ToSeek

2003-Jul-17, 09:49 PM

Looking back (http://www.spaceref.com/news/viewsr.html?pid=9797)

I like this one better than the MGS one a while back.

I like this one better than the MGS one a while back.

View Full Version : Mars Express photographs Earth and Moon

ToSeek

2003-Jul-17, 09:49 PM

Looking back (http://www.spaceref.com/news/viewsr.html?pid=9797)

I like this one better than the MGS one a while back.

I like this one better than the MGS one a while back.

Pinemarten

2003-Jul-17, 11:47 PM

Nice link. Thanx.

It is the first time I have seen a ratio of earth/moon size/distance. All the others were artists conceptions.

It is the first time I have seen a ratio of earth/moon size/distance. All the others were artists conceptions.

freddo

2003-Jul-18, 12:56 AM

Nice link. Thanx.

It is the first time I have seen a ratio of earth/moon size/distance. All the others were artists conceptions.

Not real close eh? Amazing how dark the moon looks when exposure settings are set to capture the Earth....

It is the first time I have seen a ratio of earth/moon size/distance. All the others were artists conceptions.

Not real close eh? Amazing how dark the moon looks when exposure settings are set to capture the Earth....

Vega115

2003-Jul-19, 01:08 AM

Wow...The moon almost looks...close. And it's amazing that we sent men that distance and back...and the fact that they got there in a few days!! :o

ToSeek

2003-Jul-19, 03:17 AM

Nice link. Thanx.

It is the first time I have seen a ratio of earth/moon size/distance. All the others were artists conceptions.

I think it's foreshortened. The distance from the Earth to the Moon should be about thirty times the Earth's diameter, and this appears to be much less than that.

It is the first time I have seen a ratio of earth/moon size/distance. All the others were artists conceptions.

I think it's foreshortened. The distance from the Earth to the Moon should be about thirty times the Earth's diameter, and this appears to be much less than that.

tracer

2003-Jul-19, 05:11 AM

The crescent earth is kinda eerie....

man on the moon

2003-Jul-19, 07:30 AM

I think it's foreshortened. The distance from the Earth to the Moon should be about thirty times the Earth's diameter, and this appears to be much less than that.

it could also just be a not straight on photo. the only true representation of the distance would be if the craft were at a 90 degree angle to the "line" connecting the Earthand Moon. for example, a picture from when the Moon is behind the Earth relative to the craft, you wouldn't see the Moon. that would be 180 degrees. an angle of 45 degrees would create an illusion of only half the distance, etc. in that case you would be right ToSeek. i wish i could draw a picture to illustrate...maybe i will and post it later if someone else hasn't explained better by then.

anyone know what sort of angle is held here?

it could also just be a not straight on photo. the only true representation of the distance would be if the craft were at a 90 degree angle to the "line" connecting the Earthand Moon. for example, a picture from when the Moon is behind the Earth relative to the craft, you wouldn't see the Moon. that would be 180 degrees. an angle of 45 degrees would create an illusion of only half the distance, etc. in that case you would be right ToSeek. i wish i could draw a picture to illustrate...maybe i will and post it later if someone else hasn't explained better by then.

anyone know what sort of angle is held here?

ToSeek

2003-Jul-19, 01:31 PM

I think it's foreshortened. The distance from the Earth to the Moon should be about thirty times the Earth's diameter, and this appears to be much less than that.

it could also just be a not straight on photo. the only true representation of the distance would be if the craft were at a 90 degree angle to the "line" connecting the Earthand Moon. for example, a picture from when the Moon is behind the Earth relative to the craft, you wouldn't see the Moon. that would be 180 degrees. an angle of 45 degrees would create an illusion of only half the distance, etc. in that case you would be right ToSeek. i wish i could draw a picture to illustrate...maybe i will and post it later if someone else hasn't explained better by then.

anyone know what sort of angle is held here?

That's just what I meant - the angle between the Earth and the Moon isn't perpendicular to the direction to the spacecraft.

As for the relative positions, if you're looking down at the Earth from above and consider it as the center of a clock with the Sun at the 12 o'clock position, then Mars Express is at about the 4 o'clock position, and the Moon is at about the 11 o'clock position (rather than the 1 o'clock position that would make the angle perpendicular). So there's definitely some foreshortening going on.

it could also just be a not straight on photo. the only true representation of the distance would be if the craft were at a 90 degree angle to the "line" connecting the Earthand Moon. for example, a picture from when the Moon is behind the Earth relative to the craft, you wouldn't see the Moon. that would be 180 degrees. an angle of 45 degrees would create an illusion of only half the distance, etc. in that case you would be right ToSeek. i wish i could draw a picture to illustrate...maybe i will and post it later if someone else hasn't explained better by then.

anyone know what sort of angle is held here?

That's just what I meant - the angle between the Earth and the Moon isn't perpendicular to the direction to the spacecraft.

As for the relative positions, if you're looking down at the Earth from above and consider it as the center of a clock with the Sun at the 12 o'clock position, then Mars Express is at about the 4 o'clock position, and the Moon is at about the 11 o'clock position (rather than the 1 o'clock position that would make the angle perpendicular). So there's definitely some foreshortening going on.

man on the moon

2003-Jul-19, 02:44 PM

ahh! i see! my apologies...i thought that might have been what you meant, but it was rather brief and i wasn't sure if you meant that or an optical illusion.

if the Moon were at 11 and the craft at 4, the angle would then have been what, 140 degrees? i don't feel like doing the math right now, plus i don't know how far away the craft was, and it's distance would change the angle between itself and the Moon...anyway, that's a bunch of nasty trigonometry i don't want to do! rather just enjoy the picture myself.

thanks for clarifying your thoughts. :D

if the Moon were at 11 and the craft at 4, the angle would then have been what, 140 degrees? i don't feel like doing the math right now, plus i don't know how far away the craft was, and it's distance would change the angle between itself and the Moon...anyway, that's a bunch of nasty trigonometry i don't want to do! rather just enjoy the picture myself.

thanks for clarifying your thoughts. :D

ToSeek

2003-Jul-19, 05:20 PM

If we make the simplifying assumption that the spacecraft is effectively at an infinite distance and that my clock angle estimates are correct, then the Moon is about 30° off from being directly behind the Earth. Since sin 30° = 0.5, the apparent distance is about half the actual distance. Actually, it seems closer to one-fourth, which would imply an angle more like 15°.

Pinemarten

2003-Jul-19, 08:03 PM

Can we measure the diameter of each sphere and do a Pythagoras/ triangulation thing?

man on the moon

2003-Jul-20, 11:04 AM

you could, problem is that we don't know the distance of the craft. the angle and distance to the Moon are known, that's not so difficult. the hard part is the fact that in any given triangle, even one with 15 degrees, there are an infinite number of points for the craft to be.

imagine a right triangle extending infinitely into space, the craft travelling along the hypotenuse. it can be anywhere along that line, and all three angles will remain the same. that's why in trig you have to know the values of either two angles and a leg or two legs and an angle. in this case, we may know all three angles, but if we don't know any of the leg lengths, you can't figure it out.

at whatever point the craft is located, that is one vertex. now draw one line from that vertex to the Moon. draw another line from the craft to the tangent between the Earth and Moon so that it just touches the horizon of Earth. the distance between the two points is the "apparent" distance between the two objects. at 1,000 miles you might see 10% of the total. at 1,000,000 miles you might see 70-80%. it would be [i]very[/] easy to find that length. after all, you're creating a third triangle, and with a little arithmatic mixed into your trig, it would be possible to find the lengths of the legs. one of which would be the apparent difference between the two objects.

yes, you would have to take into account to radi of the two spheres which is why i said it's nasty before. it's not so much extremely difficult thoguh as time consuming, and i'm a bit too lazy that.

i just like to look at the picture. :D :D hope you can still enjoy it too now that all this has been dumped on you. :wink:

ps--if it's still confusing, i made a diagram, i just can't figure out how to get it into the post. if anyone knows how to do that...i'd be happy to share.

imagine a right triangle extending infinitely into space, the craft travelling along the hypotenuse. it can be anywhere along that line, and all three angles will remain the same. that's why in trig you have to know the values of either two angles and a leg or two legs and an angle. in this case, we may know all three angles, but if we don't know any of the leg lengths, you can't figure it out.

at whatever point the craft is located, that is one vertex. now draw one line from that vertex to the Moon. draw another line from the craft to the tangent between the Earth and Moon so that it just touches the horizon of Earth. the distance between the two points is the "apparent" distance between the two objects. at 1,000 miles you might see 10% of the total. at 1,000,000 miles you might see 70-80%. it would be [i]very[/] easy to find that length. after all, you're creating a third triangle, and with a little arithmatic mixed into your trig, it would be possible to find the lengths of the legs. one of which would be the apparent difference between the two objects.

yes, you would have to take into account to radi of the two spheres which is why i said it's nasty before. it's not so much extremely difficult thoguh as time consuming, and i'm a bit too lazy that.

i just like to look at the picture. :D :D hope you can still enjoy it too now that all this has been dumped on you. :wink:

ps--if it's still confusing, i made a diagram, i just can't figure out how to get it into the post. if anyone knows how to do that...i'd be happy to share.

Pinemarten

2003-Jul-20, 05:55 PM

You have to post on a public site and link to it, or a bbcode thing that I haven't figured out yet.

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