Strange Increase in the Total Energy of a Rocket in the Earth-Moon system

"Matt" wrote in message
oups.com...
So to summarise, the increase in the total energy of the rocket as it
approaches the Moon is due to the Moon's kinetic and potential energy
not being considered. The thing I'm having a problem with is that the
*only* way the kinetic energy of a rocket in freefall can increase is
when there is a corresponding decrease in the gravitational potential
between the rocket and the body in question (i.e. the Moon).

Also, where does this extra kinetic energy come from? It can't be due
to a decrease in the Moon's kinetic energy, as the Moon is also being
gravitationally attracted to the rocket. The rocket's trajectory takes
it just ahead of where the Moon will be in its orbital path, so the
Moon's kinetic energy will increase rather then decrease.

Does this increase in kinetic energy for both the rocket and the Moon
come from a decrease in the Moon's gravitational potential?

Consider a fixed point in space that happens to lie on the
Moon's trajectory. The frame of reference is the barycenter
and is not rotating. If you were to look at the gravitational
potential at that location over time you'd see it change as the
Earth and Moon move, heading towards infinity as the Moon bears
down on it (point mass simplification).

So the potential associated with a location can change without
there being any related change in total kinetic energy anywhere
in the system! A test mass fixed at that location can find its
potential energy change without a change in its kinetic energy.
If the test mass is free to move then it will respond to the
potential, changing the KE of the test mass according to the
usual laws of physics but without any guarantee that the
change in KE is strictly proportional to the change in PE
over time. Interestingly, the behavior of the energies might
be more intuitive if you were to switch to a rotating frame
of reference in which the Earth and Moon are fixed! You
would need to incorporate another potential term associated
with the rotation proportional to the distance from the
origin.

All this aside, if you believe that the kinetic energy of
the spacecraft is not in accord with what you think it
should be given the geometry of the situation you could
have a problem with the integration. In this case you
might want to investigate its properties. Another poster
has suggested a simplified situation in which the Earth and
Moon are fixed. I would also suggest varying the size of
the timestep to see if numerical error is creeping in. An
RK integrator should be able to tolerate a goodly sized
timestep. Try a timestep of an hour rather than a second
to see what happens.

In article , "Greg Neill" writes:
over time. Interestingly, the behavior of the energies might
be more intuitive if you were to switch to a rotating frame
of reference in which the Earth and Moon are fixed! You
would need to incorporate another potential term associated
with the rotation proportional to the distance from the
origin.

I think you'll find that centrifugal potential goes as the square
of the radius.

force = m omega^2 r
PE = 1/2 m omega^2 r^2

As a sanity check, note that this PE matches the KE of
a "motionless" object viewed from the rotating frame.

Choose an appropriate sign convention, of course.

wrote in message ...
In article , "Greg Neill"
writes:
over time. Interestingly, the behavior of the energies might
be more intuitive if you were to switch to a rotating frame
of reference in which the Earth and Moon are fixed! You
would need to incorporate another potential term associated
with the rotation proportional to the distance from the
origin.

I think you'll find that centrifugal potential goes as the square
of the radius.

force = m omega^2 r
PE = 1/2 m omega^2 r^2

As a sanity check, note that this PE matches the KE of
a "motionless" object viewed from the rotating frame.

Choose an appropriate sign convention, of course.

D'oh! Of course. I was writing faster than I was
thinking. Again.

Cheers.