Problem: Imagine that you are orbiting a neutron star whose mass is 100 * 10^30 kilograms at a distance of 29 kilometers from its center. Give the difference between the force exerted by gravity on a one kilogram mass and the force exerted by gravity on a one kilogram mass orbiting one kilometer further from the center of the planet. By how much does the orbital velocity of the first object exceed that of the second?
From your first result, determine the average force gradient (force change per unit of distance), in Newtons per meter, for a one kilogram mass between 29 and 29+1 kilometers from the center of the neutron star. Give the approximate force difference you would therefore expect between one kilogram of your left shoulder and one kilogram of your right shoulder, assuming that one shoulder is .4500 meter further from the star than the other. Discuss in your summary the implications for your structural integrity.
From your second result, determine the average velocity gradient, in (m/s) per meter, between orbits at 29 km and at 29+1 km from the center. Give the approximate time it would therefore require the shoulder closer to the planet to lap the other, assuming that they become separated from one another.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Solution: The two orbital distances are 29 km = 2.900 * 10 ^ 4 meters and 29+1 km = 3.000 * 10 ^ 4 meters. The force on one kilogram can be determined by substituting the planet mass 100 * 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
7.931 * 10 ^ 13 Newtons,
and at the second distance it is
7.411 * 10 ^ 13 Newtons.
The difference is
force difference for 1 km separation = 5.200 * 10 ^ 12 Newtons.
This difference occurs over a difference in orbital radii of 1000 meters. The average difference per meter is therefore
ave. force gradient = force / distance = 5.200 * 10 ^ 12 Newtons / 1000 meters = 5.200 * 10 ^ 8 Newtons/meter.
This is the average force gradient between the two orbital distances. The difference in forces between the shoulders would therefore be
approx. force difference between shoulders = ( .4500 meter)( 5.200 * 10 ^ 8 Newtons/meter) = 2.340 * 10 ^ 7 Newtons.
It is safe to say that your shoulders could not withstand this force difference.
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:
condition for circular orbit: mv ^ 2/r = G m M / r ^ 2.
Solving for v, we obtain
orbital velocity = v = `sqrt(G M / r).
When r = 2.900 * 10 ^ 4 meters, using M = 100*10^30 kilograms, we obtain
velocity at orbital distance = v = 4.795 * 10 ^ 7 m/s;
when r = 3.000 * 10 ^ 4 meters, we obtain
velocity at orbital distance + 1 km = v = 4.714 * 10 ^ 7 m/s.
The difference is 8.100 * 10 ^ 5 m/s. The difference occurs over 1000 meters of orbital radius, so the average velocity gradient is
ave. vel gradient = vel difference / distance = 8.100 * 10 ^ 5 m/s / 1000 m = 810(m/s) / meter.
For two shoulders separated by .4500 meter, this implies a velocity difference of approximately
velocity difference for two shoulder chunks = ( .4500 meter)( 810 (m/s) / meter) = 364.5 m/s.
To lap the outer shoulder chunk, the inner shoulder chunk must travel one circumference, or 2 `pi ( 2.900 * 10 ^ 4 meters) = 1.822 * 10 ^ 5 meters, further than the outer. At the velocity difference, it requires
`dt = `ds / vDiff = 1.822 * 10 ^ 5 meters/ 364.5 m/s = 499.8 seconds to accomplish this.
Generalized Response: At distance r1 from a planet of mass M, and object of mass m will experience force
F1 = G M m / r1^2.
At distance r2 the force will be
F2 = G M m / r2^2.
The average force gradient between these two distances will therefore be
ave. force gradient = (F2 - F1) / (r2 - r1) = G M m (1/r2^2 - 1/r1^2) / (r2 - r1).
Between radius r1 and r1 + `dr, the distance is `dr and the approximate force difference for mass m is
approx. force difference = force gradient * `dr = [ G M m (1/r2^2 - 1/r1^2) / (r2 - r1) ] `dr = G M m (1/r2^2 - 1/r1^2) * [`dr / (r2 - r1) ].
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
velocity difference, r1 to r2 =`sqrt(G m / r2 ) - `sqrt(G m / r1).
The average `velocity gradient' will therefore be
ave. vel. gradient = vel diff. / distance = [`sqrt(G m / r2 ) - `sqrt(G m / r1) ] / (r2 - r1).
Between distance r1 and r1 + `dr, the predicted velocity difference will therefore be
approx. vel. diff. = ave. vel. grad. * `dr = [ `sqrt(G m / r2 ) - `sqrt(G m / r1) ] * `dr / (r2 - r1).
We easily obtain the `lapping time' by dividing this velocity difference into the orbital circumference 2 `pi r.
.
.
.
.
.
.
.
.
.
.
Figure Description: 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.
