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AO40 velocity

Bob said:

>I was asked how fast "those things" go.  I took a guess and probably was
>fairly correct.  That got me thinking how fast really is it.  Given the
>MA, at that part of the orbit is there a formula that would yield the
>velocity at that point?

Bob, rather than a function of MA, an easy approach is to calculate 
velocity as a function of distance from the center of the earth, given the 
semi-major axis of the orbit. This equation does that.  I've plugged in 
values for AO-40.  If I didn't make a math error, the results are shown for 
apogee and perigee, and comparison with a circular orbit at AO-40's perigee 

v^2 = G * M * ( 2/r - 1/a 

v = velocity at a point in the orbit
G = gravitational constant (6.67e-11 in mkgs units)
M = central mass (Earth masses 5.98e+24 kg)
r = radial distance from the center of the Earth
a = semi-major axis (3.6285e7 meters for AO-40)

Re = radius of the earth (6.378e6 meters, mean)
Hp = perigee height (1.155e6 meters for AO-40)
Ha = apogee height  (5.867e7 meters for AO-40)

Therefore, for AO-40:
v^2 = 6.67e-11 * 5.98e24 * (2/r - 1/3.6285e7)
     = 3.995e14 * (2/r - 2.756e-8)

AO-40 velocity at perigee:
   r = Re + Hp
   r = 7.533e6
v^2 = 3.99e14 * (2/7.533e6 - 2.75e-8)
   v =  9744 meters/sec (35,078 km/hr)

AO-40 velocity at apogee:
   r = Re + Ha
   r = 6.505e7
v^2 = 3.99e14 * (2/6.505e7 - 2.75e-8)
   v = 1138 meters/sec  (4,096 km/hr)


For comparison, consider a satellite in a circular orbit at AO-40's perigee 
height of 1155 km:
     r = a = 1.155e6 + 6.378e6
     r = a = 7.533e6
v^2 = 3.99e14 * 1/7.533e6
   v = 7277 meters/sec (26,200 km/hr)

AO-40 would zip past the "LEO" at:

9744 - 8277 = 1467 meter/sec (5,281 km/hr)

  Stacey E. Mills, W4SM    WWW:    http://www.cstone.net/~w4sm/ham1.html
    Charlottesville, VA     PGP key: http://www.cstone.net/~w4sm/key

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