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Tethered satellites (Posted on 2005-04-13) Difficulty: 3 of 5
Two identical spheres are connected by an elastic tether. The tether obeys Hooke's Law for ideal springs. At a particular moment in time, the tether is in a straight line, at its resting length, neither stretching nor contracting. This assembly is then placed into a circular orbit around the Earth, and oriented so that a line drawn from one sphere through the tether and the other sphere points directly at the Earth.

Give a qualitative description of the motion of the two spheres relative to each other over time.

See The Solution Submitted by Larry    
Rating: 3.5000 (4 votes)

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Solution No Subject | Comment 12 of 13 |

Before giving what in my opinion is the solution to this problem, I should first point out that most of the previous comments missed out on an important point.  The first part of the problem is to actually realize what "oriented so that a line drawn from one sphere through the tether and the other sphere points directly at the Earth" means.  While one could argue that this means the spheres are not rotating, anyone with any experience in these matters knows that is somewhat insane.  I would argue that it makes much more sense that this text means that the spheres, while being otherwise externally held fixed, are placed to point at the earth nearly permanently (with an angular velocity equal to its orbital angular velocity).  An easy argument to persuade one on this issue is that if the text meant the former, it might as well also potentially mean that the object is spinning at some strange arbitrary high rate, perhaps even opposite to its circular motion, but on the plane of the orbit, such that for split instances it points towards the Earth, which is clearly not what is intended.  Now that we know this, those other comments talking about different angular velocities are wrong. 

<o:p> </o:p>

In the rotating frame, we see that the inner ball, once "released", sees a stronger gravitational force stronger than the outer ball.  Thus, the inner ball falls down, while the outer object flies out slightly.  The distance it "wants" to move is small, I guess we can work it out with in the CM rotating frame for the lower ball E=mgl/2=(1/2)kx^2 yielding 2x=2sqrt(mlg/k) for the net displacement where l is the length of the tether of spring constant k, and m is each ball mass.  Nevertheless, during this time, in the rotating frame both balls experience Coriolis forces, which makes the inner ball speed up and the outer ball slow down.  The net result is a new oscillatory angular velocity initially opposite to the initial one (what a delightful sentence that is to type!).  But it is oscillatory, since it then swings back and forth like a pendulum for somewhat obvious reasons.  To get the oscillatory frequency, which should be quite small, you simply treat it as a pendulum.  <o:p></o:p>

<o:p> </o:p>

As you can see, no pen or paper was required.  So claims that this is a class 6 problem, whatever exactly that is, are clearly wrong.  With minor changes, such as putting the spheres near the event horizon, that could be rectified.  Luckily for all of us, I don’t yet have permission to post such a version.


  Posted by Jason Taylor on 2005-08-01 02:23:07
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