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AO40 velocity
- Subject: [amsat-bb] AO40 velocity
- From: "Stacey E. Mills" <w4sm@xxxxxxxxxx>
- Date: Thu, 29 Aug 2002 17:58:55 -0400
Bob said:
>Hello
>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
height.
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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