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I am wondering how to calculate the area between the 2 red dots towards the bottom and in turn calculate the time elapsed between the right red dot and the green dot at apogee. I figure if I can find the area between the 2 red dots, I can subtract this from the total area of the ellipse and use Time = A/(dA/dt). I know the radius of perigee/apogee, eccentricity, and semi major/minor axis but not the true anomaly. The blue object is earth so the horizontal distance between the red dots is 1/2 the radius of the earth.

ellipse

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  • $\begingroup$ Distance is the full radius. Half the diameter. Also I would consider the "eclipse" around apogee. $\endgroup$ Feb 24, 2019 at 12:23

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This is going to take a few steps, but here goes.. We consider just the motion from perigee to apogee. The return tobperigee is, of course, handled by symmetry.

We start with a little nomenclature.

Points:

$O$ = center of the ellipse

$G$ = focus you're orbiting around

$P$ = perigee point

$Q$ = your current point in the orbit

Distances and angles:

$a$ = semimajor axis

$e$ = eccentricity

$c=ae=|OG|$ = half the distance between the two foci

$\theta$ = angular displacement from perigee, as seen from $G$; the measure of $\angle PGQ$

$r(\theta)$=|GQ| = current distance from $G$

From one of the polar forms for the ellipse we have, in the given nomenclature:

$r(\theta)=\dfrac{r(90°)}{1+e\cos\theta}=\dfrac{ar(90°)}{a+c\cos\theta}$

We must have $r(0)=a-x$ forcing

$ar(90°)=(a+c)(a-c); r(90°)=(a^2-c^2)/a=a(1-e^2)$

We can solve for the value of $theta$ corresponding to "emergence" from a finite sized Earth. We solve the following:

$r\sin\theta=(\dfrac{(a(1-e^2)}{1+e\cos\theta})(\sin\theta)=\text{Radius of Earth}$

We get two solutions, actually. One solution, the smaller root between $0$ and $\pi$ radians=180°, corresponds to what you have drawn. The other solution corresponds to getting "eclipsed" again by Earth as you approach apogee. You may want to track both boundaries so I keep $\theta$ as a general variable below.

With $r(\theta)$ figured out Wlwe are ready to make our assault on the area. We shall do so geometrically.

Draw radii $OP$ and $OQ$, cutting off a sector $OPQ$ of the ellipse between these radii. Then draw $GQ$ to complete $\triangle OPG$. We seek the area of region $GPQ$ as a difference:

$\text{region} GPQ = \text{sector} OPQ - \triangle OPG$

First get the area of the sector. This will be given by one of two formulas (see here for the geometry behind this):

$\text{Sector Area} = \dfrac{a^2}{2}\sqrt{1-e^2}\left(\arcsin\left(\dfrac{r\sin\theta}{a\sqrt{1-e^2}}\right)\right)$ $\text{ if }\cos\theta\ge(-e)$

$\text{Sector Area} = \dfrac{a^2}{2}\sqrt{1-e^2}\left(\pi-\arcsin\left(\dfrac{r\sin\theta}{a\sqrt{1-e^2}}\right)\right)$ $\text{ if }\cos\theta\le(-e)$

The conditions $\cos\theta\ge(-e)$ and $\cos\theta\le(-e)$ correspond respectively to being on the "perigee half" and "apogee half" of the ellipse. Different formulas apply to each half because of the multivalued nature of the inverse sine function.

Now you have to subtract the area of $\triangle OPG$. This is just half the base along the major axis tines the corresponding height:

$\text{Triangle Area} = (\dfrac{a}{2})(1-e)(r\sin\theta)$

Subtract the triangle area from the sector area, and you have your net area.

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    $\begingroup$ Hi OscarLanzi have a look at the two Sector Area equations; the MathJax formatting was bad/broken. I added the larger size parenthesis, if you have a chance, have a look to make sure the equations are still what you intended. Thanks! (before and after: i.stack.imgur.com/FzVuK.png) $\endgroup$
    – uhoh
    Feb 24, 2019 at 12:04
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    $\begingroup$ Thanks. On my device it's hard to read previews so I face struggles with both typos and formatting. $\endgroup$ Feb 24, 2019 at 12:20
  • $\begingroup$ Thanks for the response, one last question I have is how to find the radius of point Q. I would need this to find θ and I only know the horizontal distance between point P and point Q. $\endgroup$
    – slader
    Feb 24, 2019 at 19:53
  • $\begingroup$ I guess I'm failing to see how to use this to solve for the radius of Q or theta. Using the radius of the earth will give θ = 0 and π and plugging that into the equation gives perigee and apogee radius. $\endgroup$
    – slader
    Feb 24, 2019 at 20:39
  • $\begingroup$ You have $(\dfrac{(a(1-e^2)}{1+e\cos\theta})(\sin\theta)=\text{Radius of Earth}$. Solve that for $\theta$, identifying two roots as I described in the answer. Then $r=(\dfrac{(a(1-e^2)}{1+e\cos\theta})$. $\endgroup$ Feb 24, 2019 at 22:17

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