Dropping a mass directly over the north pole

1. Jul 25, 2011

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.

2. Jul 25, 2011

mikeph

Why wouldn't you expect it to?

3. Jul 25, 2011

nasu

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

4. Jul 25, 2011

starz_above

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 travelling at a tangent below it?

5. Jul 25, 2011

starz_above

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

6. Jul 25, 2011

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.

7. Jul 25, 2011

Delta Kilo

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.

8. Jul 25, 2011

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.

9. Jul 25, 2011

Shoku Z

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

10. Jul 25, 2011

starz_above

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.

11. Jul 25, 2011

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.

12. Jul 25, 2011

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).

13. Jul 25, 2011

fluidistic

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

14. Jul 25, 2011

DaveC426913

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.

15. Jul 26, 2011

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.

16. Jul 26, 2011

starz_above

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 travelling at 30 km/s in orbit.

17. Jul 26, 2011

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.

18. Jul 26, 2011

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

19. Jul 26, 2011

starz_above

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.

20. Jul 26, 2011

starz_above

Your assumptions are correct. Thanks for putting it clearly.

21. Jul 26, 2011

mikeph

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.

22. Jul 26, 2011

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)?

23. Jul 26, 2011

mikeph

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.

24. Jul 26, 2011

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.

25. Jul 26, 2011

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.