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The problem arises when we do Hohmann transfer problems. Take the following
numerical example. In a transfer from a 7E6 m radius orbit to a 1E7 radius
orbit about the Earth the required delta Vs are obtained by subtracting the
lower circular orbit speed from the perigee speed of the transfer orbit (8188
m/s - 7449 m/s = 639 m/s) and subtracting the apogee speed of the transfer
orbit from the upper circular orbit speed (6316 m/s - 5731 m/s = 585 m/s).
Assume a 1 kg spacecraft and an external energy source (space - based
laser?), 2.0E5 J of kinetic energy would be added for the first delta V and
1.7E5 J of kinetic energy at the second delta V. This is a total of 3.7E5 J.
If this is done using the total energy of an orbit method where E = -GMm/2a
then a different result is obtained. The energy of the lower circular orbit
is -2.85E7 J, the energy of the transfer orbit is -2.35E7 J, and the energy
of the upper circular orbit is -2.0E7 J. A student who uses this method to
compute the transfer obtains 5.0E6 J for the first delta V and 3.6E6 J for
the second for a total of 8.6E6 J. I have got to be doing something wrong
here. I know it might be tedious to find out where I am going wrong but I
have been unable to find the flaw. I would appreciate any help.
John Mallinckrodt wrote:
On Thu, 13 May 1999, Dan Burns wrote:
... If one wants to transfer between 2 circular orbits a Hohmann
transfer is the most efficient way if there is no plane change. If the
energy required for the Hohmann transfer is compared to the difference
in the total energy (-GMm/2a) of the 2 circular orbits, it is always
greater. Where was this energy "lost"? ...
Dan,
Could you explain what you mean by this? As I see it the energy required
for insertion into the transfer orbit *is* the difference between the
energies of the transfer orbit and the original circular orbit. Similarly,
the energy required for insertion into the final circular orbit *is* the
difference between the energies of the final circular orbit and the
transfer orbit. I don't see how these two energies could possibly add up
to anything other than the difference between the initial and final
orbital energies.
John
----------------------------------------------------------
A. John Mallinckrodt http://www.csupomona.edu/~ajm
Professor of Physics mailto:ajm@csupomona.edu
Physics Department voice:909-869-4054
Cal Poly Pomona fax:909-869-5090
Pomona, CA 91768-4031 office:Building 8, Room 223