# Orbiting Pole

1. May 31, 2009

### Riemannliness

Here's a thought that was bugging me last night:

Lets say you placed a very long rigid pole (jokes kept to a minimum please) into orbit so that one end (end #1) faced the center of the earth, and the other (end #2) pointed out into space (So that the center of the earth and all the points on the pole were more or less collinear).

To simplify, lets say that the pole is sufficiently large, and end #1 is at an orbital radius of r from earth; end #2 at radius 2r.

If you investigate the necessary orbital velocities of both ends [v=sqrt(GMcentral/r)] then you'll find that: vend #2 = vend #1/sqrt2

Looking into it further you'll find that if it were geometrically possible, end #1 at this speed will complete 1 orbit for every 1/4 orbit of end#2

Assuming I haven't overlooked something trivial or made a careless mistake, how would the pole behave after it had been placed into proper orbit? What sort of mathematical tools would help me analyze the internal forces acting here? I was treating the ends as isolated as an example, but really this imbalance would occur all along the length of the object (and any orbiting object would experience this come to think of it).

I'm fairly certain that it couldn't stay in a stable orbit, in this orientation, if the only external force acting upon it was gravity. I'm just wondering if there could be a stable orbit for a rigid body like this, or if it would have some form of variable motion. There is no way to orient a straight line around a sphere so that all points on that line are equidistant from the sphere's center. Therefore there will always be some parts of this pole that are at a higher orbital radius than other parts, and thus different parts would try to move at different velocities.

Last edited: May 31, 2009
2. May 31, 2009

### D H

Staff Emeritus
You are talking about a pole, not a cloud of gas. The pole is held together as a pole by electrochemical bonds that are in general much strong that Earth's gravitational attraction. Earth's gravity will induce a huge tension in this pole. Assuming this tension doesn't pull the pole apart, the pole will merely stretch some. You might want to try to compute the tension in this long, vertically oriented pole.

Let's assume an infinitely rigid pole: It doesn't stretch at all. (The standard material used to construct such devices is called unobtanium.) Assume a vertically oriented pole of uniform density and mass m that stretches from a distance r to a distance 2r from the center of the Earth. Treating the Earth as a point mass (i.e., I am ignoring the Earth's equatorial bulge), the total gravitational force acting on the pole is

$$F_g = \frac{G M_e m_{\text{pole}}}{2r^2}$$

This is the exactly same as the gravitational force exerted on a point mass of mass m located at a distance of $\surd 2 r$ from the center of the Earth. Call this point the pole's center of gravity. The pole's orbit about the Earth will be exactly the same as if the pole was compressed to a point and collocated with the pole's center of gravity. Note well: The pole's center of gravity differs from it's center of mass. The pole's center of mass is located at a distance of 1.5 r from the center of the Earth, which is above the pole's center of gravity.

For starters, you might, as mentioned above, want to try to compute the tension in this long, vertically oriented pole. I'll leave this up to you, for now. Ignoring those tensions, you can also look into the torques acting on the rod. If the rod is in a perfectly vertical orientation and is rotating at exactly the same rate as it's orbital rate, there are no torques. This means that the vertical orientation is at a minimum a metastable orientation. There are also will be no torques if the pole is in a perfectly horizontal orientation. That orientation is also a metastable orientation. Any orientation other than perfectly vertical or perfectly horizontal will result in a gravitationally induced torque on the pole.

Qualitatively, it's easy to see which orientation is stable and which is metastable. A torque will cause the pole to rotate around its center of mass rather than its center of gravity. The configuration with the highest center of mass is the minimal energy configuration. (Another exercise that I'll leave up to you, for now.) That configuration is of course the vertical orientation. The vertical orientation is stable while the horizontal orientation is metastable. If the pole is oriented anywhere between purely vertical and purely horizontal, torques will force the pole toward a purely vertical orientation.

3. May 31, 2009

### DaveC426913

I am fairly certain that there is a straightforward answer to this.

The pole can be treated just like any other rigid body in orbit - a very long, very skinny planet.

The pole will be affected by tides, pulling on both ends, and orient itself (or in fact, stay oriented) point down. Just like the Moon, it will become a one-face satellite.

And, just like the Moon, it will be stable. Its orbit will be centred on its centre-of-mass, the outer end will be moving too fast for its orbit, which will pull it outward, the inner end will be moving too slow for its orbit, which will pull it inward/downward.

BTW, this was a trick used in Larry Niven's Descent of Anansi for a crippled shuttle to evade a competing salvage shuttle. They were carriyng a cable hundred of metres long on a reel. Extending the reel kept their CoM in the same orbit but the shuttle was out of reach!

Last edited: May 31, 2009
4. Jun 1, 2009

### D H

Staff Emeritus
It's orbit will not be centered on it's center of mass. It will be centered on its center of gravity. The pole's center of mass is at $1.5 r$. The center of gravity is lower, at $\surd 2 r$.

5. Jun 1, 2009

### Lok

The pole, if in orbit, would be spinning, reverse to it's rotation around the earth, if this makes any sense.

6. Jun 1, 2009

### D H

Staff Emeritus
As seen by an inertial observer, the pole is spinning in the same direction as it is orbiting, not the opposite.

7. Jun 1, 2009

### DaveC426913

No, the pole will stay vertical with respect to the Earth's surface. Tidal forces will hold it stable that way.

8. Jun 2, 2009

### D H

Staff Emeritus
This problem nicely illustrates where working in a rotating reference frame can be useful. Suppose the pole is in a circular orbit about the planet. To maintain a perfectly vertical (or horizontal) orientation the pole needs to be rotating with its rotational angular velocity is exactly equal to its orbital angular velocity. The pole will be stationary in a reference frame with origin at the center of the planet and that is rotating at this angular velocity.

One problem: We don't know what that angular velocity is (yet). The only forces acting on any point of mass pole in this frame are gravity, internal forces that keep the pole intact, and the (fictitious) centrifugal force. Integrating these over the entire pole yields the net apparent forces acting on the pole. The internal forces, integrated over the pole, will vanish per Newton's third law. The net apparent forces acting on the pole are gravity and centrifugal force. The total of these two forces must sum to zero for the pole to be stationary in this frame.

I'm going to generalize on the original post a bit. Suppose the pole has uniform density, a mass m, and is oriented vertically with the lower end a distance a from the center of the planet and the upper end a distance b from the center of the planet. (In terms of the original post, a=r and b=2r.) The pole's linear density is ρ=m/(b-a).

Denote the planet's gravitational coefficient as μ (μ=G*planet mass). Finally, the rotating reference frame needs to be defined. Without loss of generality, assume the pole is located on the x axis and the angular velocity vector is oriented in the +z direction. The magnitude of the angular velocity vector is some unknown quantity ω.

With this, total gravitational and centrifugal forces on the pole are

\aligned {\boldsymbol F}_g &= \int_a^b-\,\frac{\mu\rho dx}{x^2} \hat x \\ &= -\mu\rho\left(\frac 1 a - \frac 1 b\right) \hat x \\ &= -\frac{\mu m}{ab} \hat x \\ {\boldsymbol F}_c &= \int_a^b\rho \omega^2 x \, dx \hat x \\ &= \frac 1 2 \rho\omega^2(b^2-a^2) \hat x \\ &= m\omega^2\frac{a+b}2 \hat x \endaligned

The condition needed to make these forces to sum to zero is

$$\omega^2 = \frac{\mu}{ab(a+b)/2}$$

In terms of an equivalent point mass,
• The total centrifugal force on the pole is the same as that on a point mass of mass m located at the arithmetic mean of the two end points of the pole.
• The total gravitational force on the pole is the same as that on a point mass of mass m is located at the geometric mean of the two end points of the pole.
• The orbital rate of the pole is the same as that of a point mass located at the geometric mean of the two end points and the midpoint of the pole.

Call this final point (the geometric mean of a, (a+b)/2, and b) the pole's gravitational pivot point. Note that the gravitational and centrifugal forces are equal but opposite at the pivot point. The gravitational force dominates over centrifugal force for points below the pivot point; centrifugal force dominates for points above the pivot point.

What happens if the pole is oriented a bit off from a perfectly vertical? For points below and above the pivot point, the net force on the point will have a component normal to the that counters the misalignment. The net result is a torque that moves the pole back to a vertical alignment.

Next, suppose the pole is in a perfectly horizontal orientation. There is no net torque on the pole. If the pole deviates at all from this perfectly horizontal orientation, there will be a net torque on the pole. This time, however, the torque will act to increase the misalignment from the purely horizontal. The pole will quickly snap to a vertical orientation.