B Coriolis problem - Point mass movement upon release from Earth

[FactChecker]

I wouldn't try to explain the Flat Earthers except in a psychiatry forum, not in a physics forum.
Your geometry is correct, but the FEers ignore so much obvious evidence that I can never fully believe that any are serious.

 

[jbriggs444]

This makes me think of something related to this discussion. could the gravitational vectors around the Geoid create areas of parallels plumb lines, giving wide surface areas on the Earth to appear flat. From some drawings/simulations/calculations, it seems that there could be areas where water (lakes.ocean) would appear to be measured flat.straight or that you could see further than you could calculate you should see based on a spheroid if so, it might explain some of the "flat earther" communities observations and tests (however uncontrolled as they may be)

Yes, I believe that a ring-shaped mass concentration could allow a region of flatness at its center.

However, In the absence of a peer-reviewed flat earther result, there is nothing in that regard to discuss here.

 

[zanick]

I wouldn't try to explain the Flat Earthers except in a psychiatry forum, not in a physics forum.
Your geometry is correct, but the FEers ignore so much obvious evidence that I can never fully believe that any are serious.

Now that's funny! I know, they see flat, so therefore its flat. I think they are serious too. They had one observation on the salton see of a guy holding a mirror (tipping up and down) and then a person on the opposing shore at 19miles away who saw the reflection and caught it on camera. refraction didn't seem like it could account for 150+ft of something being hidden by calculated curve... so , maybe it was one of the "flat spots" on the earth. ;)

Yes, I believe that a ring-shaped mass concentration could allow a region of flatness at its center.

However, In the absence of a peer-reviewed flat earther result, there is nothing in that regard to discuss here.

What "ring shaped" concentration? if I'm reading the chart correctly , would that be one side view where the south east corner has a relatively gravitationaly flat area a few 1000 miles north to south?

 

[jbriggs444]

Now that's funny! I know, they see flat, so therefore its flat. I think they are serious too. They had one observation on the salton see of a guy holding a mirror (tipping up and down) and then a person on the opposing shore at 19miles away who saw the reflection and caught it on camera. refraction didn't seem like it could account for 150+ft of something being hidden by calculated curve... so , maybe it was one of the "flat spots" on the earth. ;)
What "ring shaped" concentration? if I'm reading the chart correctly , would that be one side view where the south east corner has a relatively gravitationaly flat area a few 1000 miles north to south?

Lay a steel hoop on the ground with a radius of a few hundred meters. Adjust the thickness of the hoop until, for a few meters in the middle, you have the local "vertical" lined up all nice and parallel.

Good luck detecting the difference between this and no hoop at all.

 

[zanick]

I see what you are talking about. that wasn't what I was asking or indicating... a hoop, could be analogous to the horizon , which always will be at the same angle from a vertical center "z" non 0 axis point. vs a truly flat surface. The difference would be the angle from that axis that the horizon/hoop would lie. If there was an anomalous
area of equipotential gravity , where plumb lines were all vertical or possibly angled outward, the water would appear and be flat and maybe even concave in some regions, right?

Yes, I believe that a ring-shaped mass concentration could allow a region of flatness at its center.

However, In the absence of a peer-reviewed flat earther result, there is nothing in that regard to discuss here.

 

[Jeff Root]

I am replying after reading only the first page (25 posts). The thread
goes to three pages, and I will read them after I have committed myself
by posting this.

I agree with the original poster, zanick, that an object at the 45th parallel
(which is where I happen to live) would move toward the equator.

Rather than hovering on jets, I assume that the Earth has a perfectly
smooth, frictionless surface, and the point mass is free to slide across
this surface frictionlessly. So once it is released, the rotation of the Earth
no longer has any first-order effects. It can be ignored. All that matters
is the Earth's shape, it's centrally-directed gravity, and the object's initial
motion to the east at 330 m/s, or 0 m/s relative to the local surface of the
rotating Earth.

The Earth's shape is an oblate spheroid, where every point on the surface
is perfectly level. So unlike a sphere, it is not gravitationally downhill to
the equator. Gravity is always perpendicular to the surface. It is always
straight down relative to the nearby surface, even if it is straight toward
Earth's center only at the equator and poles. So gravity does not pull the
point mass across the surface. As jbriggs444 said in post #16, this "level"
surface is called the geoid.

With gravity keeping the point mass on the surface of the geoid, but no
friction making it stick to the surface or move with the surface, it "tries" to
continue traveling in a straight line. If we imagine the Earth is no longer
rotating, we can consider the point mass sliding frictionlessly across the
Earth's surface to the east at 330 m/s. Try to draw a "straight line" around
a sphere, and the result is a great circle, centered on the center of the
sphere. Doing the same thing on an oblate spheroid results in an ellipse,
centered on the center of the spheroid. The point mass will coast from the
45th parallel toward the equator, cross the equator, reach the opposite
45th parallel, and then coast back toward the equator again. It is in orbit.
It does not require "orbital velocity" to stay in orbit because it instead
rides on the frictionless surface.

Again considering the Earth as rotating, as the mass follows the surface of
the oblate spheroid toward the equator, it rises away from Earth's center,
thus losing speed. Also, the speed to the east of the surface of the rotating
Earth immediately underneath it increases. For both reasons, it falls back to
the west relative to the Earth's rotating surface. After crossing the equator
it gains speed and moves ahead to the east relative to the rotating surface.
The part of these motions to the east and west caused by the change in the
speed of the surface immediately under the mass is the Coriolis effect.

Now I'll read the other posts in which I expect this has all already been
fully explained, with maybe even an answer to the original question.

-- Jeff, in Minneapolis

 

[Jeff Root]

I see that a question has come up about what an "orbit" is.

I say that an object sliding over the surface of a frictionless
planet is orbiting the panet, because it will coast around the
planet on a path determined by the object's momentum and
the planet's gravity. Add enough friction, and the object will
stop coasting freely, so it will no longer be orbiting.

It is the same with the object hovering on jets, but the jets
make it much, much, much more difficult to tell for sure that
there are no significant sideways forces. Using jets will always
create suspicion that they are the cause of unexpected motion.

The principal "force" that causes the object to move toward
the equator is the centrifugal force. Which of course is not
really a force, but simply the object continuing to move in a
straight line-- or as nearly straight as possible. In this case, an
approximation of a great circle. It is the same "force" that
makes the water of Earth's oceans pile up around the equator.
The water stops moving and piles up because of friction.

-- Jeff, in Minneapolis

 

[Jeff Root]

If you question whether an object on Earth's surface would
move toward the equator from a point away from the equator,
just get a globe and a straightedge. Put the straightedge on
a point on a line of latitude, like Minneapolis on the 45 degrees
north line, so that it is oriented east-west. The straightedge is
tangent to the globe and tangent to the line of latitude.

Notice that away from the point of contact, the straightedge is
directly above locations south of the point of contact. East of
Minneapolis, the straightedge goes approximately over Boston,
which is at 42.5 degrees north. It crosses the Atlantic, and
passes over the equator south of Liberia, Africa.

-- Jeff, in Minneapolis

 

[A.T.]

The Earth's shape is an oblate spheroid, where every point on the surface
is perfectly level. So unlike a sphere, it is not gravitationally downhill to
the equator. Gravity is always perpendicular to the surface.

Not gravity is perpendicular to the surface, but the vector sum for gravity and centrifugal force in the geoid's rotating frame.

Doing the same thing on an oblate spheroid results in an ellipse, centered on the center of the spheroid. The point mass will coast from the 45th parallel toward the equator, cross the equator, reach the opposite 45th parallel, and then coast back toward the equator again.

No. Your step from sphere to geoid is wrong. Here is what happens on the geoid :

In the rotating frame of the geoid all forces (gravity, centrifugal, normal) balance and the mass remains at rest. That defines a geoid.

In the inertial frame, there is no centrifugal force, and thus the forces are not balanced. The resultant force points perpendicularly towards the axis (opposite to the now missing centrifugal force), not towards the center. Thus the mass moves on a circle around the axis at constant latitude, not on an ellipse around the center.

 

[FactChecker]

Not gravity is perpendicular to the surface, but the vector sum for gravity and centrifugal force in the geoid's rotating frame.

This is dependent on what the applied definition of the term "gravity" is. It often includes the centrifugal force.

 

[A.T.]

This is dependent on what the applied definition of the term "gravity" is. It often includes the centrifugal force.

Yes, and that's the likely source of the error in Jeff's analysis, so I explicitly kept them separate.

 

[FactChecker]

Yes, and that's the likely source of the error in Jeff's analysis, so I explicitly kept them separate.

Good. But in threads this long, it is not possible to keep everyone's definition in mind. That is probably the source of a lot of disagreement in this thread.

 

[Jeff Root]

Not gravity is perpendicular to the surface, but the vector sum
for gravity and centrifugal force in the geoid's rotating frame.

Yes. As FactChecker said, that is what I meant, and as you replied,
I should have noticed and kept them separate. I just didn't think
of doing that.

No. Your step from sphere to geoid is wrong. Here is what happens
on the geoid :

In the rotating frame of the geoid all forces (gravity, centrifugal, normal)
balance and the mass remains at rest. That defines a geoid.

In the inertial frame, there is no centrifugal force, and thus the forces
are not balanced. The resultant force points perpendicularly towards the
axis (opposite to the now missing centrifugal force), not towards the
center. Thus the mass moves on a circle around the axis at constant
latitude, not on an ellipse around the center.

I was trying to avoid discussing forces in different reference frames.

My argument is that an object on Earth's surface, moving with Earth's
surface, and suddenly released from all friction,
(1) will move in a straight line away from the point of release,
(2) the straight line passes over locations progressively closer to
the equator than the point of release, and
(3) the straight line is bent by Earth's gravity into a path approximating
a great circle, so that the object slides along the surface.

Since the Earth is an oblate spheroid rather than a sphere, the path is
a great ellipse instead of a great circle. It is an ellipse for a different
reason than why a satellite in orbit follows an elliptical path, and in fact
is sort of opposite in some ways, but they are similar in other ways,
such as losing and gaining speed as they rise and fall. (The surface of
the Earth moves to the east fastest at the equator, but the object sliding
frictionlessly across the surface slows from its already-lower initial speed
as it approaches the equator and rises out of the gravity field, so the
Earth will be moving rapidly to the east underneath it.)

My argument has the serious shortcoming that it doesn't say anything
about how the initial speed of the object relates to the path it takes.
If the argument is wrong, that is probably where it fails. My argument
is mostly just geometric.

-- Jeff, in Minneapolis

 

[A.T.]

Since the Earth is an oblate spheroid rather than a sphere, the path is a great ellipse instead of a great circle.

Repeating that claim doesn't make it less wrong. The issue was explained several times in this thread:

The geoid is per definition an equipotential surface in its rest frame. So a mass placed at rest relative to the surface of a frictionless geoid will not move relative to it, and thus cannot move to the equator.

 

[Jeff Root]

Since the Earth is an oblate spheroid rather than a sphere,
the path is a great ellipse instead of a great circle.

Repeating that claim doesn't make it less wrong.
The issue was explained several times in this thread:

The geoid is per definition an equipotential surface in its
rest frame. So a mass placed at rest relative to the surface
of a frictionless geoid will not move relative to it, and thus
cannot move to the equator.

I take it that what you disagree with is my assertion that the
mass moves, not my description of the shape of the Earth or my
description of the imaginary path as a great ellipse, correct?

Your argument seems to be that there is a net gravitational
force on the mass toward a point on Earth's axis on the same
side of the equator as the mass, pulling the mass toward the
nearer pole, into a path along the line of latitude it starts
at, and maintaining it at that latitude, instead of moving in
a straight line away from the point of release.

When I analyze the situation, I get the opposite: There is a
net gravitational force on the mass toward a point on Earth's
axis that is on the opposite side of the equator from the mass,
pulling the mass toward the equator, due to the pull of the
equatorial bulge. The shift of direction of the net force away
from Earth's center is very small (zero at the equator and poles)
because the gravitational effect of the bulge is small compared
to the effect of the whole Earth. It is not part of my argumet,
just a minor adjustment. I only mention it because it appears
to be opposite to your assertion.

The very small shift caused by the gravity of the bulge is in
addition to the larger shift caused by the "centrifugal force"
toward the celestial equator, away from Earth's axis, which
causes the bulge. This "force" is strongest at the equator,
where it is straight up, and drops to zero at the poles, where
it is horizontal. The shift in the direction of "down" between
the equator and poles caused by this "centrifugal force" is
always larger than the shift caused by the gravity of the bulge.
This shift is not part of my argument either. My agument refers
to Newton's first law instead of "centrifugal force".

Both shifts are in the same direction, toward the opposite side
of the equator. The net result is that "down" at any point
between the equator and poles is toward a point on the axis in
the opposite hemisphere, such that it is always perpendicular
to the surface, which is "level".

So my question to you is where the gravitational force comes
from that causes a mass on the frictionless surface to turn
toward the nearer pole and stay at the latitude it started at,
instead of continuing on in the direction it was moving when
the friction was removed.

If the mass were in orbit around the Earth, directly above the
release point on the surface and traveling in exactly the same
direction (due east), it would be on a trajectory that would take
it toward the equator. Even if was moving at the same speed as
the surface, it would travel toward the equator as it fell to the
ground along a narrow elliptical orbit, because either way it
would orbit Earth's center, not a point on Earth's axis closer
to the nearer pole.

Why would a mass sliding across the surface be different? Why
would a mass on the surface be pulled sideways to circle around
the nearer pole, while a mass in orbit travels straight ahead,
on a great circle or great ellipse around Earth's center?

This is exactly what the original poster was asking. He was
mistaken, though, in confusing it with Coriolis effect in the
thread title. That is a change in relative east-west motion,
not north-south motion.

-- Jeff, in Minneapolis

 

[A.T.]

So my question to you is where the gravitational force comes
from that causes a mass on the frictionless surface to turn
toward the nearer pole and stay at the latitude it started at,
instead of continuing on in the direction it was moving when
the friction was removed.

On a sphere Newtonian gravity is perpendicular to the surface everywhere. On the geoid the surface is oriented differently between the poles and equator, and Newtonian gravity has a component that is parallel to the local surface, and points towards the pole.

I explained the forces in both reference frames in post #59.

 

[Jeff Root]

On a sphere Newtonian gravity is perpendicular to the surface everywhere.
On the geoid the surface is oriented differently between the poles and
equator, and Newtonian gravity has a component that is parallel to the
local surface, and points towards the pole.

I explained the forces in both reference frames in post #59.

Yes, that is why I asked you where this mysterious gravity force comes
from. It seems to be something that you made up. I've never seen any
reference to anything like it before you brought it up here. Nobody
else posting in this thread suggested such an idea.

On the surface of the geoid, the gravitational force produced by
the equatorial bulge is toward the equator, not toward the poles.
Between the equator and the poles, the net effect of gravity is toward
a point on the rotational axis on the opposite side of the equator.
The opposite of what you said.

-- Jeff, in Minneapolis

 

[jbriggs444]

Yes, that is why I asked you where this mysterious gravity force comes
from.

There is no mysterious gravitational force here. There is simply the gravitational force, the angle of the surface and the motion of the point on that surface. There are two reasonable ways of analyzing the situation.

1. From the inertial frame: The object is in motion. It is following a circular path that moves with a point on the Earth's surface. The gravitational force is toward the Earth's center. The centripetal acceleration is toward the Earth's rotational axis. The two are not in parallel directions. However, the Earth's surface is sloped just right so that the resultant of the upward normal force and the downward gravitational force accounts for the centripetal acceleration.

2. From the rotating frame: The object is at rest. There is no Coriolis force. There is still a gravitational force toward the Earth's center. There is a centifugal force outward from the Earth's axis. The two are not parallel. However, the Earth's surface is sloped just right so that the resultant of the upward normal force, the downward gravitational force and the outward centrifugal force is zero.

On the surface of the geoid, the gravitational force produced by
the equatorial bulge is toward the equator, not toward the poles.

The effect on gravitation of the equatorial bulge is nearly negligible. The bulge is not caused by the bulge. It is caused by centrifugal force. In any case, the shape of the geoid is, by definition, enough to cancel any net force along its surface, as measured in the rotating frame.

If you hang a plumb line from a point at a temperate latitude, it will not point directly down toward the Earth's center. It will, however, be perpendicular to the geoid.

 

[A.T.]

The opposite of what you said.

Look up the definition of a geoid:

https://en.wikipedia.org/wiki/Geoid

It's an equipotential surface in its rest frame. So if you place a friction-less mass on the geoid at rest relative to geoid, the mass will stay at rest relative to the geoid. Therefore the mass cannot change latitude, and move towards the equator.

What about this simple and straightforward explanation is confusing you?

 

[Jeff Root]

I agree that the effect on gravitation of the equatorial bulge
is nearly negligible. As I said, it is not part of my argument,
just a minor adjustment. However, it is opposite the direction
of the gravitational force needed to make the path of the mass
curve away from a great circle into a smaller circle around the
nearer pole.

Rotating planet assumes oblate ellipsoid shape. Cannon fires
cannonball from a point between the equator and a pole, heading
straight east. It goes into an orbit around the center of the planet,
crossing the equator.

Rotating planet assumes oblate ellipsoid shape, and freezes in
that shape. The planet stops rotating and the surface becomes
frictionless. Put a mass on the surface at a point between the
equator and a pole, heading straight east at the speed the surface
had at that point when the planet was rotating. The mass keeps
moving in the same direction, but gravity pulls it into a great ellipse
around the center of the planet, heading for the equator.

This is what the original poster was arguing.

-- Jeff, in Minneapolis

 

[A.T.]

Rotating planet assumes oblate ellipsoid shape, and freezes in
that shape. The planet stops rotating and the surface becomes
frictionless. Put a mass on the surface at a point between the
equator and a pole, heading straight east at the speed the surface
had at that point when the planet was rotating. The mass keeps
moving in the same direction, but gravity pulls it into a great ellipse
around the center of the planet, heading for the equator.

Wrong, for the reason explained in the post right above yours. Stopping the rotation of the planet doesn't matter if it is friction less and keeps the flatten shape. It's just a pointless obfuscation.

 

[Jeff Root]

Can you explain why the path of the mass curves around
the pole instead of continuing in the same direction?

-- Jeff, in Minneapolis

 

[A.T.]

Can you explain why the path of the mass curves around
the pole instead of continuing in the same direction?

-- Jeff, in Minneapolis

See post #66.

 

[Jeff Root]

That's what I asked you to explain. What is this mysterious
component of Newtonian gravity that is "parallel to the local
surface"? Where does it come from?

-- Jeff, in Minneapolis

 

[jbriggs444]

That's what I asked you to explain. What is this mysterious
component of Newtonian gravity that is "parallel to the local
surface"? Where does it come from?

It comes from the fact that the surface is tilted. It is not perpendicular to a line drawn to the gravitational center of the Earth.

Because the surface is tilted, a force directed toward the center of the Earth has a component parallel to the surface.

Note that the small discrepancy between the "gravitational center of the Earth" and the "geometric center of the Earth" is unimportant here. The tilt of the surface is the important bit.

 

[russ_watters]

Now that's funny! I know, they see flat, so therefore its flat. I think they are serious too. They had one observation on the salton see of a guy holding a mirror (tipping up and down) and then a person on the opposing shore at 19miles away who saw the reflection and caught it on camera. refraction didn't seem like it could account for 150+ft of something being hidden by calculated curve... so , maybe it was one of the "flat spots" on the earth. ;)

The allowable distance between two objects of a certain height is twice the distance to the horizon. Two observers need only to be about 60' off the ground to see each other at 19 miles (horizon distance: 9.5mi).

Anyone who has ever been out in the ocean or a large lake with a decent pair of binoculars can see this phenomena, so there's just no excuse for not accepting it.

 

[A.T.]

That's what I asked you to explain. What is this mysterious
component of Newtonian gravity that is "parallel to the local
surface"? Where does it come from?

It comes from the orientation of the geoid surface relative to Newtonian gravitation. As already explained, the geoid is shaped this way per definition .

 

[Jeff Root]

That's what I asked you to explain. What is this mysterious
component of Newtonian gravity that is "parallel to the local
surface"? Where does it come from?

It comes from the orientation of the geoid surface
relative to Newtonian gravitation.

I don't see it. The net gravitational force is toward the
opposite hemisphere. That is the opposite of what you are
claiming.

The diagram shows a cross-section through the Earth with the
axis of rotation vertical and the equator horizontal. The blue
arrows are vectors of gravitational strength at the surface.
Red arrows are vectors of "centrifugal force". Green arrows are
vectors of the net downward force. This net force is everywhere
perpendicular to the surface, so the surface is "level". There
is no gravitational force parallel to the local surface. No
force that can make a loose, frictionless object travel around
the nearest pole instead of around Earth's center.

Notice that the green vectors point toward locations on the axis
in the opposite hemisphere. The net force is deviated away from
Earth's center toward the equator, not toward the nearer pole.

-- Jeff, in Minneapolis

 

[jbriggs444]

View attachment 253614

You forgot to draw the normal force. If you are going to do a free body diagram, draw all the forces.

 

[FactChecker]

The surface of the Earth has always been under the influence of gravitational attraction and centrifugal force. The Ocean surface and the land surface (especially while still hot and molten) was free to flow toward the Equator if there was a net force in that direction. It piled up toward the Equator to form the shape that we have today. When the centrifugal force component toward the Equator was exactly balanced by the increased height toward the Equator, the surface shape stopped changing. I am ignoring local density differences and Lunar tides. Today, something that is constrained to follow the shape of the Earth's surface will, likewise, not want to move toward the Equator.

 

[A.T.]

View attachment 253614

There is no gravitational force parallel to the local surface.

The blue arrows are Newtonian gravitation, and they are not perpendicular to the surface, so they do have a component parallel to the local surface.

If you want to talk about net force and movement, you have to state the reference frame. See posts #59 and #68 on how to do this.

The net force in either frame includes the normal force from the surface, which is missing in your diagram.

 

[A.T.]

I don't see it.

This might help:

The description was already posted:

From the rotating frame: The object is at rest. There is no Coriolis force. There is still a gravitational force toward the Earth's center. There is a centifugal force outward from the Earth's axis. The two are not parallel. However, the Earth's surface is sloped just right so that the resultant of the upward normal force, the downward gravitational force and the outward centrifugal force is zero.

From the inertial frame: The object is in motion. It is following a circular path that moves with a point on the Earth's surface. The gravitational force is toward the Earth's center. The centripetal acceleration is toward the Earth's rotational axis. The two are not in parallel directions. However, the Earth's surface is sloped just right so that the resultant of the upward normal force and the downward gravitational force accounts for the centripetal acceleration.