# RE: Orbital velocity question

```Hi Ken:

This equation appears to be derived when you set centripetal force (mv^2/r)
equal to gravitational force (GMm/r^2), Mu= GM (gravitational constant*earth
mass). For the simpler circular orbit, you end up with v^2 = GM/r, which of
course is what I learned in high school to solve orbital velocity.

What apparently is being done in the AMSAT journal is a proportion where
g`/g o< R/r and then g`= gR/r, where g` is acceleration of gravity at
orbital height. As orbital height(r) approaches the surface of earth(R),
g`=g. As orbital height increases, g`< g. You can think of it as the
"weight" of the satellite (mg`)at orbital height will equal its centripetal
force (mv^2/r) - this will give g`= v^2/r AT orbital height. On the ground,
it would be g = v^2/R, where R (not r) is the radius of the earth.

Substituting g` in the proportion, we get gR/r = v^2/r (NOT v^2/R as the
equation in the journal would suggest). This is where I get stuck - probably
just looking at it too long :)

73, Jamie
WB4YDL

-----Original Message-----
From: Ken Ernandes [mailto:n2wwd@mindspring.com]
Sent: Monday, May 30, 2005 7:22 AM
To: James C. Hall, MD; amsat-bb@amsat.org
Subject: RE: [amsat-bb] Orbital velocity question

Jamie -

I my copy of the Journal, but a generic equation for orbital speed is:

v^2/Mu = 2/r - 1/a

v  = speed (i.e., velocity magnitude)
Mu = gravitational parameter for attracting body (Earth = 398600.5
km^3/sec^2)
r  = instantaneous orbital radial position magnitude
a  = orbital semi-major axis

Does this help?

73, Ken Ernandes, N2WWD

-----Original Message-----
From: owner-AMSAT-BB@amsat.org [mailto:owner-AMSAT-BB@amsat.org]On
Behalf Of James C. Hall, MD
Sent: Sunday, May 29, 2005 7:22 PM
To: amsat-bb@amsat.org
Subject: [amsat-bb] Orbital velocity question

Hello all:

I've been out of Physics for some time, but while reviewing Gould Smith's
excellent article on orbital basics in the Jan/Feb issue of the AMSAT
journal, I was confused as to the derivation of the orbital velocity
equation. I've fooled with it for a few hours but I can't seem to figure out
the R^2/r factor in the equation. I understand that the acceleration due to
gravity is less at the orbital height and that, due to conservation of
angular momentum, v`r = vR (where v` is the orbital velocity), but when I
square both sides and divide by r^2, I get stuck . because g = v^2/R, not
v^2/r.

So I can't get the v`^2 = g R^2/r.  I'm missing something .

I know, I know . it's been a slow day !!

73, Jamie

WB4YDL
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