justcurious33 said:
would there be a way to "catch" or harness it if it was aimed earth-ward on purpose
You would be layering impracticality on top of impracicality. But let us set that to one side and consider a particular (impractical) proposal.
We will erect a parabolic mirror with a cross-sectional area equal to that of the moon. That is about 3500 kilometers across. Or 1750 km in radius. This is impractical because a structure that large would collapse even under the weak gravity of the moon. But we are ignoring such practical concerns.
The sun delivers light energy at a rate of about 1380 watts per square meter. Our mirror will reflect light with a total of$$P = 1380 \times \pi r^2 \approx 13 \text{ quadrillion watts}$$By contrast, total world power generation is about 3 terawatts.
Great news, right?!
But there may be a problem. The sun is a big light source. We cannot focus it down to a point. The size of the sun's focused image will be related to the sun's actual size. The two will be in a ratio corresponding to the distance from moon to earth compared to the distance from the moon to the sun. The sun is about 150 million km away. The moon is a bit about 384 thousand km away. That ratio is about 390 to one. Meanwhile, the sun has a diameter of 1.4 million km. If we divide that by our 390 to one ratio, we get an image size of about 3600 km.
By no coincidence, this is the same as the diameter of the moon. As we already knew, the moon and the sun have the same angular size when viewed from Earth!
What does this mean for the image intensity? We took the light from the sun striking the moon. Now it is striking a moon-sized region on the Earth instead. The effect is double intensity sunlight over that region.
This is great if you want to grow a field of popcorn fast. Don't even need to put it in a popper! It is not so good if you are trying to propel a space craft. You double the propulsion that you can get from a [photon driven] solar sail. But that is still not much thrust.
The thrust that you get from sunlight is given by $$E=pc$$Here ##E## is the energy of the sunlight. ##p## is the momentum imparted and ##c## is the speed of light.
Suppose that we double the sun's illumination from 1380 watts per square meter to 2760 watts per square meter. We set up a solar sail that is a square 1000 meters by 1000 meters. Now in one second we will have received 2.76 billion Joules.
This is about twice the capacity of the flux capacitor on a Delorean (1.21 gigawatts).
If we divide this by the speed of light (##3 \times 10^8## meters/second). We will have about 9.2 newton-seconds of momentum. Or a continuous thrust of 9.2 newtons. Versus 4.6 newtons without the moon-sized mirror dedicated to accelerating this craft. [Figures updated after a sanity check]
Economics suggest that it is cheaper to make the sails 1.4 times bigger rather than erecting a mirror on the moon. [You can also buy yourself a factor of two by reflecting instead of simply absorbing].
If I have not dropped too many decimals anyway.
Edit: Cross checking propulsion numbers with the Wiki article on solar sails: "F = 4.54 μN per square metre". Multiply by one million square meters and we have 4.54 N. Versus my original figure of 460. Clearly I slipped a couple digits... Yep. There it is. I had fumbled when I divided by the speed of light.