Imagine that you are orbiting a neutron star whose mass is 55 * 10^30 kilograms at a distance of 27 kilometers from its center.
From your first result, which gives the force change per kilometer, determine the average force gradient (force change per unit of distance), in Newtons per meter, for one kilogram masses between 27 and 27+1 kilometers from the center of the neutron star.
From your second result, determine the average velocity gradient, in (m/s) per meter, between orbits at 27 km and at 27+1 km from the center.
The two orbital distances are 27 km = 27000 meters and 27+1 km = 28000 meters. The force on one kilogram can be determined by substituting the planet mass 55 * 10^30 kg, the hypothesized one kilogram mass, and the orbital distance r into the expression F = G m M / r ^ 2, with G = 6.67 * 10^-11 N m ^ 2/kg ^ 2.
At the first distance the force is
and at the second distance it is
The difference is
This difference occurs over a difference in orbital radii of 1000 meters. The average difference per meter is therefore
This is the average force gradient between the two orbital distances. The difference in forces between the shoulders would therefore be
The orbital velocity of an object will be related to its distance from the center by the fact that gravitational force is equal to the centripetal acceleration:
Solving for v, we obtain
When r = 27000 meters, using M = 55*10^30 kilograms, we obtain
when r = 28000 meters, we obtain
The difference is 6700000 m/s. The difference occurs over 1000 meters of orbital radius, so the average velocity gradient is
For two shoulders separated by .43 meter, this implies a velocity difference of approximately
To lap the outer shoulder chunk, the inner shoulder chunk must travel one circumference, or 2 `pi ( 27000 meters) = `vbl10 meters, further than the outer. At the velocity difference, it requires
to accomplish this.
At distance r1 from a planet of mass M, and object of mass m will experience force
At distance r2 the force will be
The average force gradient between these two distances will therefore be
Between radius r1 and r1 + `dr, the distance is `dr and the approximate force difference for mass m is
In the latter form we see the difference as the product of the change in gravitational force over the original separation, multiplied by the ratio of the present separation to the original.
From the condition for a circular orbit we obtain the expression v = `sqrt(G M / r). So the velocity difference between orbital distances r1 and r2 will be
The average `velocity gradient' will therefore be
Between distance r1 and r1 + `dr, the predicted velocity difference will therefore be
We easily obtain the `lapping time' by dividing this velocity difference into the orbital circumference 2 `pi r.
The figure below shows the velocities v1 and v2 at distances r1 and r2 from the neutron star (depicted in blue, which is not realistic: a neutron star is probably the most perfectly smooth object in the universe and in fact constitutes a nearly perfect mirror; we might assume that this neutron star is illuminated by a blue source).
The velocity doesn't fall off linearly with distance, but for small velocity changes we can make the approximating assumption that it does. In this case the rate at whichvelocity changes with distance, or the velocity gradient, is
velocity gradient = (v(r2) - v(r1)) / (r2 - r1).
Over the distance `dr between r1 and r1 + `dr, the velocity will therefore change by approximately the product of `dr and the velocity gradient.
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