B Coriolis problem - Point mass movement upon release from Earth

[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.