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At a space station located at Lagrange point 2 how would we see the Earth and Moon? Would the Earth appear to be stationary? How would the Moon look when it passes in front of the Earth from that persepective? Tiny? See diagram.

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http://upload.wikimedia.org/wikipedia/commons/8/88/Lagrange_points.jpg
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http://upload.wikimedia.org/wikipedia/commons/4/4d/Lagrangepoint1.png
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2007-09-04 07:24:44 · 5 answers · asked by ericbryce2 7 in Science & Mathematics Astronomy & Space

5 answers

From the discussion of the L2 point that includes these diagrams, I have to conclude that, yes, the earth would appear stationary. The L2 point represents an orbital station around the sun with a period that, because of earth's extra gravity, exactly matches the earth's orbital period, so the earth would always be between the object and the sun.

Since the point exists at approximately 1.5 million km (about 930,000 miles) from the earth, which is slightly less than 4 times the distance of the moon from the earth, I would imagine the moon would appear to be about 1/4 of its apparent size from earth. However, since the L2 point is coplanar with the moon's orbit, one would never be able to observe the full moon, only various partial phases (with the best being a slightly gibbous phase before the moon went behind the earth).

2007-09-04 07:47:32 · answer #1 · answered by dansinger61 6 · 2 1

Do you mean the Earth-Moon L2? Or the Earth-Sun L2?

I will assume you mean the Earth-Moon L2, which is 61,500 km from the moon. As I understand it, at L2 the moon would always be between L2 and Earth -- L2 orbits at the same rate as the moon. Also, it's so close to the Moon that the Moon would subtend 3.23 degrees, while the Earth subtends 1.62 degrees. In other words, the Moon would completely cover the Earth.

If you mean the Earth-Sun L2 point, on the other hand, then ...
It's about 1.5 million km away from the Earth. At this point, the moon (at closest approach) appears about 21.4 arc-minutes diameter, and the Earth about 58.4 arc-minutes. So the Moon appears about 1/3 the size of the Earth. For reference, the Moon appears from Earth's surface to be about 31 arc-minutes diameter.

2007-09-04 08:42:26 · answer #2 · answered by morningfoxnorth 6 · 0 0

Well, assuming the station didn't rotate, then it would be in constant eclipse by the Earth, and we'd always see it turning below. The L2 position is about a million miles from Earth, on a line extendinging from the sun through Earth. The moon would look tinier than when viewed from Earth - because you'd be about 3 times as far away from it at it's closest approach.

2007-09-04 08:09:00 · answer #3 · answered by quantumclaustrophobe 7 · 0 0

Hi. I don't think the Lagrange points 1 and 2 would be stable due to the shifting gravity of the Moon. If a station was in point 2 however, the Moon would change from quite large when it was opposite point 1 and would shrink in apparent size as it receded to the other part of its orbit.

2007-09-04 07:35:20 · answer #4 · answered by Cirric 7 · 1 0

Lagrange factors are factors the place gravity reasons an merchandise to stay in a similar place relative to the earth, in different words it follows the earth around the sunlight, despite the fact that if it does not revolve around the earth.

2016-12-31 12:18:55 · answer #5 · answered by tray 3 · 0 0

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