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*Subject*: RE: [amsat-bb] Orbital velocity question*From*: "James C. Hall, MD" <heartdoc@xxxxxxxxx>*Date*: Mon, 30 May 2005 10:06:12 -0500*In-reply-to*: <HJEOIEMJILBJEACBOHKFGEKDCFAA.n2wwd@mindspring.com>*Thread-index*: AcVlElUCqmI1p02xQ76GeLO6wY9vNgAE0/Mg

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 ---- 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

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