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If Earth had Rings..
By Tobiah Waldron on Sunday, September 23, 2001 - 08:48 pm:

If the earth had low density rings around it, but the mass of these rings plus the mass of the earth with those rings equaled the present mass of the earth without rings:

1>How would this affect the length of the year?

2>How would this affect the length of the month?

By Brad Rodgers on Sunday, September 23, 2001 - 11:05 pm:

This sounds as though it's a trick question. Let E=mass of Earth without rings, and R=mass of rings. Then we are given

"the mass of these rings plus the mass of the earth with those rings equaled the present mass of the earth without rings"

R+(E+R)=E

subtracting E from both sides

2R=0

so R=0. So, with this in mind the answer to 1 would be: it wouldn't. Can you guess the answer to number 2?

I'm not sure how hard the problem is supposed to be, and if this problem was for a class doing gravitational physics or something, I'm pretty sure I've not understood a point...

Brad

By Jim Oldfield on Monday, September 24, 2001 - 12:52 am:

I think she was wondering if the different distribution of mass would make a difference i.e. would it matter if the Earth's mass was more spread out?

The answer is no. We can treat bodies (like the Earth) as if they were point masses, with all their mass concentrated at their centre of mass. Although they obviously aren't, the maths still works, so if the rings were even (even thickness and material) then the Earth's centre of mass would be the same as in the real situation, and so the maths would give the same result. Obviously their is no atmosphere in space, so the rings wouldn't cause any sort of 'air resistance'.

Depending on how deep you want to take the problem, I'm not sure how rings would have affected the Earth's route during the formation of the solar system. I don't know much about astrophysics, but I believe the system was a big cloud of gas at that point, so a sort of 'air' resistance might have occured (?)

Also, forgeting about be the system's formation, I don't actually think an increase in mass would affect the path anyway. Circular motion equations don't depend on mass, (you might understand this: a = v2/r = GM/r2, where M is the mass of the sun), so I doubt elliptical paths would.

If I have misunderstood what you are asking please post back.

Jim

By Arun Iyer on Monday, September 24, 2001 - 07:45 pm:

By kepler's third law,
T2 proportional to r3
T = time period of earth
r = radius of orbit of earth with sun as center

In your problem,
since r remains unchanged,T will remain unchanged which i suppose answers your question.

love arun

please do correct me if i missed anything...

By Jim Oldfield on Monday, September 24, 2001 - 09:34 pm:

What Arun and I have answered (assuming this question isn't a trick one) is really question 1). For 2), assuming "month" means lunar month (i.e. time for moon to orbit the earth) if the moon orbits outside of the rings our answers still hold.

BUT: if the moon's orbit is lower than the rings or in between the rings then the length of the month will change. I will explain why if anyone wants.

Jim

By Tobiah Waldron on Wednesday, September 26, 2001 - 03:36 pm:

No trick question, just the distribution of mass, as Jim stated.

Is there a particular mass for the earth and moon that would make the lunar month and solar year coincide in a simpler ratio?

By Tom Hardcastle on Wednesday, September 26, 2001 - 08:18 pm:

Wouldn't increased tidal forces have a small (probably negligible) effect on the orbits?

By Jim Oldfield on Thursday, September 27, 2001 - 12:14 am:

In my last post when I said "...the length of the moon will change" I was of course referring to the length of the moon's orbit.

To answer Tobiah's most recent post, I believe the answer is... yes. Ok, so you want a more detailed answer. Well, you would have to start worrying about Netwon's Law of Gravitation ( F=Gm1m2/r2) and the equations of bodies moving in an ellipse. I have studied circular motion but not elliptical, so I get a bit stuck here (though I could always model the motion as circular).

Tom, you're probably right. I don't know.

Jim