I've been playing with formulas for circular orbits and characteristic orbital energy hoping to ovoid the need to remember the value of the gravitational parameter, mu. It turns out that the equations have been and can be simplified to some surprising results, (to me). But the accepted value of the radius of the Earth seems to be in error. Please follow my logic and let me know where I erred, or not.

Start by finding velocity of a circular orbit, V

_{o} around Earth.

eqn. 1) V

_{o}^{2} = mu/r , but I don't want to use mu, so I go the long way by setting gravitational acceleration equal to centrifugal acceleration and solving for V

_{o}^{2}.

eqn. 2) g

_{h} = a

_{c}eqn. 3) g = mu/r

^{2} eqn. 4) a

_{c} = V

_{o}^{2}/r where

r is the Earth centered distance of the circular orbit, equals r

_{e} + h , that is, Earth's radius plus orbit altitude.

Write eqn. 3) twice and divide equations. After substituting values gives

eqn. 5) g

_{h} = g

_{surface} * r

_{e}^{2}/(r

_{e} +h)

^{2}where g

_{surface} = 9.80665 m/s

^{2} by definition

Substitute eqn. 4) and eqn. 5) into eqn. 2), rearranging and using r = r

_{e} + h, gives

V

_{o}^{2} = g

_{surface}*r

_{e}^{2}/(r

_{e} + h)

So I now have the equation I was after, circular orbital velocity without resorting to the use of mu.

Now I looked at characteristic energy, C

_{3} of the orbit. From Wikipedia I find

eqn. 6) V

_{o}/2 - mu/r = C

_{3}/2 where r is again equal to r

_{e} + h.

Just out of curiosity and to avoid using mu, I substitute eqn. 1) into eqn. 6) and multiply through by 2. The result

eqn. 7) C

_{3} = - V

_{o}^{2}Characteristic energy equals the square of circular orbital velocity. This did surprise me so I checked numerically over a range of heights using Excel. With

mu = 398600.4418 km^3/s^2 and

r

_{e} = 6,371,000 meters

I got good agreement, but not perfect agreement between the C

_{3} values calculated using eqn. 6) and eqn. 7) . The difference was a constant 0.2766% over the altitude range from 0 to 55,000 km. So what is wrong? The only common input parameter in both calculations for C

_{3} is Earth radius. A bit of variation on r

_{e} results in 6,375,417 meters giving 0.0000% difference to four decimal places in the values of C

_{3} calculated by eqn. 6) and eqn. 7).

I will need to read your comments before I'll know if they are welcome.

add: Looking at

http://en.wikipedia.org/wiki/Earth_radius I see a lot of different values for the radius of the Earth, ranging from 6,353 km to 6,384 km.

I will bravely conjecture that the Astrodynamic radius of the earth = 6,375.417 km.