Coriolis problem - Point mass movement upon release from Earth

• B
• zanick
In summary: If released in a non-rotating reference frame, it would have no initial velocity and would be deflected due to the centrifugal force.
zanick
TL;DR Summary
How long would a point mass take to reach the equator if released at a position on the 45th parallel?
If there was no atmosphere and a point mass was released at the 45th parallel and able to counteract the centripetal force of gravity (hovering 10ft off the surface), how long would it take before it ended up at the equator.

What are its initial velocity and ecxact initial position(also height)?

10ft off the ground... and no added momentum.

Never. It would just fall to ground. effect of corialisi force is very small and is not directed towars equator. it only affects pointamsses longitude not lattitude.

zanick said:
Added to what? State the initial velocity and the reference frame it is measured in.

zanick said:
Summary: How long would a point mass take to reach the equator if released at a position on the 45th parallel?

If there was no atmosphere and a point mass was released at the 45th parallel and able to counteract the centripetal force of gravity (hovering 10ft off the surface), how long would it take before it ended up at the equator.
Forever. It will stay in place. The geoid is not a sphere.

vanhees71
olgerm said:
Never. It would just fall to ground. effect of corialisi force is very small and is not directed towars equator. it only affects pointamsses longitude not lattitude.
I mentioned that the object was hovering... not in contact with the earth. (say it was hovering with small jets only counteracting the centripital force due to gravity)
actually, coriolis can effect objects in motion east or west toward the equator. (makes objects deflect to the right in the n-hemisphere)
jbriggs444 said:
Forever. It will stay in place. The geoid is not a sphere.
why will it stay in place? when it leaves the ground it will be retaining the momentum (as measured from the inertial reference frame) it will remain in a curved path due to the centripetal force / thrust balance (i.e. maintained altitude) but will take a southbound path (as seen from the non inertial frame of reference) . If we assume a sphere and not a geoid, how does that change the thought experiment? (let's ignor elevation changes and gravitational variations and assume a sphere)

A.T. said:
Added to what? State the initial velocity and the reference frame it is measured in.
from the inertial reference frame, its initial velocity is 500mph. in the non inertial reference frame, its has no velocity.

zanick said:
I mentioned that the object was hovering... not in contact with the earth. (say it was hovering with small jets only counteracting the centripital force due to gravity)
actually, coriolis can effect objects in motion east or west toward the equator. (makes objects deflect to the right in the n-hemisphere)
why will it stay in place? when it leaves the ground it will be retaining the momentum (as measured from the inertial reference frame) it will remain in a curved path due to the centripetal force / thrust balance (i.e. maintained altitude) but will take a southbound path (as seen from the non inertial frame of reference) . If we assume a sphere and not a geoid, how does that change the thought experiment? (let's ignor elevation changes and gravitational variations and assume a sphere)
It is motionless in the rotating frame. Zero coriolis force. The hovering force exactly counters the force of "gravity" which is not radial but is, by definition, vertical -- normal to the geoid.

vanhees71
jbriggs444 said:
It is motionless in the rotating frame. Zero coriolis force. The hovering force exactly counters the force of "gravity" which is not radial but is, by definition, vertical -- normal to the geoid.
correct, BUT we know that gravity (and the balance mentioned for hovering) will allow the object to be "forced" to rotate ….we know this because its path is circular and there for has an acceleration due to the change of velocity (not magnitude but direction) and that force id 90 degree inward... centripetal. Now, how does a point mass traveling to the right on the sphere in the non inertial reference frame, not travel southward because it would now be traveling in a straight line from the non inertial reference frame? I guess the same reason might be due to why you can't orbit around the tropic of cancer .

zanick said:
in the non inertial reference frame, its has no velocity
Then how does Coriolis come into this?
zanick said:
...on the sphere ...
It's not a sphere.

vanhees71 and jbriggs444
It's a thought discussion. assume a sphere. also, Coriolis would effect that point mass if it had an apparent deflection (verse its previous path) based on the sphere's spin.

zanick said:
It's a thought discussion. assume a sphere. also, Coriolis would effect that point mass if it had an apparent deflection (verse its previous path) based on the sphere's spin.
A point mass which is motionless in the rotating frame experiences zero Coriolis force in the rotating frame.

A point mass which is not subject to gravity will fly away in a straight line, escaping to infinity.

A marble rolling on the surface of a hypothetical perfectly spherical Earth will roll downhill toward the equator.

Thanks Jbriggs.. I'm having a hard time with the point mass or marble that is not touching the sphere, not starting to move toward the equator even if it has a 0 velocity in the rotating frame. could it be related to the impossibility of an artic orbit?

zanick said:
I'm having a hard time with the point mass or marble that is not touching the sphere, not starting to move toward the equator even if it has a 0 velocity in the rotating frame.
On a sphere it will drift towards the equator.

zanick said:
Thanks Jbriggs.. I'm having a hard time with the point mass or marble that is not touching the sphere, not starting to move toward the equator even if it has a 0 velocity in the rotating frame. could it be related to the impossibility of an artic orbit?
Centrifugal force on a spherical Earth causes an equator-ward drift. The impression from the point of view of a person standing on such a hypothetical planet is that it slopes [very slightly] downward toward the equator. On the real Earth, the geoid bulges upward at the equator, exactly cancelling this, so that the marble stays in place.

zanick said:
correct, BUT we know that gravity (and the balance mentioned for hovering) will allow the object to be "forced" to rotate ….we know this because its path is circular and there for has an acceleration due to the change of velocity (not magnitude but direction) and that force id 90 degree inward... centripetal. Now, how does a point mass traveling to the right on the sphere in the non inertial reference frame, not travel southward because it would now be traveling in a straight line from the non inertial reference frame? I guess the same reason might be due to why you can't orbit around the tropic of cancer .
If you cancel gravity completely than the object travels in a straight line and fly off into space. From the frame of the person standing on the surface, this will appear to be a curved trajectory(Coriolis effect).
To have it "hover" a fixed distance above the ground (assuming a spherical Earth), Then you need to just cancel enough gravitational pull to allow it to "orbit" the Earth at the tangential speed it has at the 45th parallel.
You would also have to insure that the thrust is always oriented towards the center of the Earth.
Such a "forced orbit" would follow a great circle.
The orbit would have a period of just under 34 hrs, and the object would take just under 8.5 hrs to reach the Equator.
Coriolis effect comes into play when you trace this trajectory against the rotating surface of the Earth; the object will tend to "drift" relative to to surface of the Earth.

jbriggs444
Janus said:
If you cancel gravity completely than the object travels in a straight line and fly off into space. From the frame of the person standing on the surface, this will appear to be a curved trajectory(Coriolis effect).
To have it "hover" a fixed distance above the ground (assuming a spherical Earth), Then you need to just cancel enough gravitational pull to allow it to "orbit" the Earth at the tangential speed it has at the 45th parallel.
You would also have to insure that the thrust is always oriented towards the center of the Earth.
Such a "forced orbit" would follow a great circle.
The orbit would have a period of just under 34 hrs, and the object would take just under 8.5 hrs to reach the Equator.
Coriolis effect comes into play when you trace this trajectory against the rotating surface of the Earth; the object will tend to "drift" relative to to surface of the Earth.
Yes, those were the conditions of the thought experiment. the object would have vectored thrust to only be in the plumb line direction Interesting that the period of the orbit is the same as a Foucault pendulum for that latitude. also, hard to imagine that if you had such an object leave the ground, that there would be enough change of direction to reach the equator in 8 hours. I guess one way to look at this is to think about how fast through the air that the object would have to fight against (if there was atm).

The Coriolis effect would create a pretty interesting path as viewed from a point of reference in the NI frame.

jbriggs444 said:
Centrifugal force on a spherical Earth causes an equator-ward drift. The impression from the point of view of a person standing on such a hypothetical planet is that it slopes [very slightly] downward toward the equator. On the real Earth, the geoid bulges upward at the equator, exactly cancelling this, so that the marble stays in place.
the difference between the geoid and the spherical Earth is a number with a decimal point and lots of zeros, so I don't think that is much to consider. I don't think the analogy works, because the object would continue to meet the equator and continue on southward eventually making a full orbit, and the minimal difference between the geoid and spheroid shape wouldn't be enough to cancel out the effect.. If it could , can you describe how?

zanick said:
the difference between the geoid and the spherical Earth is a number with a decimal point and lots of zeros, so I don't think that is much to consider. I don't think the analogy works, because the object would continue to meet the equator and continue on southward eventually making a full orbit, and the minimal difference between the geoid and spheroid shape wouldn't be enough to cancel out the effect.. If it could , can you describe how?
The difference does not have a lot of zeroes. Only about three. And the effect is enough to get to the equator the requisite time frame. One gee times several hours times 10-3 adds up.

zanick said:
...a number with a decimal point and lots of zeros, so I don't think that is much to consider...
Convincing analysis.

jbriggs444 said:
The difference does not have a lot of zeroes. Only about three. And the effect is enough to get to the equator the requisite time frame. One gee times several hours times 10-3 adds up.
yes, 3 decimals.. I don't think that's enough of a factor to stop the object in its path as you said initially. if it is, tell me how. and if its thought of as going "downhill", then how would It continue passed and orbit the earth?

jbriggs444 said:
It is motionless in the rotating frame. Zero coriolis force. The hovering force exactly counters the force of "gravity" which is not radial but is, by definition, vertical -- normal to the geoid..
can the object orbit around the tropic of cancer? so there must be another acceleration component, no?

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weirdoguy
A.T. said:
Convincing analysis.
So, one g over several hours x 10^-3 adds up is "convincing"?

weirdoguy
zanick said:
if it is, tell me how.
The geoid is an equipotential surface in the rotating frame, so there is no drift along its surface per defintion.

A.T. said:
The geoid is an equipotential surface in the rotating frame, so there is no drift along its surface per definition.
so that 0.5% change in radius, keeps the surface equipotential? Its been a while since I've dabbled into this , but this sounds like it would allow for a higher latitude orbit.

zanick said:
yes, 3 decimals.. I don't think that's enough of a factor to stop the object in its path as you said initially.
I never said "stop the object in its path". Reference, please.

But let us find the raw data and do some math. My go to reference for this is an essay by Isaac Asimov, "The Relativity of Wrong" [a fun read]. We are talking about 44 kilometer discrepancy in diameter between equator and poles. So let us estimate an 11 kilometer discrepancy in radius between the 45th parallel and the equator.

The distance from 45th parallel and equator is about 1/8 th of a circumference. So call it 5000 kilometers.

So the effective downward slope of our hypothetical spherical Earth is 11 kilometers rise in 5000 kilometers run. Round down to 10 kilometers in 5000 or 1 kilometer in 500. So sin theta is one part in five hundred.

Let's round g up to 10 meters per second squared. And multiply by sin theta. That's about 2 centimeters per second squared.

Now let's wait an hour (and ignore the rotational aspect of our hypothetical marble). The marble is now rolling at 0.02 meters per second2 times 3600 seconds = 72 meters per second. It has covered about 130 kilometers in the first hour.
if it is, tell me how. and if its thought of as going "downhill", then how would It continue passed and orbit the earth?
Now that the marble has significant southward velocity, Coriolis has an effect. If the marble is in the northern hemisphere, it is deflected rightward as it rolls downhill. The deflection rate will be reduced as it approaches the equator and reverse once beyond it in a complicated pattern.

Edit: It seems clear after considering the analysis from the inertial frame that the effect is a surface-relative trajectory that looks vaguely sinusoidal, but with the tops and bottoms of the sine waves being pointy instead of arched. The Coriolis deflection has the effect of creating westward motion when the marble is near the equator and cancelling to no east-west motion when the marble reaches the 45th South parallel and 45th North parallel over and over again.

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I think there is an entirely different reason that the point mass ("object) would move toward the equator. I earlier gave the hint of a "artic circle orbit being impossible". do you know why?

this is not about the geioid vectors of gravity, this is about a tangential velocity of the object when "hovering". (assuming it is high enough to clear any Earth shape obstacles on the spheroid)

jbriggs444 said:
I never said "stop the object in its path". Reference, please.

But let us find the raw data and do some math. My go to reference for this is an essay by Isaac Asimov, "The Relativity of Wrong" [a fun read]. We are talking about 44 kilometer discrepancy in diameter between equator and poles. So let us estimate an 11 kilometer discrepancy in radius between the 45th parallel and the equator.

The distance from 45th parallel and equator is about 1/8 th of a circumference. So call it 5000 kilometers.

So the effective downward slope of our hypothetical spherical Earth is 11 kilometers rise in 5000 kilometers run. Round down to 10 kilometers in 5000 or 1 kilometer in 500. So sin theta is one part in five hundred.

Let's round g up to 10 meters per second squared. And multiply by sin theta. That's about 2 centimeters per second squared.

Now let's wait an hour (and ignore the rotational aspect of our hypothetical marble). The marble is now rolling at 0.02 meters per second2 times 3600 seconds = 72 meters per second. It has covered about 130 kilometers in the first hour.

Now that the marble has significant southward velocity, Coriolis has an effect. If the marble is in the northern hemisphere, it is deflected rightward as it rolls downhill. The deflection rate will be reduced as it approaches the equator and reverse once beyond it in a complicated pattern.

Edit: It seems clear after considering the analysis from the inertial frame that the effect is a surface-relative trajectory that looks vaguely sinusoidal, but with the tops and bottoms of the sine waves being pointy instead of arched. The Coriolis deflection has the effect of creating westward motion when the marble is near the equator and cancelling to no east-west motion when the marble reaches the 45th South parallel and 45th North parallel over and over again.

zanick said:
I think there is an entirely different reason that the point mass ("object) would move toward the equator.
The explanation of movement depends on the reference frame (rotating vs. inertial).

zanick said:
this is not about the geioid vectors of gravity,
It very much is, because it won't happen on a geoid, but would on a sphere.

A.T. said:
The explanation of movement depends on the reference frame (rotating vs. inertial).It very much is, because it won't happen on a geoid, but would on a sphere.
help me understand... can you have an orbit around the tropic of cancer or artic circle? wouldn't an object released from the Earth at some altitude above the spheroid, (from a point of reference in the inertial frame) take a southeast path and establish an orbit?

zanick said:
help me understand... can you have an orbit around the tropic of cancer or artic circle?
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).

jbriggs444
A.T. said:
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).

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.

jbriggs444
A.T. said:
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.

zanick said:
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.

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

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