Title pretty much says it all.
If the Earth was perfectly round, perfectly smooth, and I placed a likewise perfectly round ball on it (Let’s say the ball is the size of a baseball) would it roll? Or would it stay in place?
There are certain factors involved however I’m assuming that your large glass ball is just like the Earth and if you place a ball on the ground on the earth it doesn’t really roll away. I’m assuming the same for your scenario. Is that what you wanted to hear?
Assuming it is perfectly round sphere; there was not friction; either from the earth or your ball; as long as you gave your ball some momentum; yes it should roll.
A perfectly spherical Earth would have a perfectly gravitationally “level” surface, i.e. the same gravitational potential everywhere. So the ball would not roll of its own accord, but it would roll if you gave it a push.
It would act just like a ball on a hypothetical flat plane, with uniform gravity everywhere on it.
Hmmm, considering the Earth is an oblate spheroid, I assume that this is partly due to Earth’s centrifugal force. So therefore if the Earth were perfectly round, I’d image that there’d be a potential of 10 miles or so between the equator and the poles, causing the ball to roll given no friction at all. The earth in this situation would “want” to deform to the “correct” shape but would be prevented from this by the rigidity of the glass.
If, for some reason I am mistaken in this, then this means that if the Earth were a completely featureless glass oblate spheroid with equal rotation, then the ball would move. I assume that this is the more unlikely scenario since this would mean there would currently be an average potential between the equator and the poles and so have to assume that my above scenario, with a centrifugal-force caused potential difference, to be the more likely one.
Yes, I guess I should state the sphere would be in a vacuum (No atmosphere) but still spinning on its axis around the sun. I was thinking if I delicately place it on the sphere though, as opposed to giving it momentum.
Agreeing with the posters above -
The force of gravity will be perpendicular to the surface of the earth-globe, so the ball will stay in place unless there is any sideways momentum given to it.
If it receives any force except straight down, it will roll. In the absence of friction it would roll indefinitely, I think.
ETA: I am not considering angular momentum imparted by the rotation of the earth.
Are you planning on accelerating the ball up to the speed the earth is spinning before setting it down on the surface? The surface of the earth at equator is moving at roughly 1000 MPH due to spin.
Would the Moon’s gravity be strong enough to act on the ball and move it in the same way the tides move?
It seems to me that, since Earth is rotating, the Coriolis force would cause the ball to roll towards the equator. Consider the extreme case where you placed the ball six inches away from the North pole. If the ball were glued to that spot, it would trace a six-inch-radius circle every 23 hours and 56 minutes. But without the glue, momentum would carry it away from the pole and it would roll towards the equator. But then, by conservation of angular momentum, it would drift towards the West. Ignoring wind resistance, it would eventually end up at the equator, rolling West at very high speed (from Earth’s point of view)
Did you do it yet?
Sure, much as it makes the level shift in your teacup.
I think your description is correct. Centrifugal force would propel it toward the equator, and once moving Coriolis Force would turn it right, i.e. West. I think maybe momentum carries it past the equator, and into a reducing N/S oscillation to either side of the equator, asymptotically just moving west on the equator?
I disagree. Contrary to popular believe, the net result of tidal forces is mostly horizontal, not vertical. In an ocean, this causes the water to move East and West by quite a large amount. But in your teacup, the walls of the cup are preventing it from moving East and West. The exact same thing would happen in the ocean if something were preventing East-West motion. For example, if you placed North-South baffles in the ocean, spaced close together, the oceans wouldn’t slosh East and West so the tidal effect would largely disappear. Conversely, if you allowed your teacup to slide East and West on a frictionless surface, you’d find that yes it would move with the tides.
ETA: a couple more folks said much the same thing while I was typing.
Ludovic got it.
On a non-rotating perfectly spherical perfectly uniform planet the gravitational potential is exactly perpendicular to the surface everywhere. No force would act along the surface of the planet to cause a stationary ball to move or a ball moving on a great circle to change course = 3D plane of motion.
Any ball placed in motion not on a great circle will describe an interesting spiraling motion.
OTOH On a *rotating *perfectly spherical perfectly uniform planet everything is more complicated.
Centripetal force creates a force gradient that partly offsets gravity at the equator but has no effect at the poles. At intermediate latitudes the sum of the gravity + centripetal forces is such that there’s a net pull equatorward.
Assuming a very, very, very low friction environment you could place a ball stationary relative to the surface on the exact equator or the poles and it’d sit still forever. But place it anywhere else and it’ll immediately start accelerating poleward.
As sbunny8 says, the moment something is moving north / south it begins to experience Coriolis effects and will experience acceleration east or west in the reference frame of the planet.
And since by definition a curving path on the surface isn’t a great circle, it’ll end doing an interesting spiral across the surface of the planet.
From the number of detailed responses so far, clearly we got the ball rolling on this topic.
There’s a net pull equatorward, but it immediately starts accelerating poleward?
There’s an important difference between poles and equator in that the poles are unstable, the equator is (I think) stable.
I’m interested in the path and final velocity, given a stationary start. It seems to me that it turns westward and oscillates N/S either side of the equator, tending toward westward movement straight along the equator? Is that right? Is the final velocity at the equator a function of initial latitude?
Only if it’s magic glass. The rigidity of rock isn’t enough to prevent the Earth from assuming the “correct” shape (except on very small scales like mountains), and glass is less rigid than rock.
EDIT:
With no friction, it’ll continue oscillating. With friction, it’ll tend to eventually come to rest somewhere on (or very close to) the Equator.
In the zero-friction case, one can very easily find the westward component of its motion by applying conservation of angular momentum. It has some amount of angular momentum dependent on the latitude where it was placed, and it will keep it, since there won’t be any torques on it.
The ball is, from a gravitational point of view, no different from a liquid. The tidal force on the ball, which would ultimately make it move, is the exact same force that would, in a confined space, make the level of a liquid rise in the same direction.