dharmesh dave
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Yesterday a rocket launched from Cape Canaveral in Florida carrying the LightSail into space!  It's a small spacecraft with a big shiny screen that's pushed by the light of the sun.

It's just a test - it won't go far.   It will fall to the Earth and burn up.  But next year there will be a more serious test.  And someday, solar-powered space flight may become a force to be reckoned with.

One cool thing is that all this is being paid for private donations, by a Kickstarter campaign!

The LightSail is carried to space in a cute little CubeSat.  It looks like a big toaster, and it weighs just 10 kilograms.   But it holds a sail 32 square meters in area,  made of a shiny plastic called Mylar, just 4.5 microns thick.  This unfolds in a clever way - watch the movie! - to form a big square.

The Sun will push on this with a tiny force.

Puzzle: How tiny is this force?

Someone named Bill Russell answered this over on Yahoo.  Let me go through his calculation so we can check it.

The momentum of light is given by

p = E/c

where E is the energy of the light, p is the momentum, and c is the speed of light.

In outer space near earth the sunlight provides 1370 watts per square meter - that's energy per area per time.  We can use the formula above to convert this to momentum per area per time, better known as force per area... or pressure

Russell calculates

(1370 watts / meter²) / c = 9.13 micronewtons / meter²

and concludes the pressure is 9.13 micronewtons per square meter.  His arithmetic checks out, but I think he's neglecting some physics: when the light bounces back off the mirror its momentum completely reverses, so I think we get an extra factor of 2.

Puzzle 2:  Am I right or am I wrong?

The area of the LightSail is about 32 square meters.  Russell says this gives a total force of

9.13 micronewtons/meter² x 32 meter²

or about 300 micronewtons.   I'd double this and get 600 micronewtons.

Puzzle 3: Once it's out of the box, the LightSail weighs about 4.5 kilograms.  How much will it accelerate due to sunlight?

Here we use Newton's good old

F = ma

and solve for the acceleration a.   But at this point Russell seems to make a serious mistake.  I'll let you see what you think, and fix it if necessary!  Here is his calculation: