Dropping a mass directly over the north pole

[starz_above]

Hi all,

I'd like to understand what would happen to a mass which is dropped at the north/south pole.

Let's say you raised a mass to a certain height over a certain point, which we will call A. Point A is at the center of the north pole. The height at which the mass is raised is such that it will take over 10 mins to hit the ground, let's say it's 6km high above point A.

When the mass is dropped would it land on point A? Let's assume that there no external forces to alter the descent in anyway. We can assume that the mass falls in a vacuum.

Thanks for any comments.

 

[mikeph]

Why wouldn't you expect it to?

 

[nasu]

If you neglect air resistance, 6km is way too low to take 10 min.

 

[starz_above]

Why wouldn't you expect it to?

I was thinking whether the Earth's orbit around the sun would have any effect.

If the mass falls directly onto point A after 10 mins, what causes it to stay above point A as it falls given that the Earth is traveling at a tangent below it?

 

[starz_above]

If you neglect air resistance, 6km is way too low to take 10 min.

You're right. Let's say it takes about 35 seconds to land on point A.

 

[fluidistic]

The biggest effect that you'd have to take into account would be Corriolis effect. But since you are over a pole, I don't think it applies. It would have been quite different over the equator I think.

 

[Delta Kilo]

I was thinking whether the Earth's orbit around the sun would have any effect.

If the mass falls directly onto point A after 10 mins, what causes it to stay above point A as it falls given that the Earth is traveling at a tangent below it?

Acceleration due to Sun/Moon gravity will be pretty much the same for the mass and for the whole Earth as long as the distance between the centres of mass of the body and the Earth is small compared to the distance to the Sun and/or Moon. Assuming that the mass is at rest relative to the Earth at the moment of release, the mass and the Earth will be in the same orbit.
If the mass is moved significant distance from the Earth, gravitational influences due to other bodies will no longer cancel out and you'd have to calculate the motion using orbital mechanics.

 

[Shoku Z]

The object initially was on the surface of Earth and so the velocities of Earth and the object are same.We carry it up and then drop it.observing from earth,the trajectory will be a straight line.This is because the velocity of the object is same as Earth.The only other force acting upon them is that gravitational force of sun.Hence,the trajectory appears to be a straight line from Earth.

 

[Shoku Z]

In short,Delta is right.I didnt see his post!

 

[starz_above]

Acceleration due to Sun/Moon gravity will be pretty much the same for the mass and for the whole Earth as long as the distance between the centres of mass of the body and the Earth is small compared to the distance to the Sun and/or Moon. Assuming that the mass is at rest relative to the Earth at the moment of release, the mass and the Earth will be in the same orbit.
If the mass is moved significant distance from the Earth, gravitational influences due to other bodies will no longer cancel out and you'd have to calculate the motion using orbital mechanics.

I understand that the mass and the Earth will be in the same orbit and hence have the same velocity. However if the mass lands on point A, then doesn't that mean the mass must have a higher velocity than that of the Earth in order to land on point A since it takes x amount of time descend. If the mass has the same velocity as that of the Earth then it should land "behind" point A in the path of the Earth's orbit around the sun.

 

[fluidistic]

Delta and Shoku z, you guys are missing that the Earth isn't an inertial frame. There is a Coriolis effect and I can assure you that if you drop an object say from a skyscrapper and if you remove all air (so that there's no air resistance), the object trajectory WON'T be a straight line, despite what you said. However I think that the Coriolis effect is inexistant in both poles of the Earth. I have no idea about the centrifugal force though. I'll wait for some expert.
I'm assignated a problem in classical mechanics where I have to throw upward an object and calculate at what distance it will fall from me. I'll have to take into account Coriolis effect.

 

[Delta Kilo]

The Earth is influences by the gravitation of Sun, Moon and other bodies. As a result, the Earth moves with acceleration let's say $\vec{a}$ (pointing roughly towards the Sun). The mass will receive the same acceleration $\vec{a}$ from Sun, Moon etc, plus acceleration $\vec{g}$ towards the centre of the Earth. So the acceleration of the mass relative to the Earth will be just $\vec{g}$ towards the centre. Since initial relative velocity is 0, the resulting relative velocity and hence the relative trajectory will be directed toward the centre of the Earth. When I say relative, I mean non-inertial reference frame with the origin at the centre of the Earth and axis pointing in fixed direction in space (moving with acceleration $\vec{a}$ but not rotating).

 

[fluidistic]

Look, here is a problem I see in the assignments of the classical mechanics from my university:

An object is thrown up vertically from the Earth surface with an initial velocity $V_0$.
The place where the object is thrown is at colatitude L. Demonstrate that when returning over Earth's surface, the object falls displaced to the west at a distance of $4 \Omega V_0 ^3 \cos \left ( \frac{L}{3g^2} \right )$ from the point where it has been thrown up.
Omega is the angular velocity of the Earth and g is the gravitational acceleration over the surface of the Earth.
Disregard the motion of the Earth around the Sun and also any friction of the object with the atmosphere. Assume that the maximum height reached by the object is much less than the radius of the Earth.

I'm unsure of the expression of the distance. It's written $4 \Omega V_0 ^3 \cos L/3g^2$.

So that I'm almost 100% sure that the Coriolis effect is the most important effect that can effect the object. However as I said twice, right over the poles I think it's worth 0 (maybe L is worth pi/2 and the right expression is $\frac{4 \Omega V_0 ^3 \cos (L)}{3g^2}$?) So that your object would indeed fall over the north pole. The answer depends where you are on Earth and the 2 poles are the special cases where there's no appreciable displacement of objects thrown upward.

 

[DaveC426913]

I understand that the mass and the Earth will be in the same orbit and hence have the same velocity. However if the mass lands on point A, then doesn't that mean the mass must have a higher velocity than that of the Earth in order to land on point A since it takes x amount of time descend. If the mass has the same velocity as that of the Earth then it should land "behind" point A in the path of the Earth's orbit around the sun.

starsz_above has a point. If we set aside the pull of the Earth for a moment, once the object is released, it is essentially in orbit. But it is not quite in the same orbit as the Earth. Its orbital plane is tilted to cross Earth's.

This effect will be enhanced or reduced, depending on what season it is. If it is summer, then the object's initial position is slightly inside the Earth's orbit (because the axial tilt is causing the north pole to tilt inward); if in the winter, the object's initial position is outside the Earth's orbit.

These are small effects, but they're there.

 

[mikeph]

If you're ignoring the sun, then the problem can be reduced to a single dimension (object ---- Earth's COG) since the object is placed on the axis of rotation of the Earth.

If you're including the sun, it's a circular restricted three body problem (chaotic), and it's not too clear from the question what the initial conditions are (no external forces = ??, if you let the particle go, does that mean giving it zero velocity? Then it won't even land on the Earth which is moving at 30 km/s in orbit.

 

[starz_above]

If you're ignoring the sun, then the problem can be reduced to a single dimension (object ---- Earth's COG) since the object is placed on the axis of rotation of the Earth.

If you're including the sun, it's a circular restricted three body problem (chaotic), and it's not too clear from the question what the initial conditions are (no external forces = ??, if you let the particle go, does that mean giving it zero velocity? Then it won't even land on the Earth which is moving at 30 km/s in orbit.

The mass has inital zero velocity wrt the earth. To clarify, the mass is placed at a certain height directly above the north pole, then it is released to fall without any external forces being added and falls just due to the Earth's gravity.

So initially both the mass and the Earth are traveling at 30 km/s in orbit.

 

[mikeph]

So the Earth is subject to the sun's gravity but the mass isn't? Aside from breaking physical laws, it wouldn't land at the same point because that point is following a circular trajectory (or some kind of 3D spirograph pattern if you account for the seasons) and the mass isn't.

 

[FireStorm000]

Let us make the following assumptions:
Gravity from all sources apply
Zero influence due to air resistance
Zero influence due to Earth's magnetic field

That leaves us with the following forces which have a component perpendicular to the acceleration due to Earth's gravity:
-tidal stresses:
--Acceleration due to the moon; the moon is a different distance from Earth's CoM than the object, because the Earth's axis of rotation isn't perpendicular to the moon's orbital plane.
--Acceleration due to the sun; again, the sun is a different distance from Earth's CoM than the object, because the Earth's axis of rotation isn't perpendicular to the sun's orbital plane.

Assuming the object was placed exactly distance X above the north or south pole, defining above as being normal to the surface of a perfect sphere, and the velocity of the object is the same as the instantaneous rate of change of position of a point positioned x distance from the surface along the axis, then those are the only other forces applying

 

[starz_above]

So the Earth is subject to the sun's gravity but the mass isn't? Aside from breaking physical laws, it wouldn't land at the same point because that point is following a circular trajectory (or some kind of 3D spirograph pattern if you account for the seasons) and the mass isn't.

The mass is also subject to the sun's gravity. When I said no external forces I meant to say no air resistance and it isn't pushed in any direction when it's released. Apologies for not being clear.

As i understand it, the trajectory of the mass would also be some kind of 3D spirograph pattern as you suggested.

What is puzzling me is that if the Earth was not orbiting the sun then you would get the same result ie. the mass landing directly on top of the north pole. Initial thoughts tell me that if the Earth is orbiting the sun then the mass should not fall directly on top of the north pole because in order to do so the mass must increase it's horizontal velocity component in order to "catch up" with the north pole as it is falling.

 

[starz_above]

Let us make the following assumptions:
Gravity from all sources apply
Zero influence due to air resistance
Zero influence due to Earth's magnetic field

That leaves us with the following forces which have a component perpendicular to the acceleration due to Earth's gravity:
-tidal stresses:
--Acceleration due to the moon; the moon is a different distance from Earth's CoM than the object, because the Earth's axis of rotation isn't perpendicular to the moon's orbital plane.
--Acceleration due to the sun; again, the sun is a different distance from Earth's CoM than the object, because the Earth's axis of rotation isn't perpendicular to the sun's orbital plane.

Assuming the object was placed exactly distance X above the north or south pole, defining above as being normal to the surface of a perfect sphere, and the velocity of the object is the same as the instantaneous rate of change of position of a point positioned x distance from the surface along the axis, then those are the only other forces applying

Your assumptions are correct. Thanks for putting it clearly.

 

[mikeph]

What is puzzling me is that if the Earth was not orbiting the sun then you would get the same result ie. the mass landing directly on top of the north pole. Initial thoughts tell me that if the Earth is orbiting the sun then the mass should not fall directly on top of the north pole because in order to do so the mass must increase it's horizontal velocity component in order to "catch up" with the north pole as it is falling.

That is not the same result. If there is no sun the problem is one-dimensional and it will land on the pole. If there is a sun it's a three body restricted problem with an entirely different (chaotic) trajectory and will not land on the pole. You could calculate it with arbitrary accuracy if you have a powerful enough computer and some decent numerical equations.

 

[olivermsun]

You could also posit the calculation to be arbitrarily accurate by calling the mass a "small test mass." :) That still leaves the problem interesting in the usual sense of the non-inertial frame.

But even if you allowed the small mass to be (slightly) massive, would you really expect any interesting chaotic trajectory between the mass starting a small distance above the pole and striking the Earth (therefore ending the trajectory)?

 

[mikeph]

You could also posit the calculation to be arbitrarily accurate by calling the mass a "small test mass." :) That still leaves the problem interesting in the usual sense of the non-inertial frame.

But even if you allowed the small mass to be (slightly) massive, would you really expect any interesting chaotic trajectory between the mass starting a small distance above the pole and striking the Earth (therefore ending the trajectory)?

Even with a small test mass it's chaotic (restricted three body problem) because the sun also moves around the barycentre of the Earth-sun system. If you consider that the body has a non-negligible mass (general three body system) it's even worse. In both cases the trajectories are, for general initial conditions, chaotic.

 

[olivermsun]

But my point is, are these "general" initial conditions? The test mass does not seem to continue along its path for very long in the problem as stated.

 

[mikeph]

The starting coordinates are not the Lagrange points (the only five analytical solutions to the problem), so yeah it's a general initial condition. You're right, it will probably meet the Earth's surface soon after it's released, but where exactly on the surface I believe is impossible to compute.

 

[fluidistic]

I feel like my interventions has been totally ignored in this thread so far. Here is a last attempt of pointing an important effect that people seem to ignore. Even without Sun, the mass wouldn't land on point A (despite what MikeyW said), I believe.
Take Goldstein's book, 1st edition on pages 136 to 140. (138 and 139 to be more precise for the next quote)

Another classical experiment where Coriolis force plays an important role is in the deflection from the vertical of a freely falling particle (...) Thus, in the Northern Hemisphere, a body falling freely will be deflected to the east. (...) $x=-\frac{\omega }{3} \sqrt {\frac{(2z)^3}{g} } \sin \theta$. An order of magnitude of the deflection can be obtained by assuming $\theta = \pi /2$ (corresponding to the Equator) and z=100ft. The deflection is then, roughly $x \approx 0.15$ inch.

$\theta$ is the colatitude. This effect is by far the greater I think (that would also agree with the assumptions the problem I mentioned in my last post in this thread makes allusion).
I might be totally wrong (it wouldn't surprises me), but I'd like a comment on both the Coriolis and centrifugal force that I think you have to take into account.

P.S.:On page 137, one reads

Physically, this result means that a projectile shot off at the North Pole has no initial rotational motion and hence its trajectory in the inertial space is a straight line, the apparent deflection being due to the Earth rotating beneath it.

which implies, I think, that the body wouldn't land over the north pole. Unlike what I previously thought.

 

[willem2]

I feel like my interventions has been totally ignored in this thread so far. Here is a last attempt of pointing an important effect that people seem to ignore. Even without Sun, the mass wouldn't land on point A (despite what MikeyW said), I believe.
Take Goldstein's book, 1st edition on pages 136 to 140. (138 and 139 to be more precise for the next quote)

I might be totally wrong (it wouldn't surprises me), but I'd like a comment on both the Coriolis and centrifugal force that I think you have to take into account.

I think most people have been ignoring the Coriolis force, because it is zero for an
object, moving parallel to the axis of rotation. The centrifugal force is also 0 at or directly above the north pole.
The most important deflection will be the tidal effects from the moon and the sun. If the Earth's axis were perpendicular to a line between the Earth and the moon (or the sun), the tidal force would be vertical, and there would be no horizontal deflection. Because of the angle of the Earth's axis with the ecliptic, this does rarely happen. (you need a full or a new moon at one of the equinoxes)

 

[mikeph]

The Coriolis and centrifugal forces are psuedo-forces that arise when you're working in a rotating coordinate system (I haven't been in the previous discussion, so I didn't mention them). Coriolis is proportional to the cross product of velocity with angular rotation of the system, and for an object falling over the north pole, these two vectors are parallel, so it's zero. Centrifugal is also zero because the object is always on the axis of rotation.

The physics of a problem is independent of our choice of coordinate system, and if we chose a rotating coordinate system (one in which rotating objects remain at fixed points) then we must accept the presence of these "forces". But they are just mathematical manifestations of the coordinate system, not actual forces.

 

[nasu]

I feel like my interventions has been totally ignored in this thread so far. Here is a last attempt of pointing an important effect that people seem to ignore. Even without Sun, the mass wouldn't land on point A (despite what MikeyW said), I believe.
Take Goldstein's book, 1st edition on pages 136 to 140. (138 and 139 to be more precise for the next quote)

$\theta$ is the colatitude. This effect is by far the greater I think (that would also agree with the assumptions the problem I mentioned in my last post in this thread makes allusion).
I might be totally wrong (it wouldn't surprises me), but I'd like a comment on both the Coriolis and centrifugal force that I think you have to take into account.

P.S.:On page 137, one reads

Physically, this result means that a projectile shot off at the North Pole has no initial rotational motion and hence its trajectory in the inertial space is a straight line, the apparent deflection being due to the Earth rotating beneath it.

which implies, I think, that the body wouldn't land over the north pole. Unlike what I previously thought.

I don't have the book now to check the context. However the last quote may refer to a projectile launched from the north pole along a meridian, southbound (initial velocity perpendicular to the axis of rotation).
The question under discussion here is about a body moving along the axis, as I understand.
In this later case the Coriolis force is zero, as you mentioned yourself in an earlier post.

PS. The quote from Goldstein won't snot show up in my answer, I had to add it by hand.

 

[fluidistic]

Oh I see guys. Thanks a lot, this makes things clearer to me!

 

[starz_above]

Thanks to all for their comments. Much appreciated.