# RE: Orbital velocity question

• Subject: RE: [amsat-bb] Orbital velocity question
• From: "Ken Ernandes" <n2wwd@xxxxxxxxxxxxxx>
• Date: Mon, 30 May 2005 08:21:54 -0400

```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
----
Sent via amsat-bb@amsat.org. Opinions expressed are those of the author.
Not an AMSAT member? Join now to support the amateur satellite program!
To unsubscribe, send "unsubscribe amsat-bb" to Majordomo@amsat.org
----
Sent via amsat-bb@amsat.org. Opinions expressed are those of the author.
Not an AMSAT member? Join now to support the amateur satellite program!
To unsubscribe, send "unsubscribe amsat-bb" to Majordomo@amsat.org

```

AMSAT Home