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*Subject*: RE: [amsat-bb] Orbital velocity question*From*: "James C. Hall, MD" <heartdoc@xxxxxxxxx>*Date*: Mon, 30 May 2005 19:48:36 -0500*In-reply-to*: <000201c56578$95894790$6501a8c0@N4NAB>*Thread-index*: AcVleJbGisUgZZ5qSfuEbYTnnOojIgAAZY6w

That's it !! Well done John. I see you were staring at it longer than I. HI HI I've printed it off so I can keep it - I knew it was the correct equation, I just couldn't quite get to it. Thanks for your expertise ! 73, Jamie WB4YDL _____ From: John Henderson [mailto:jah@ec.rr.com] Sent: Monday, May 30, 2005 7:35 PM To: 'James C. Hall, MD' Cc: AMSAT-BB Subject: RE: [amsat-bb] Orbital velocity question Jamie, I finally derived the equation in the Journal and it is correct. Hope I can type this clearly.... The balancing force on a satellite at distance r is: (GMm)/r**2 = mw**2r G = gravity at surface, M = Earth mass m=sat mass r=distance from center of Earth w= angular velocity of sat The Orbital Velocity found from using the above and orb velocity : v = rw and g = GM/R**2 where g is gravity at sat and R is earth radius. Solving the forced equation for wr we get SqRT(GM/r) =wr Therefore v = SqRT (GM/R); solve g=GM/R**2 for GM we get g(R**2) Substitute in v= SqRT (GM/r) GM=g(R**2) and you get: V = SqRT (( gR**2)/r) 73, John N4NAB/ FM14lq AMSAT # 32411 ---- 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

**References**:**RE: Orbital velocity question***From:*John Henderson

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