Re: Orbit Transfer

[Dan Burns <kilroi@LGHS.NET>]

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
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