The parts of the moon in direct sunlight are blown out too. Where we see the terrain's texture, it's illuminated from the back; with what must be light reflected from the moon's surface.
The dark side of Earth is illuminated by a full moon.
> The dark side of Earth is illuminated by a full moon.
Yes it is, but it's still going to be extremely dark. Think of how dark it is outside even when it's a full moon. You can see things, but it's still orders of magnitude less than daylight brightness.
And if the camera were exposed for minutes in order to not just capture that but overexpose it, we'd see stars everywhere. But we don't. So no. Earth is going to be black in this picture, period, except for a possible crescent.
I think we're agreeing -- you're using a shorter exposure that captures the dark part of the moon, but isn't long enough to capture stars.
If you want to capture stars, it needs to be longer. Looking it up says that 20 seconds is roughly the minimum. Which is also around what would overexpose the entire moon in your example. So that's what I'm saying -- the dark side of the earth doesn't seem like it could be overexposed because we can't see any stars.
You propose that the sun is above the horizon and that the bright light on the horizon is a reflection of sunlight. That seems off to me: With the sun in the sky, why would its reflection have the strongest diffraction spikes? Why would there be a blown out fringe along the horizon, and no blown out surface close to the craft?
I agree that I'd expect to see stars when the moonshine on earth is overexposed. Maybe it's just glare that makes the light from earth's crescent orb-shaped. In an image of a different camera where the sun is obscured[0], the earth crescent is well visible, and no moonshine on earth. Firefly's page on the mission has enough detail[1] to put your and my speculation to rest :-)
The dark side of Earth is illuminated by a full moon.