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By how much, if at all, is a ray of light bent when it passes by the sun a million miles from center?
The bottom line is, do you have an order-of-magnitude feel for the sun's gravity? Can you judge the rough size of the bending angle well enough to spot it in the line-up?
In natural units (c=G=hbar=1) the answer is simply 4M/R radians, where M is sun's mass and R the distance of closest approach----a million miles being E44.
The sun's mass? Well the Earth's orbit speed is E-4 and its distance from the sun is 93E44 (93 million miles). So the sun's mass is 93E44 multiplied by the square of E-4.
The mass of a central body equals the radius of any circular orbit multiplied by the square of the orbit speed. This assumes the orbiter is small compared with the central body, and the Earth is that.
M, as you can see, is easy to find and the angle is simply 4M divided by E44.
The corresponding thing is messy in metric----distance 15E10 meters, orbit speed 3E4 meters per second, then divide by
G = 6.673E-11 cubicmeters/squaresecond per kilogram,
which gives the mass M in kilograms. Then the angle is 4GM/c2R, where R is the distance of closest approach.
What natural units do for you in this case is get rid of the Gees and the cees, making things a little less bother. The answer comes out in radians either way---doesn't matter whether you work in natural units or in metric.
The bottom line is, do you have an order-of-magnitude feel for the sun's gravity? Can you judge the rough size of the bending angle well enough to spot it in the line-up?
In natural units (c=G=hbar=1) the answer is simply 4M/R radians, where M is sun's mass and R the distance of closest approach----a million miles being E44.
The sun's mass? Well the Earth's orbit speed is E-4 and its distance from the sun is 93E44 (93 million miles). So the sun's mass is 93E44 multiplied by the square of E-4.
The mass of a central body equals the radius of any circular orbit multiplied by the square of the orbit speed. This assumes the orbiter is small compared with the central body, and the Earth is that.
M, as you can see, is easy to find and the angle is simply 4M divided by E44.
The corresponding thing is messy in metric----distance 15E10 meters, orbit speed 3E4 meters per second, then divide by
G = 6.673E-11 cubicmeters/squaresecond per kilogram,
which gives the mass M in kilograms. Then the angle is 4GM/c2R, where R is the distance of closest approach.
What natural units do for you in this case is get rid of the Gees and the cees, making things a little less bother. The answer comes out in radians either way---doesn't matter whether you work in natural units or in metric.
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