POBOnline
  Home
  Subscribe
  RPLS.com Bulletin Board
  eNewsletter
  Online Updates
  Latest News
  Calendar of Events
  Call for Papers
  Fun + Games
  New Products
  Surveying in the Headlines
  Top RPLS Bulletin Board Posts
  Current Issue
  Cover Story
  Features
  Columns
  Resources
  Archives
  Classified Ads
  Classified Listings
  Digital Edition Archives
  Distance Education
  Geo Locator Buyers Guide
  Industry Links
  Information Showcase
  Market Research
  POB Store
  Product Showrooms
  Product Surveys
  POB Info
  Special Collections
  GIAA Mailbag/Tool Tips
  GPS Observer Series
  Site Prep
  Surveyor + the Law
Search in: EditorialProductsCompanies
Email this Article Print View
Gravitational Bending of Light
by Lawrence Paul Lopresti
April 28, 2008



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
 
Diameter of Moon = Dm = 3,746 Km
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
 
Click here to read the rest of this thread:
 


Lawrence Paul Lopresti


Did you enjoy this article? Click here to subscribe to the magazine.








BNP Media
© 2008 BNP Media. All rights reserved. | Privacy Policy