# Need help with gravitational lensing issue!

Dr Wu
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.

Staff Emeritus
I don't understand. Maybe the analogy will help - how would you use ordinary lenses as a means of transmitting signals over terrestrial distances?

Dr Wu
I'm afraid the analogy is lost on me. Perhaps this may help to explain what I was attempting to convey in my post:

https://www.centauri-dreams.org/?p=10123

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.

Staff Emeritus
Thanks. So the lenses aren't transmitting anything.

Staff Emeritus
Thanks. So the lenses aren't transmitting anything.

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.

"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.

• Dr Wu