I got quite a bit of bicker because of the last calculations, and rightly
so. If you want to see the old working
out click here to show it.
It's my holidays so I'm not going to come up with a perfect model for this,
but I think I have a better way to estimate this pressure, its not pretty, but it will do fine unless someone wants to come up
with something better...
Sun rays are not going to hit the earth in a perpendicular fashion, so the
force projected is not going to be in one direction but there still will be
a bit of push downwards, but just not as great (the horizontal forces will
be negated by opposing forces on the other side of the earth). I'll slice
the north hemisphere up into 3 cross-sections. 1st cross-section will
reflect at 0 degrees to the normal, the 2nd: 30 degrees, 3rd: 60 degrees.
Yeah, I could do the 4th cross section at the equator, but why bother? On
average the light is going to either skim off the face or bounce off with such
a negligible force.
Next, we'll figure out the force of the sun's light rays for each 'disk'.
Calculating circular area for each disk (sapincher
spotted my spherical area error first)
Also, thanks to
for pointing out sun ray angle error too.
Global Radius: 6,378 km
This is assuming the earth is a perfect sphere, which it's not... but it
will do for an estimation.
Time to find the light radiation pressure for each disk :D
|Here I'm calculating the radius for each
θ = arcsin ( 6,378 * 0.75 / 6,378 ) = 49°
r = 6,378 * cos ( 49 ) = 4,217 km
a1 = π * 4,217 = 55,865,870 km^2
j1 = 55,865,870 km^2
θ = arcsin ( 6,378 * 0.5 / 6,378 ) = 30°
r = 6,378 * cos ( 30 ) = 5,523 km
a2 = π * 5,523 = 95,827,322 km^2
j2 = a2 - a1 = 39,961,452 km^2
θ = arcsin ( 6,378 * 0.25 / 6,378 ) = 14.5°
r = 6,378 * cos ( 14.5 ) = 6,175.3 km
a3 = π * 6175.3^2 = 119,804,193 km^2
j3 = a3 - a2 = 23,976,871 km^2
Conversion to square centimetres:
= 5,586,587,000,000 cm^2
= 3,996,145,200,000 cm^2
= 2,397,687,100,000 cm^2
For every 6.45 cm^2, 0.0000000005 Kg of force is applied in the direction of
the travel of light... so, we'll divide 6.45 into each disk area
= 5,586,587,000,000 / 6.45 = 866,137,519,380
= 3,996,145,200,000 / 6.45 = 614,791,569,231
= 2,397,687,100,000 / 6.45 = 368,874,938,462
Then we should apply these solar forces to each 6.5 cm^2
= 866,137,519,380 * 0.0000000005 = 433 Kg
= 614,791,569,231 * 0.0000000005 = 307 Kg
= 368,874,938,462 * 0.0000000005 = 184 Kg
And, finally we take into account the angle of light rays and then add the
= 433 * cos ( 0 ) = 433 Kg
= 307 * cos ( 30 ) = 266 Kg
= 184 * cos ( 60 ) = 92 Kg
Which leaves us with a total of 791 Kg
The real life value would be slightly smaller seeing as not all light is
reflected back (and therefore not contributing any push to the earth), but
all in all I think this is a fair estimate as opposed to my last :)