

It might be also good to think of each cone in terms of volume.
 
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  • #4
DaveC426913
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Howdy. I'm not a physics expert, but I am checking for solutions to this problem.

For an individual standing on the surface of a sphere (assume a perfectly spherical earth),
gravity of all atoms "pull" as per their fixed relation to that individual and their distance.
While the only perceivable force is "down", there is a pull nearly sideways and and all angles downward. The question is, will this force (negated for its even-on-all-sides pull) continue
at a gradient from weak to strong until it reaches its maximum gradient at straight down, or
will it have "spikes" of stronger pull at definite angles downward?
It is important o point out that you cannot experience "gross" pull of gravity. You only experience net gravity.

It'll be a vector, having a magnitude and a direction. One direction, not many. It'll be straight down.
 
  • #5
I appreciate your answer, and agree with the comment on net pull. I'd like to make this a hypothetical question. Assuming that I was standing on the north pole, and the earth suddenly became a stable (and perfect) hemisphere from north to south, at what angle would the pull from gravity be the strongest?
 
  • #6
DaveC426913
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Assuming that I was standing on the north pole ... at what angle would the pull from gravity be the strongest?
What do you think?
 
  • #7
Guess answer is 45 degrees to the equator. I don't have the math to discover whether
a 1/2 conical section at 45 degrees will have more area/volume than one at 55 or 65.

OK, here is a way to solve for volume: http://www.karlscalculus.org/pr_conesp.html [Broken]
I'm not a calculus expert but I might be able to get through it. Still looking for surface area.
 
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  • #8
DaveC426913
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Guess answer is 45 degrees to the equator.
Wait. Do you mean if the Earth were sliced in half? Like the entire Western Hemisphere vanished in a puff of smoke? Ah. I see. That makes more sense.
 
  • #9
DaveC426913
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One thread mentioned that gravity is null in the center of the sphere (earth) but 109 percent
at 55 percent of the distance towards the surface (from the center). At a certain angle
from a fixed point on the surface there might be a greater concentration of mass (like a giant doughnut) where distance and the essential pull of gravity is actually larger than at a further point on the sphere (closer to straight down). I imagine drawing a series of straight lines (chords) through the sphere to make a cone. While each line individually cuts through fewer points on the line (atoms) than a straight line to the bottom, the wider circumference allows for
a greater number of these lines, and so more points exerting gravitational pull. A longer cone (to the perpendicular surface) has longer lines, but fewer of them (assuming a point distance between each line origination). There must be a solution where the number of lines and their combined distance allows for the greatest number of points (assuming some fixed length for each point). Assuming the above is correct, what is the angle of the cone? I'm curious about what I observe in nature, and am looking for reasons why things appear as they do.

It might be also good to think of each cone in terms of volume.
You should look up Newton's Shell Theorem. The answer you seek is much simpler than this.

The net force of gravity at any depth on or in a sphere is
1] always straight toward the centre of mass
2] directly a result of the mass below the point of observation*

i.e if you are at a depth of 1000 miles standing on a planet that is 4000 miles in radius, It is exactly as if you were standing on a planet 3000 miles in radius.
 
  • #10
Assuming that they were peering off onto the hypothetical flat section, they would be standing
at an angle that was not directly north-south. Were they to step over on the flat section, they would be at some angle (not necessarily 45 degrees) initially. As they walked toward the center of the flat section some 4000 miles away, "down" would become more and more perpendicular to the ground. It would be like walking down a steep slope to a flat level plain.

Actually this thought picture does reinforce 45 degrees as a guess.
 
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  • #11
DaveC426913
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Assuming that they were peering off onto the hypothetical flat section, they would be standing
at an angle that was not directly north-south. Were they to step over on the flat section, they would be at some angle (not necessarily 45 degrees) initially. As they walked toward the center of the flat section some 4000 miles away, "down" would become more and more perpendicular to the ground. It would be like walking down a steep slope to a flat level plain.

Actually this thought picture does reinforce 45 degrees as a guess.
You are right about what they would experience. Walking on the flat surface would be like an extremely steep mountainside.

But the angle will be much more vertical than 45 degrees. Attached is a diagram that spitballs it.

A quick Google reveals this:
A solid hemisphere of radius r has its centre of mass 3r/8 from the centre of the circular face, on the axis of symmetry.
That will tell you exactly.


What I don't understand is what this has to do with your initial question. We're kind of going off-topic.
 

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  • #12
I'm creating a hypothetical to get around the net effect of gravity as might be found in Newton's shell theorem. While the net effect is straight down, a section of the sphere would reveal the angle at which that (eventually) cumulative effect might "pull" the strongest.

3r/8. So with a radius of ~4000 miles, the center of mass would be 1500 miles from the center of the circular face. I'm not sure where this would show up on the spherical surface, or the angle of the chord from the north pole to that spot.


Appreciate this forum's help.
 
  • #13
DaveC426913
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I'm creating a hypothetical to get around the net effect of gravity as might be found in Newton's shell theorem. While the net effect is straight down, a section of the sphere would reveal the angle at which that (eventually) cumulative effect might "pull" the strongest.
I am still not sure it makes any sense. But whatever.

3r/8. So with a radius of ~4000 miles, the center of mass would be 1500 miles from the center of the circular face. I'm not sure where this would show up on the spherical surface,
Well that would depend on where you were standing. But since you now have measurements, it's pretty simple to work out the geometry. Take the diagram I mocked up and apply a few right triangles to it.

or the angle of the chord from the north pole to that spot.
Easy. It forms a right triangle 4000 mi by 1500 mi. Angle is then trivial to work out.
 
  • #14
Thanks for your help. This is a great forum. Signing out.
 
  • #15
DaveC426913
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Thanks for your help. This is a great forum. Signing out.
Don't be a stranger. :smile: That was an intriguing question.
 
  • #16
An interesting, semi-related thing, I was trying to figure up some stuff about gravity, using two dimenstions: If you had a sector of a circle, whole angle was a constant α, and that circle had uniform density d(kg/m2, the area of the sector would be:

A = (α/360)∏r2

The center of gravity is located at r/sqrt(2) (*)

So given that the mass is determined by A*d and A is determined by r, the mass increases and center of gravity increase in such a fashion that no matter the r chosen, the newtonian gravitation pull will always be the same at the point that would be the center of the circle

Acceleration, a = mG/r12

m = dA = d∏r2α/360

a = (Gd∏r2α/360)/(r/sqrt(2))2

which simplifies to

a = 2Gd∏α/360



(*) the area of an annulus between r1 and r2 is: ∏r22(α/360) - r12(α/360)

the shape resulting from cutting an equal area annulus off the original sector should be another small sector using r1 as it's radius, the center of gravity should be r1 in this equation for a given r and α

∏r12(α/360) = ∏r2(α/360) - r12(α/360)

Multiplying everything by 360/∏α gives

r12 = r2 - r12

add r12 to each side, and then divide and take the square root, you get

r1 = r/sqrt(2)
 

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