> To achieve the energy transition, you need SO MANY solar panels and they will take up a lot of land.

This isn't true.

> According to Jesse Jenkins, to power America, "the solar farms are an area the size of Connecticut, Rhode Island and Massachusetts.

The estimates are that it's anywhere from 10000 to 30000 square miles which is compared to the 3.8 million square miles of the continental USA an absolute pittance. Doing this by tightbeaming the output of an equivalent solar installation in space is ridiculous.

The roof of every single new big box store should be covered in solar panels and we should start covering large parking lots in them too. Shaded parking that powers the area would be sweet.

There is even a potential convergence: https://en.wikipedia.org/wiki/Agrivoltaics

Getting rid of waste heat in space is harder than it is on earth!

If we used the area that is currently producing ethanol and "biodiesel" for solar panels we would be 80% done with the energy transition.

One standard design had a 10km contiguous(!) diameter receiver for a 1km diameter transmitter in GEO. You can lower the transmitter to about 6371 km[0] before having to worry about night, but that's only going to give you a factor of about 6 improvement. Different wavelengths can do better antenna gain, but start to need to care about atmospheric absorption. You could in principle also relay the power around the same sort of way we do for data, though that way you get a percentage loss from each relay (might still be worth it, might not, I've not tried to cost it).

You're also limited by the maximum safe power density; from what I've heard this is surprisingly close to what you'd get from the yearly average output of PV, which on the one hand means that although it can be an alternative to storage when the question is "what about night and winter?" it is fairly unhelpful when the question is "can we use less land"?

> That land is most expensive near the very population centers which need the most power.

Yes, but wire-based transmission is better than most give it credit for. And even if it was that bad, if you perfectly solved wireless transmission by any mechanism you like, it's a shorter route from the ground on the opposite side of the planet, to low orbit, around the planet, and back down again (~20e6m), than from geostationary orbit (~36e6m).

> According to Jesse Jenkins, to power America, "the solar farms are an area the size of Connecticut, Rhode Island and Massachusetts."[1] (this isn't the most precise source I could find, but it's reputable)

Sounds about right: https://www.wolframalpha.com/input?i=%28Connecticut%2C+Rhode... (20% efficient cells, 10% capacity factor).

Those are 2.9e11, 6.3e10, 5.5e11 W, compared to recent US use of 4.6e11 W.

But they're also fairly small states, with 1%, 0.3%, and 2% of the US total population: https://www.wolframalpha.com/input?i=%28Connecticut%2C+Rhode...

[0] If I've done the maths right:

1. The radius of the inscribed circle of an equilateral triangle is r = (sqrt(3)/6)*a

2. The altitude of the triangle from any side is h = (sqrt(3)/2)*a = 3*r

3. So the closest you can get the satellites while being sure at you never have a period where the only visible satellites are themselves not in shadow is:

triangle_altitude - diameter_earth = 3*r - 2*r = r

I now think my assumption is wrong, that the closest you can get without worrying about night is minimal bounding square, not a minimal bounding triangle, so the orbital altitude should be:

(sqrt(2) * radius) - radius = (sqrt(2) - 1) * radius ~= 2639 km

so a factor of about 14 better than GEO.