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In my previous question @JackD'Aurizio helped me to solve the problem I had. Now I have the following identity:

$$\frac{A}{\text{n}}\int_{0}^{2\pi}\frac{1}{2}\left(\frac{p}{1+\varepsilon\cos\left(\theta\right)}\right)^2\,d\theta = \int_{0}^{\theta_A}\frac{1}{2}\left(\frac{p}{1+\varepsilon\cos\left(\theta\right)}\right)^2\,d\theta$$

QUESTION: And I need to solve that equation for $\theta_\text{A}$.


What I did, is try to evalute the integral:

$$\int\frac{1}{2}\left(\frac{p}{1+\varepsilon\cos\left(\theta\right)}\right)^2\,d\theta=\frac{p^2}{2}\int\frac{1}{\left(1+\varepsilon\cos\left(\theta\right)\right)^2}\,d\theta=$$ $$\frac{p^2}{2}\int\frac{1}{1+2\epsilon\cos\left(\theta\right)+\epsilon^2\cos^2\left(\theta\right)}\,d\theta$$

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  • $\begingroup$ I'm not sure whether calculating that antiderivative is the best way to go. It certainly doesn't look very easy. $\endgroup$
    – rubik
    Commented Mar 4, 2017 at 14:17
  • $\begingroup$ @rubik The problem is, I do not know another way to go :( $\endgroup$
    – treq
    Commented Mar 4, 2017 at 14:22
  • $\begingroup$ Also, I doubt wolframalpha's primitive could even be applied straight forward to evaluate those quantities: it is apparent that the integral needs, in order to be defined, that $\lvert\varepsilon\rvert<1$. However, wolframalpha's primitive has some nasty $\sqrt{(\varepsilon^2-1)^3}$ and $\sqrt{\varepsilon^2-1}$, which are probably meant to cancel out "somehow", according to the thoughts of its mechanical head. $\endgroup$
    – user228113
    Commented Mar 4, 2017 at 14:27
  • $\begingroup$ @G.Sassatelli The thing I know about $\epsilon$ is: $$0\space<\space\epsilon\space<\space1$$ $\endgroup$
    – treq
    Commented Mar 4, 2017 at 14:32
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    $\begingroup$ The antiderivative of $1/(1+k\cos x)^2$ is $$\frac{2 \arctan\left(\sqrt{\frac{1-k}{1+k}}\,\tan\frac x 2\right)} {(1-k^2)^{3/2}} - \frac{k\sin x} {(1-k^2)(1+k \cos x)}$$ $\endgroup$ Commented Mar 4, 2017 at 20:51

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QUESTION: And I need to solve that equation for $θ_A$.

No, you don't, for two reasons. One is that it is incorrect. The other is that even if it was correct, you would not need to do so in the context of your other question. The formula

$$\text{G}_\text{sc}\left(\text{n}\right)=1367\cdot\left(1+0.03\cos\left(\frac{360}{365}\cdot\text{n}\right)\right)\space\space\space\space\space\space\space\space\space\left[\text{W}/\text{m}^2\right]\tag1$$

is an approximation. It assumes

  • An anomalistic year of 365 days. The anomalistic year is actually 365.259636 days long.
  • That Earth perihelion falls on December 31. Currently, perihelion falls somewhere between the second and fifth of January, depending on the year.
  • That the Earth's eccentricity is 0.015 (0.03/2). Actually, it's 0.01671, which would mean using 0.0334 would be better than 0.03, but 0.03 is good for this somewhat rough approximation.
  • That second order terms in eccentricity can be ignored.

Given the somewhat rough nature of this approximation, assuming that mean motion and true motion are the same is quite fine. This, coupled with a year of 365 days and perihelion on December 31 leads to $\theta(n) = 360^\circ \frac n {365}$.

Regarding my egregious charge that $\frac{A}{\text{n}}\int_{0}^{2\pi}\frac{1}{2}\left(\frac{p}{1+\varepsilon\cos\left(\theta\right)}\right)^2\,d\theta = \int_{0}^{\theta_A}\frac{1}{2}\left(\frac{p}{1+\varepsilon\cos\left(\theta\right)}\right)^2\,d\theta$ is incorrect: The two integrals have solutions in the elementary functions. Kepler's equation is transcendental, meaning it does not have a solution in the elementary functions. Try as hard as you want, but you will not be able to integrate the correct expression.

What you can do is come up with a somewhat ugly relationship between eccentric anomaly and true anomaly and a seemingly simple relationship between eccentric anomaly and mean anomaly, the last of which marches uniformly with time. The somewhat ugly relationship between eccentric anomaly and true anomaly, $\sqrt{1-e}\tan\frac\theta2=\sqrt{1+e}\tan\frac E2$, is invertible. The seemingly simple relationship between eccentric anomaly and mean anomaly, $M = E - e\sin E$, is not invertible (in the elementary functions).

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  • $\begingroup$ Why is it wrong (the integral equality)? Because when we assume that $\epsilon\approx0$ we get the thing we expected: $$\theta_\text{A}=360^\circ\cdot\frac{\text{n}}{365}$$ Because: $$\text{G}_\text{sc}=\sigma\cdot\text{T}^4\cdot\left(\frac{\text{R}}{\frac{\text{p}}{1+\epsilon\cos\left(\theta_\text{A}\right)}}\right)^2=\sigma\cdot\text{T}^4\cdot\left(\frac{\text{R}}{\frac{\text{p}}{1+\epsilon\cos\left(360^\circ\cdot\frac{\text{n}}{365}\right)}}\right)^2$$ $\endgroup$ Commented Mar 5, 2017 at 12:23

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