Would an Object Fall Straight Through a Hole in a Spinning Planet?

[Dmstifik8ion]

Ignoring the intense heat and pressure of subterranean environments, the difficulties of constructing a hole under such conditions, drag and friction, etc:

Only a hole through the axis of rotation will provide a straight trajectory or one the passes through the exact center of a spinning planet, however;

Would the trajectory (determined entirely by the planets gravity and surface rotational momentum alone) of an object released at any point on the surface of a rotating planet bring it invariably to rendezvous with the opposite (half way around) side of the sphere?

In other words:

If a hole of suitable size to permit my passage suddenly opened up beneath me, would I travel to the opposite side of the Earth?

 

[DaveC426913]

I don't have a rigorous answer but my take on it is no.

Consider that two planets of the same mass and radius could easily have very different rotational speeds. So, while the object is falling at the same rate in both, one planet might turn only .1 degree while another might turn 20 degrees. Without bothering to do the math, we can see that the falling body will not be affected by this rotation, thus they cannot possibly both come out in the same place.

 

[Dmstifik8ion]

I don't have a rigorous answer but my take on it is no.

Consider that two planets of the same mass and radius could easily have very different rotational speeds. So, while the object is falling at the same rate in both, one planet might turn only .1 degree while another might turn 20 degrees. Without bothering to do the math, we can see that the falling body will not be affected by this rotation, thus they cannot possibly both come out in the same place.

My thought was that the object when released would have the component "horizontal" momentum of the surface from which it was released. This horizontal component of its initial vector would cause the object in free-fall to "orbit" the center of gravity of the planet.

(Rotational velocities exceeding gravitational accelleration would not be permitted as this would change the figure of the planet accordingly, that is a planets surface can not withstand an angular momentum greater than the acceleration of gravity without becoming a satellite).

I hope you can appreciate that I am not trying to change the rules as this discussion progresses. This is only intended as a clarification of my original hypothesis. With apologies I have edited the original question in italics.

However I do have another but related question:
A faster rate of rotation would increase the perigee of the "orbit" but would the time required to return to the (opposite?) surface remain constant for any rate of rotation within planetary limits?

 

[LURCH]

My thought was that the object when released would have the component "horizontal" momentum of the surface from which it was released. This horizontal component of its initial vector would cause the object in free-fall to "orbit" the center of gravity of the planet.

That's exactly correct. If you don't hit the sides fo the hole (let's say the hole continues to "open up" under you, or ahead of you as you fall), then you will drop into an orbit lower than the surface. You would return to about the same place you fell through, though planetary rotation would effect your location of return to the surface. Let's say you fall through from the equator, at midnight. You would be movign about 1000 m/hr when you begin infall. This would drop you into a highly elliptical orbit which would bring you back to nearly the same location from which you fell; midnight (the point on the surface farthest from the sun). But, if the trip took you a couple of hours, you would find yourself popping up in a town about 2000 miles west of where you fell through.

(Rotational velocities exceeding gravitational accelleration would not be permitted as this would change the figure of the planet accordingly, that is a planets surface can not withstand an angular momentum greater than the acceleration of gravity without becoming a satellite).

I hope you can appreciate that I am not trying to change the rules as this discussion progresses. This is only intended as a clarification of my original hypothesis. With apologies I have edited the original question in italics.

However I do have another but related question:
A faster rate of rotation would increase the perigee of the "orbit" but would the time required to return to the (opposite?) surface remain constant for any rate of rotation within planetary limits?

No, the faster orbit is in fact a "higher" orbit, having the same apphelion but a higher semimajor axis (or lower foci), so it would take longer, (because the square of the period is proportional to the cube of the semimajor axis).

 

[Janus]

My thought was that the object when released would have the component "horizontal" momentum of the surface from which it was released. This horizontal component of its initial vector would cause the object in free-fall to "orbit" the center of gravity of the planet.

Yes, but it would not be a typical Orbit. in a standard orbit, the object follows an ellipse with the centrer of gravity of the planet at one focus of the ellipse. In this case the center of gravity will be at the center of the ellipse. What you get is harmonic motion.

However I do have another but related question:
A faster rate of rotation would increase the perigee of the "orbit" but would the time required to return to the (opposite?) surface remain constant for any rate of rotation within planetary limits?

First off, this "orbit" would have two "perigees". As it will pass the center of the Planet twice, once on its trip to the "opposite" side and again during the return trip.

And yes, the time would remain constant. If fact, it stays the same right up to the point where the rate of rotation causes a speed at the surface equal to orbital speed.

 

[Dmstifik8ion]

Thanx Janus, LURCH, and DaveC426913,
Obviously a subterranean "orbit" wound not be geosynchronous, what was I thinking? . . . don't answer that! I'm off with my shovel now; wish me luck!