Welcome to our community

Be a part of something great, join today!

Taylor's question at Yahoo! Answers regarding celestial mechanics

  • Thread starter
  • Admin
  • #1

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Here is the question:

The moon is a sphere with radius of 959 km.?


Determine an equation for the ellipse if the distance of the satellite from the surface of the moon varies from 357 km to 710 km.
I have posted a link there to this topic so the OP can see my work.
 
  • Thread starter
  • Admin
  • #2

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hello Taylor,

We know the center of the moon must be one of the foci of the ellipse. I would choose to orient the coordinate system so that this focus is at the origin, and the apogee is on the positive $x$ axis and the perigee is on the negative $x$-axis.

The center of the ellipse is therefore at the point:

\(\displaystyle (c,0)=\left(\sqrt{a^2-b^2},0 \right)\)

The major axis, will then be given by:

\(\displaystyle 2a=357+2\cdot959+710=2985\)

\(\displaystyle a=\frac{2985}{2}\)

We also must have:

\(\displaystyle a-c=959+357=1316\)

Hence:

\(\displaystyle c=\frac{353}{2}\)

\(\displaystyle b^2=\left(\frac{2985}{2} \right)^2-\left(\frac{353}{2} \right)^2=2196404\)

And so the equation describing the orbit of the satellite is:

\(\displaystyle \frac{\left(x-\frac{353}{2} \right)^2}{\left(\frac{2985}{2} \right)^2}+\frac{y^2}{2196404}=1\)

Simplify a bit:

\(\displaystyle \frac{(2x-353)^2}{8910225}+\frac{y^2}{2196404}=1\)

Here is a plot of the moon and the orbit of the satellite:

taylor.jpg