# Gravitational Bending of Light

April 28, 2008

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Posted By Lawrence Paul Lopresti on 4/27/2008 at 12:43 PM

Gravitational Bending of Light

From an article by Laurence R. Doyle at the SETI Institute

“Albert Einstein predicted that the suns gravity should bend the light of background stars, so they would appear to move outward during a solar eclipse.”

“This deflection -- about 1/1000th the width of a full moon -- was first measured by the famous British astronomer, Sir Arthur Eddington in 1919. Several years later a group from the Lick Observatory proved that Einstein was correct, by measuring to the required accuracy.”

Based on the above and with the following information:

Diameter of Earth = De = 12,742 Km

Radius of Earth = Re = 6,371 Km

Mass of the Earth = Me = 5.9736 * 10^24 Kg

Radius of Earth = Re = 6,371 Km

Mass of the Earth = Me = 5.9736 * 10^24 Kg

Diameter of Moon = Dm = 3,746 Km

Radius of Moon = Rm = 1,738 Km

Mass of the Moon = Mm = 7.349 * 10^22 Kg

Radius of Moon = Rm = 1,738 Km

Mass of the Moon = Mm = 7.349 * 10^22 Kg

Gravitational Equation

Fg = G (M’M”)/R^2

The force of gravity is the Gravitational Constant times Mass 1 times Mass 2 divided by the Radius (distance between the centers of mass).

I want to solve for the deflection due to gravity on earth.

M’ is the Mass of a Light Particle, M” is the Mass of the Earth/ Moon.

Me/Mm = 5.9736 * 10^24 Kg / 7.349 * 10^22 Kg = 81.28

Me = 81.28 Mm

Re/Rm = 6,371 Km / 1,738 Km = 3.666

Re = 3.666 Rm

Solving the Gravitational Equation for Fg of the Earth relative to the Moon, G and M’ being constant:

Fge = 81.28 Mm /( 3.666 Rm^2) = 6 Fgm

If the force is 6 times greater the deflection should be 6 times greater, therefore the deflection to light as effected by Earth’s gravity is 6/1000 or 0.006

The number used to account for refraction is 0.011 Km^2.

Let us assume the full effect of gravity occurs when light travels the distance of one Earth diameter. In reality is is over a greater distance abd one would then have to differentiate over that greater distance, I believe one Earth diamter is a fair approximation.

The amount due to gravity alone is 0.006 Km^2 or a bit more than half. That leaves 0.005Km^2 to be caused by the effect of the atmospheric density difference.

I believe the above to be a sound first assumption upon which further studies may refine the actual value but would not change the scale. That is I am within +/- 50%.

Paul in PA

Modified By Lawrence Paul Lopresti on 4/27/2008 at 12:43 PM

Modified By Lawrence Paul Lopresti on 4/27/2008 at 12:43 PM

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