If you could put yourself out in space so you were lined up behind the Moon, then Earth, then the Sun........?

Question

Would it line up perfectly? Like if you were at that point where the moon perfectly fit over the Earth, would the Earth and Moon, then also fit perfectly over the Sun?

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Answer 1

No, I re-did the math and it will not work.

Using similar triangles and the known sizes of the Earth, Sun and Moon, combined with the known distance between the Earth and Sun, I calculate you would need to be 1,377,338 km behind Earth to see it just cover the Sun, and the Moon would have to be 275,736 km from you to just cover the Earth. That would put the Moon 1,001,606 km from Earth, and in reality it is only 384,400 km from Earth.

And when you think about how lunar eclipses work, this should be obvious. The Shadow of Earth on the Moon is much larger than the Moon. The whole Moon fits inside the cone of darkness cast by the Earth with plenty of room to spare. For your perfect fit arrangement to work, the Earth's shadow would have to just barely cover the Moon, or be the exact same width as the Moon at the distance of the Moon. And we know from looking at lunar eclipses that Earth's shadow is larger than that. This tells you that either the Moon is too close to Earth or too small for your perfect fit view.

Answer 2

No, but they would completely cover it.

In the absence of diagrams, you'll have to take my word for it that the calculations show that in order to make the Moon perfectly cover the Earth you need to be about 144353km behind it. At that distance, both Moon and Earth will appear to subtend an angle of 1.38 degrees. The Sun, however, is only 0.53 degrees across at that distance. Earth and Moon would cover the Sun entirely, but not perfectly.

Answer 3

To make the Earth the same angular diameter as the Moon, you would need to be six times as far from the Earth than the Moon. This would be 300, 000 miles from the Earth. That would make the Earth and the Moon 1.5 degrees of the vision. The Sun would still be 0.5 degrees of the vision. .