Planet Insolation

[Hungry4info]

How is planetary insolation calculated?

Is there an equation that takes into account only the luminosity of the star and the distance of the planet?

Is there an equation that also takes into account the radius of the star?

L / d^2 doesn't seem to work for me... =\

[bdm]

I would guess you need bolometric luminosity and orbital radius.

Insolation = luminosity / orbital_radius ^ 2.

This assumes a circular or near-circular orbit. You would also need to take the eccentricity of the orbit into account. With an elliptical orbit, the body spends more time farther away from the star than it spends up close, so the mean insolation is lower for eccentric orbits.

So you will need to modify the formula as follows:

Insolation = luminosity / (orbital_radius * (1 + ((eccentricity ^ 2) / 2)) ^ 2.

Units could be solar luminosities, AU for the orbit and solar units for the insolation. Or you could use SI units: luminosity in watts, orbital radius in metres, and insolation in watts per square metre.

I don't know if the solar radius comes into play.

[bdm]

Oops - If you are computing the value in watts, you need to divide the result by 4?pi in the second formula to get the correct answer (4?pi/r^2 is the formula for a sphere).

In the first formula you don't need to do this if the answer is in insolation relative to the Earth.

The revised formula is:

Insolation = luminosity / (4 ? pi ? (orbital_radius ? (1 + ((eccentricity ^ 2) / 2)) ^ 2).

For Earth, this gives a mean insolation:
Solar output = 3.824e+26 watts
Earth orbit = 149,600,000,000 metres
Eccentricity = 0.01671
Insolation
= 3.824e+26 / (4 ? pi ? (149,600,000,000 ? (1 + ((0.01671 ^ 2) / 2)) ^ 2)
= 1359 watts / m^2
(The eccentricity of the orbit decreases mean insolation by about 0.4 watts)

[ajtribick]

Actually bdm's got it wrong: if you do the mathematics to work it out, a planet on an eccentric orbit receives greater orbit-averaged flux than one on a circular orbit, despite being located most of the time further from the star.

Equation should be [tex]\left<f\right> = \frac{L}{4\pi a^2} \, \frac{1}{\sqrt{1 - e^2}}[/tex]

[bdm]

Actually bdm's got it wrong: if you do the mathematics to work it out, a planet on an eccentric orbit receives greater orbit-averaged flux than one on a circular orbit, despite being located most of the time further from the star.

Equation should be [tex]\left<f\right> = \frac{L}{4\pi a^2} \, \frac{1}{\sqrt{1 - e^2}}[/tex]

Since I made this mistake, and others may not know the mechanism, can you provide a brief explanation?

Not that it has a great effect for a near-circular orbit; the simpler formula is a good approximation in that case.

[ajtribick]

Ok, you asked for it...

The time averaged insolation is given by [tex]\left<f\right> = \frac{L}{4\pi} \left<\frac{1}{r^2}\right>[/tex].

To proceed we can express r in terms of the semimajor axis a, eccentric anomaly E and eccentricity ? (using epsilon to distinguish it from the irrational number e, which makes an appearance later on): [tex]r = a(1 - \epsilon \cos E)[/tex]

To average over time is the same as averaging over a complete cycle of mean anomaly M. The relationship between M and E is given by the (in)famous Kepler equation [tex]M = E - \epsilon \sin E[/tex].

So: [tex]\left<\frac{1}{r^2}\right> = \frac{1}{2\pi} \int_0^{2\pi} \! \frac{dM}{r^2} = \frac{1}{2\pi a^2} \int_0^{2\pi} \! \frac{dE}{1-\epsilon \cos E}[/tex]

To solve the integral we can make the substitution [tex]z = e^{iE}[/tex], which transforms it into a complex path integral around a circle of radius 1. Using this substitution, [tex]\cos E = \frac{1}{2}\left(z + z^{-1}\right)[/tex]

After a bit of manipulation, we get: [tex]\left<\frac{1}{r^2}\right> = \frac{i}{\pi a^2 \epsilon} \oint \limits_{\left|z\right| = 1} \! \frac{dz}{z^2 - \frac{2}{\epsilon} z + 1}[/tex]

So how did this help? We can now use the residue theorem to evaluate the integral. The residue theorem states [tex]\oint_C f(z) = 2 \pi i \sum_{k = 1}^n \mathrm{Res}(f, a_k)[/tex], where [tex]a_k[/tex] are the singularities of the function enclosed by the curve C.

The function in our integral has singularities where the denominator is equal to zero: solving the quadratic reveals the singularities are at [tex]z_\pm = \frac{1 \pm \sqrt{1 - \epsilon^2}}{\epsilon}[/tex]

Of these, for the range of ? we are interested in, only the singularity at [tex]z_-[/tex] is inside the unit circle. Since the singularity is a simple pole, the residue at that pole is given by [tex]\mathrm{Res}(f, z_-) = (z - z_-)f(z_-) = \frac{1}{z_- - z_+}[/tex]

When put into the residue theorem this then gives:

[tex]\left<\frac{1}{r^2}\right> = \frac{i}{\pi a^2 \epsilon} \left( 2\pi i \ \frac{-\epsilon}{2\sqrt{1 - \epsilon^2}} \right) = \frac{1}{a^2 \sqrt{1 - \epsilon^2}}[/tex]

therefore

[tex]\left<f\right> = \frac{L}{4\pi a^2 \sqrt{1 - \epsilon^2}}[/tex]

Wasn't that fun?

[t00fri]

While the calculation of this integral by means of the Residue Theorem is the correct "academic" way of getting the result, Maple or Mathematica give the same result far less instructively, with a single click. Provided one restricts -1 < epsilon <1, of course.

Fridger