Spaghettification near event horizon

[PeterDonis]

Is this wrong?

Yes. Look at it; it's not the same as what I wrote, is it? That means it's wrong.

But I didn't change c to 1.

Neither did I in the equation you referred to here. In the post you quoted, I did the entire derivation without changing ##c## to 1, leaving it as ##c## in all the formulas. And I got ##c^4 / 4 G M \sqrt{1 - 1}##; not ##c^4 / 4 G M \sqrt{1 - c^2}##. So what you wrote is wrong. Go back and check your algebra again. Or look carefully at what I posted and see how it goes, step by step.

I imagine we divide mass of the Earth with this unit, not multiply it.

Why would you imagine that, since you just got the right answer by multiplying? You just figured out the right answer, and now you're throwing it away?

I think G/c^2 is half of Schwarzschild radius.

No, it isn't. It is a conversion factor from ordinary units to geometric units. Think of it as ##G / c^2## meters per kilogram. You have a number in kilograms, like the mass of the Earth; you want to know what it is in meters; so you multiply the number in kilograms by the conversion factor, ##G / c^2## meters per kilogram, to get the number in meters.

If you want to say that, for a given mass ##M##, the distance ##GM / c^2## is half the Schwarzschild radius of a black hole of mass ##M##, that would be correct. But the distance ##GM / c^2## can be used much more generally than just in a scenario with black holes. In GR, that distance is the mass in geometric units, the same way the distance ##ct## is the time ##t## in natural relativistic units.

If we multiply the mass of an (planetary) object with this unit, we'll get acceleration at that distance.

I have no idea where you're getting this from. It's wrong. You don't get an acceleration, you get a distance. See above.