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The cartesian heliocentric ecliptic coordinates for an orbit's aphelion are:
x = +0.5877971
y = -2.098983
z = -0.008080998

The unit vector normal to the orbit is:
Xn = -0.0004498044
Yn = -0.003975890
Zn = +0.9999920

The orbit's semimajor axis:
a = 1.319533

Distances are in astronomical units.

2006-07-20 05:38:34 · 3 answers · asked by David S 5 in Science & Mathematics Astronomy & Space

This isn't a homework problem, buddy. I graduated in 1982.

2006-07-20 07:36:23 · update #1

3 answers

The velocity would be a little harder, but the speed is easy.
I calculated the distance of aphelion from your Cartesian coordinates as follows: sqrt(x^2+y^+z^2)=2.179747776 AU
Multiply that by 149,597,871 km/AU to get the aphelion distance of 326,085,626.6 km.
Also multiply a by km/AU to get a=197,399,328 km.
According to Wikipeda, the Sun's gravitational parameter (which is basically just GM) is 132,712,440,018
Then just plug those values into the vis viva equation (also from Wikipedia) to get a speed of 11.90246 km/s.

(EDIT) OK, I added this direction part after some thought.
The direction would be tangent to the ellipse at that point. Since you have specified that the point in question is the aphelion, you know that the direction is normal to the direction to the Sun. Of course it would also be normal to the unit vector normal to the orbit . I would think you could get a unit vector to (from?) the Sun by dividing the Cartesian coordinates by the magnitude of that vector. Then you could take the cross product of that vector and the unit normal vector to the orbit to get a unit vector specifying the direction. Multiply that unit vector by the speed to get velocity. When I did that I got Vx=11.46153, Vy=3.209606 and Vz=0.017917. A couple sanity checks follow. Since z is near 0 and Zn is near 1, the plane of the orbit is near the X-Y plane. The Z velocity I got is small, so check. The ratio of x to y is the same as the ratio of Vy to Vx, meaning the velocity in the X-Y plane is at right angles to the direction to the Sun, so check. Looks OK to me.

2006-07-20 06:16:44 · answer #1 · answered by campbelp2002 7 · 4 0

man, this place is not for people to do your hw for you...

crack a book, this is an easy orbital mechanics problem

you already have the sun-relative coordinates, you know the orbit normal, and you know a... there is a very simple equation for velocity at any point in a given elliptical orbit, and the velocity's direction can be found using what you know about the orbit itself.

2006-07-20 12:48:10 · answer #2 · answered by AresIV 4 · 0 0

no

2006-07-20 12:43:47 · answer #3 · answered by Michael G 2 · 0 0

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