Need help with gravitational lensing issue

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SUMMARY

The discussion centers on the challenges of using gravitational lensing for interstellar signal transmission, specifically focusing on calculating the correct focal distance. The formula discussed, R²/M, where R is the radius and M is the mass of an astronomical object, is crucial for determining the focal distance. The user tested this with the Sun, arriving at a focal distance of approximately 550 AU, but encountered difficulties in applying the formula correctly. A refined calculation using the formula dfocal = 0.25c²r²G⁻¹M⁻¹ yielded a result of 543 AU, confirming the validity of the approach.

PREREQUISITES
  • Understanding of gravitational lensing principles
  • Familiarity with astronomical units (AU) and their conversions
  • Basic knowledge of physics formulas involving mass and radius
  • Proficiency in using constants such as the speed of light (c) and gravitational constant (G)
NEXT STEPS
  • Research advanced gravitational lensing techniques for signal transmission
  • Study the application of the formula dfocal = 0.25c²r²G⁻¹M⁻¹ in various astronomical contexts
  • Explore the limitations and potential of using gravitational lensing in communication systems
  • Investigate alternative methods for interstellar communication, such as laser-based systems
USEFUL FOR

Astronomers, astrophysicists, and engineers interested in the application of gravitational lensing for communication and signal transmission over vast distances.

Dr Wu
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I'm afraid I've come to a problem which I cannot solve. It concerns using gravitational lensing as a means of transmitting signals over interstellar distances. The real issue is finding the correct focal distance to make this possible. Now the only information I've been able to glean from the web is this: "The ratio of a planet's radius squared to its mass calculates the distance a spacecraft must reach to take advantage of its gravitational lensing."

On the same scrap of paper I copied the above quote, I also jotted down the following formula: R2/M (meaning radius squared divided by the mass of the object). Now I'm not sure if this formula actually belongs to the quote, but I've tried using it anyway, applying it in this instance to the Sun as a test case. I already know that the focal distance for the Sun tallies out at 550 AU. Unfortunately, I can't make the formula work. All I get are impossible answers - anything from between tens of thousands of light-years down to a measily 27 AU. . . not the required 550 AU. (I assume all measurements are supposed to be in metres and kilograms).

I would therefore be extremely grateful if someone knowledgeable on this subject can point me in the right direction. Many thanks.
 
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I don't understand. Maybe the analogy will help - how would you use ordinary lenses as a means of transmitting signals over terrestrial distances?
 
Vanadium 50 said:
I don't understand. Maybe the analogy will help - how would you use ordinary lenses as a means of transmitting signals over terrestrial distances?

I can send you a Morse code signal with my flashlight. If I am standing in daylight you will have a hard time seeing the light from far away. If you have a pair of binoculars then you can see the flashes at a longer distance.
 
Thanks. So the lenses aren't transmitting anything.
 
Thanks. So the lenses aren't transmitting anything.
 
Vanadium 50 said:
Thanks. So the lenses aren't transmitting anything.
They could be. The binocular analogy might break down. Recall someone who is nearly blind wearing thick glasses. When you look at their eyes the iris/pupil/lashes are much larger than when (s)he takes the glasses off.

I am not sure about the communications. The signal would look like it originated in a ring around the star. Might be possible to filter out noise by only taking signals that appear all around the ring. I do not see how it would work better than redundant lasers.
 
Dr Wu said:
"The ratio of a planet's radius squared to its mass calculates the distance a spacecraft must reach to take advantage of its gravitational lensing."

On the same scrap of paper I copied the above quote, I also jotted down the following formula: R2/M (meaning radius squared divided by the mass of the object). ..

dfocal = 0.25c2r2G-1M-1

c = 2.99 x 108 m s-1
r = 6.96 x 108 m
G = 6.67 x 10-11 m3kg-1s-2
M = 1.99 x 1030 kg

I got 543 au after I figured out which number to use for G.

quick check the exponents are (8x2)+(8x2)-(-11)-30 = 13
meters: m2 x m2 /m3 = m
time: s-2/s-2 = 1
mass: kg x kg-1 = 1
Answer should be in meters. Since au is around 1.5x 1011 meters that looks right.
 
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