David Sims
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Removed from YouTube on 7 Oct 2017 because it didn't fit the mainstream media narrative by which James Alex Fields was judicially lynched.
klokeid
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Tommy Robinson has defended the weak at his own peril. He has confronted evil repeatedly, at personal cost. He has rescued children and damsels in distress by his courageous activism, when the bureaucracy was too corrupt and the police were too cowardly to do their jobs. His state has put him in prison for his exertions. His Queen can't be bothered to pardon him, let alone give him the knighthood for which his actions have qualified him. May God sweep all of this dirtiness out of Britain, and may all of the white people in the world be His broom.
The pessimist says that a box is empty. An optimist says it contains doughnut holes.
Find the probability of an eclipsing habitable zone exoplanet as a function of star's effective temperature?

For a planet in a circular orbit in its star's habitable zone, find the probability that the inclination of the planet's orbit with respect to the observer's line of sight will be within required bounds for an eclipse of the star to occur. State the probability as a function of the star's effective temperature.

The probability, P, that an exoplanet in a circular orbit will eclipse its star, as seen from Earth, is

P = (2/π) { arcsin(R/r) − arcsin(R/d) }

where R is the star's radius, r is the radius of the planet's orbit, and d is the distance between the observer and the star's center. Since it is always true that R«d, we can write

P = (2/π) arcsin(R/r)

If you knew the radius, R, and effective temperature, T, of the star, you could find its luminosity from

L/L๏ = (R/R๏)² (T/T๏)⁴

The nominal distance, r, between a star and its habitable zone is

r = 1.496e11 meters √(L/L๏)

And the radius of the star, converted to MKS units, is

R = 6.96e8 meters (R/R๏)

So, given a planet in a circular orbit in the star's habitable zone, the probability for an eclipse becomes

P = (2/π) arcsin{0.0046524 (R/R๏) / √(L/L๏)}

Which the Stefan-Boltzmann relation converts to

P = (2/π) arcsin{0.0046524 / (T/T๏)²}

Or, in terms of the star's effective temperature in Kelvins,

P = (2/π) arcsin{155644 K² / T²}

So you could make a table of probability for a planet in a circular orbit in the habitable zone of a star to eclipse the star, versus the effective temperature of that star.

For a star as hot as the sun, T=5784K, and P=0.0029618. For a red dwarf star with an effective temperature of T=3000K, P=0.0110101. For each exoplanet in the habitable zone that eclipses there are about 150 or so that do not eclipse.