|Feb4-09, 05:07 PM||#1|
I am trying to figure out the speed of Halley's Comet in km/sec. at aphelion and perihelion... my answers just don't seem to be practical.. please help thanks...
I am using
v² = (4Π²a³) / (P²) * ((2/r) - (1/a))
where a = mean distance from the sun (semimajor axis of the ellipse)
P= sidereal period (75 years)
r = distance of the object from the Sun at a given instant
a=16.8 A.U. c=16 A.U.
"r" at perihelion = a-c "r" at aphelion = a + c
|Feb6-09, 10:14 PM||#2|
Well, what are your answers?
I see that you have quoted units in terms of years and astronomical units. Did you convert these units to seconds and meters, respectively?
As a reference point for your answers, know that the earth moves through space in its orbit about the sun around 30 km/s, which is quite fast relative to what we are accustomed to in our daily experiences living on earth.
|Feb9-09, 12:31 AM||#3|
Maybe this will add another equation for a simultaneous solution? Conservation of angular momentum. Perihelion and aphelion are the two points where the radius vector from the sun to the comet is perpendicular to the tangential velocity of the comet, so the sin theta drops out of the expression for angular momentum and it's just mvr. Then mass cancels out, so v_perihelion r_perihelion = v_aphelion r_aphelion.
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