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Are you sure it will be probably less than 1 order of magnitude? Unless there is some restriction that limit the plane of the orbit around a star with relation

Are you sure it will be probably less than 1 order of magnitude? Unless there is some restriction that limit the plane of the orbit around a star with relation to the galaxy, which is unlikely... Assume the plane needs to be within 1 degree from the line joining the star and kepler. I would guess it probably less. It would be a function of size of the planet, size of the star, radius of the orbit and distance of star from kepler. That would give about 179 planes that we cannot see, which is two orders of magnitude, but the probability calculation won't be that simple.

darkmighty 11y ago
Oh yea I just realized it obviously has to be quite low, shouldn't be too hard to estimate for earth (the problem is, it varies with planet size). I would guess two orders of magnitude or more actually, which would indeed put the expected number close to 1. A naive calculation for Earth would be Theta ~= 2 * (Earth diameter) / (Mean distance to Sun), which is ~2*10^-4. Since most planets found by kepler were much larger than earth, it seems 10^-2 is a reasonable guesstimate.
greeneggs 11y ago
> Assume the plane needs to be within 1 degree from the line joining the star and kepler... That would give about 179 planes that we cannot see... Assuming you mean 1 degree in either direction, or 2/360 = 1/180, this is off by about a factor of pi/2 ~= 1.57. The correct answer is sin(pi/360) ~= pi/360 ~= 1/115. Still two orders of magnitude, but slightly better. The reason is that the degrees are not equally likely (small degrees, that you want, are more likely, and large degrees less likely). Here's the calculation: Consider the unit vector normal to the equator of the orbit, using say the right-hand rule. This vector is uniformly distributed over the unit sphere (surface area 4pi). You want the vector to lie within +-1 degree of the rim of the sphere. Call the region of such vectors R; the probability of a good planet is P = area(R)/(4pi). R is an annulus going around the equator and up/down by 1 degree. In area, this is close to a rectangle with width 2pi (circumference of sphere) and height 2pi/360 (i.e., 2 degrees), so P ~= (pi^2/90) / (4pi) = pi/360. The area of R (an equatorial annulus) can be computed exactly using calculus---or looked up on Wikipedia [1]: area(R) = 4pi - 2(2pi(1-sin(pi/360))) = 4pi sin(pi/360). Thus, P = sin(pi/360). [1] https://en.wikipedia.org/wiki/Spherical_cap#Volume_and_surface_area https://en.wikipedia.org/wiki/Spherical_cap#Volume_and_surfa...
- darkmighty 11y ago
  Ah that's some good analysis. I would love seeing this calculation done precisely, for that 30 light year figure (and other distances). There's other factors to put in too whose significance idk like the size of the sun -- if it's much larger this increases significantly the range of angles in which occlusion occurs.
  - deleted 11y ago
    [deleted]
- darkmighty 11y ago
  Indeed taking into account that the sun is much larger than earth I got the value of P=0.00465 by taking P=sin(arctg(solar radius/1 au)). So the planet size actually shouldn't matter significantly on this particular probability, since the planetary radius is like 1% of the solar radius.

darkmighty 11y ago

Oh yea I just realized it obviously has to be quite low, shouldn't be too hard to estimate for earth (the problem is, it varies with planet size). I would guess two orders of magnitude or more actually, which would indeed put the expected number close to 1. A naive calculation for Earth would be Theta ~= 2 * (Earth diameter) / (Mean distance to Sun), which is ~2*10^-4. Since most planets found by kepler were much larger than earth, it seems 10^-2 is a reasonable guesstimate.

greeneggs 11y ago

> Assume the plane needs to be within 1 degree from the line joining the star and kepler... That would give about 179 planes that we cannot see... Assuming you mean 1 degree in either direction, or 2/360 = 1/180, this is off by about a factor of pi/2 ~= 1.57. The correct answer is sin(pi/360) ~= pi/360 ~= 1/115. Still two orders of magnitude, but slightly better. The reason is that the degrees are not equally likely (small degrees, that you want, are more likely, and large degrees less likely). Here's the calculation: Consider the unit vector normal to the equator of the orbit, using say the right-hand rule. This vector is uniformly distributed over the unit sphere (surface area 4pi). You want the vector to lie within +-1 degree of the rim of the sphere. Call the region of such vectors R; the probability of a good planet is P = area(R)/(4pi). R is an annulus going around the equator and up/down by 1 degree. In area, this is close to a rectangle with width 2pi (circumference of sphere) and height 2pi/360 (i.e., 2 degrees), so P ~= (pi^2/90) / (4pi) = pi/360. The area of R (an equatorial annulus) can be computed exactly using calculus---or looked up on Wikipedia [1]: area(R) = 4pi - 2(2pi(1-sin(pi/360))) = 4pi sin(pi/360). Thus, P = sin(pi/360). [1] https://en.wikipedia.org/wiki/Spherical_cap#Volume_and_surface_area https://en.wikipedia.org/wiki/Spherical_cap#Volume_and_surfa...

darkmighty 11y ago
Ah that's some good analysis. I would love seeing this calculation done precisely, for that 30 light year figure (and other distances). There's other factors to put in too whose significance idk like the size of the sun -- if it's much larger this increases significantly the range of angles in which occlusion occurs.
- deleted 11y ago
  [deleted]
darkmighty 11y ago
Indeed taking into account that the sun is much larger than earth I got the value of P=0.00465 by taking P=sin(arctg(solar radius/1 au)). So the planet size actually shouldn't matter significantly on this particular probability, since the planetary radius is like 1% of the solar radius.