Taylor's question at Yahoo Answers regarding celestial mechanics

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SUMMARY

The discussion focuses on deriving the equation for the elliptical orbit of a satellite around the moon, which has a radius of 959 km. The satellite's distance from the moon's surface varies between 357 km and 710 km. The center of the moon is positioned at one of the foci of the ellipse, leading to the major axis calculation of 2985 km. The final equation for the satellite's orbit is presented as $$\frac{(2x-353)^2}{8910225}+\frac{y^2}{2196404}=1$$.

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MarkFL
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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.
 
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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:

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

The major axis, will then be given by:

$$2a=357+2\cdot959+710=2985$$

$$a=\frac{2985}{2}$$

We also must have:

$$a-c=959+357=1316$$

Hence:

$$c=\frac{353}{2}$$

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

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

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

Simplify a bit:

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

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

View attachment 1507
 

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