B Coriolis problem - Point mass movement upon release from Earth

[jbriggs444]

A mass constrained by the surface (sliding/rolling/hovering along the surface) is not "in orbit". What path it will take depends on the shape of the surface (geoid vs. sphere).

Further, the path taken by a mass constrained by a magical "hovering force" will depend critically on the angle at which the hovering force is applied. In particular, it will depend on whether the force is applied opposite to the direction of a plumb line or opposite to the direction to the center of the planet.

If opposite to a plumb line, the mass will remain in place above the rotating surface. If opposite to the radial direction, the mass will slide down toward the equator (likely impacting the surface first unless the non-spherical surface is hand-waved away).

 

[FactChecker]

Consider that the ocean water is free to level out so that the surface is exactly an equipotential. So enough water has drifted toward the equator to exactly account for the centrifugal force of the Earth's rotation. An object floating on that surface, free to float in any direction, would not move.

As a "thought experiment" we can suppose that the surface of the Earth is a sphere and is frictionless (like ice). Then an object sitting on the surface would slide toward the Equator. But that is not reality.

PS. I once wrote simulation code for an airplane on the runway. I kept wondering why it would start to slide South, toward the Equator when it was supposed to be stationary on the runway. (In a simulation, it is easy to turn off frictional forces, or to forget them as I did.) I finally realized my mistake was that I had the wrong "level" Earth surface.

 

[zanick]

A mass constrained by the surface (sliding/rolling/hovering along the surface) is not "in orbit". What path it will take depends on the shape of the surface (geoid vs. sphere).

that's the point of the thought experiment... to remove the sliding, rolling and surface shapes. If the point mass is hovering above the surface...… then does the object move toward the equator in a straight line as viewing from the inertial reference frame.

 

[jbriggs444]

that's the point of the thought experiment... to remove the sliding, rolling and surface shapes. If the point mass is hovering above the surface...… then does the object move toward the equator in a straight line as viewing from the inertial reference frame.

As pointed out, it depends the direction of the force that causes it to hover. If constrained to a spherical surface, it drifts southward. If constrained to an equipotential surface it does not.

 

[zanick]

I mentioned that the thrust would be assumed to be vectored , plumb line

 

[jbriggs444]

I mentioned that the thrust would be assumed to be vectored , plumb line

Plumb line -- that means that the force is at right angles to the equipotential surface. So the mass stays in place on the equipotential surface above the oblate ellipsoid Earth.

 

[Couchyam]

This sounds a bit like a homework question, but here are a few hints:
(i) What quantity (or quantities) are conserved?
(ii) From a corotating reference frame, what fictitious force determines the magnitude of the longitudinal angular acceleration of the particle?
(iii) At a more advanced level, this example is an instructive exercise in Lagrangian mechanics.

 

[FactChecker]

As pointed out, it depends the direction of the force that causes it to hover. If constrained to a spherical surface, it drifts southward. If constrained to an equipotential surface it does not.

Exactly. If it is forced to hover over an equipotential surface, then the tilt of the surface is exactly what is required to oppose any drift toward the Equator. That is the actual shape of the Earth (at least the sea level and the overall shape of the land.) Without that tilt -- with a spherical Earth, there would be a drift toward the Equator. That drift would accelerate to a high velocity. The actual calculation of the drift while constrained to hover over a spherical surface would probably be complicated.

 

[jbriggs444]

The actual calculation of the drift while constrained to hover over a spherical surface would probably be complicated.

@Janus did that computation in #17 using the inertial frame where it turns out to be quite simple.

 

[FactChecker]

@Janus did that computation in #17 using the inertial frame where it turns out to be quite simple.

Your interpretation might be the correct one, but I interpret the original question differently. I interpret the question to be what would happen if a stationary, hovering object were allowed to start accelerating toward the Equator, how long would it take to get there. An orbit is not a "hover". The first relies on speed and centrifugal force, while a "hover" has some other force. So the speeds can be very different.

 

[jbriggs444]

Your interpretation might be the correct one, but I interpret the original question differently. I interpret the question to be what would happen if a stationary, hovering object were allowed to start accelerating toward the Equator, how long would it take to get there. An orbit is not a "hover". The first relies on speed and centrifugal force, while a "hover" has some other force. So the speeds can be very different.

An object constrained to a spherical shell by a force normal to that shell and which starts at relative rest above a point on the rotating spherical Earth beneath that shell counts in my book as a "hover". And is an orbit as described in the inertial frame.

Note that this is "orbit" in the linguistic sense of "motion in a circle about". Not in the usual astronomy sense of "a closed ballistic trajectory about".

 

[A.T.]

that's the point of the thought experiment... to remove the sliding, rolling and surface shapes. If the point mass is hovering above the surface...…

If you remove the surface, you cannot be "hovering above the surface"

then does the object move toward the equator in a straight line as viewing from the inertial reference frame.

In a straight line? Do you mean a great circle?

 

[FactChecker]

An object constrained to a spherical shell by a force normal to that shell and which starts at relative rest above a point on the rotating spherical Earth beneath that shell counts in my book as a "hover". And is an orbit as described in the inertial frame.

Note that this is "orbit" in the linguistic sense of "motion in a circle about". Not in the usual astronomy sense of "a closed ballistic trajectory about".

Ok. But what is the velocity of that "orbit". It seems like you are switching to the usual meaning of "orbit" due to centrifugal force to determine the velocity.

 

[jbriggs444]

Ok. But what is the velocity of that "orbit". It seems like you are switching to the usual meaning of "orbit" due to centrifugal force to determine the velocity.

The velocity of the orbit is the velocity of a point on [actually 10 meters above] the Earth's surface compared against a non-rotating spherical earth.

We know that to be the case since that is its initial velocity and, in the inertial frame, it never experiences any tangential force.

 

[zanick]

Exactly. If it is forced to hover over an equipotential surface, then the tilt of the surface is exactly what is required to oppose any drift toward the Equator. That is the actual shape of the Earth (at least the sea level and the overall shape of the land.) Without that tilt -- with a spherical Earth, there would be a drift toward the Equator. That drift would accelerate to a high velocity. The actual calculation of the drift while constrained to hover over a spherical surface would probably be complicated.

That makes sense.

 

[zanick]

Your interpretation might be the correct one, but I interpret the original question differently. I interpret the question to be what would happen if a stationary, hovering object were allowed to start accelerating toward the Equator, how long would it take to get there. An orbit is not a "hover". The first relies on speed and centrifugal force, while a "hover" has some other force. So the speeds can be very different.

yes, the hovering tangential velocity would only be (cos) latitude x 25,000mi / 24 hours in the inertial reference frame. ( vs what you (point mass) would see if it was "orbiting") So what you are saying is that centrifugal acceleration is not enough to overcome vector change due the geoid characteristics.

 

[zanick]

If you remove the surface, you cannot be "hovering above the surface"

In a straight line? Do you mean a great circle?

I didn't mean "remove the surface", but remove its interaction . that's why I made the point mass hover... with a vector force/thrust to plumb. (10m above the surface) and yes, straight line , great circle ( straight- non Euclidian)

 

[A.T.]

... with a vector force/thrust to plumb. ...

The plumb-line on a spinning planet is along the effective gravity (gravity + centrifugal). If your thrust exactly cancels that, then the mass obviously won't move relative to the planet, regardless what its shape is.

 

[zanick]

The plumb-line on a spinning planet is along the effective gravity (gravity + centrifugal). If your thrust exactly cancels that, then the mass obviously won't move relative to the planet, regardless what its shape is.

Makes sense. Simply, I used the wrong example, in using "thrust vectored to plumb". we should have stayed with the ball rolling, but i wanted to remove friction and local surface shape characteristics. So basically, with thrust vectoring to plumb of this hypothetical object, it will not move when placed anywhere on the Geoid.
thanks.

 

[zanick]

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)

Plumb line -- that means that the force is at right angles to the equipotential surface. So the mass stays in place on the equipotential surface above the oblate ellipsoid 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.