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Center of Mass between the Earth and Sun

 
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Oct18-07, 11:11 AM   #1
 

Center of Mass between the Earth and Sun


The ratio of the mass of the earth to the mass of the moon is 84.5. Assume that the radius of the earth is about 6578.0 km and that the distance between the center of the earth and the moon is 385815.0 km. Determine the distance of the center of mass of the earth-moon system from the center of the earth.

I was about to start this and I had a few questions:

Can I just assumed that it is 385815 km is the distance from the center of the earth to the center of the moon?

If so, what do I do about the ratio? Would my equation be:

(6578.1*1+385815*1/84.5)/84.5
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Oct18-07, 12:11 PM   #2
 
Solve as follows : i) Masses of earth and moon are assumed to be concentrated at the centres of the respective bodies. Hence, only the centre to centre distance between the two is relevent (radius of earth is given just to confuse). ii) Assume the coordinate system as a striaght line with the earth situated at the origin. iii) Apply formula for determination of the coordinate of the centre of mass. [As the masses of earth and moon are not given, only their ratio is given, modify the formula suitably so as to have in it the mass ratio instead of the actual masses.]
Oct18-07, 12:16 PM   #3
 
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Quote by chaotixmonjuish View Post
Can I just assumed that it is 385815 km is the distance from the center of the earth to the center of the moon?
Yes.

If so, what do I do about the ratio?
You'll need it to compute the center of mass. (You don't need the actual masses.)

What's the generic definition of center of mass between two masses?
Oct18-07, 02:36 PM   #4
 

Center of Mass between the Earth and Sun


(M1X1+M2X2....MnXn)/(M1+M2)

I'm guessing I made it too complicated?
Oct18-07, 02:40 PM   #5
 
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Quote by chaotixmonjuish View Post
(M1X1+M2X2....MnXn)/(M1+M2)
That's all there is to it.
Oct18-07, 05:28 PM   #6
 
So are all the Ms unknown, or do I use the 84.5 as a mass
Oct18-07, 05:42 PM   #7
 
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All that matters is their ratio, so try this: If you call the mass of the moon "M", what's the mass of the earth?
Oct18-07, 06:29 PM   #8
 
84.5m?
Oct18-07, 06:31 PM   #9
 
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Quote by chaotixmonjuish View Post
84.5m?
Right!
Mass of moon = M
Mass of earth = 84.5M

Now just crank out the center of mass. Use the center of the earth as the origin.
Oct18-07, 11:14 PM   #10
 
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Quote by chaotixmonjuish View Post
(M1X1+M2X2....MnXn)/(M1+M2)
Minor correction:

If you have a system of n masses, then the X(CoM) = (M1X1+M2X2+...+MnXn)/(M1+M2+...+MnXn)

For a 2-body system, this reduces to X(CoM) = (M1X1+M2X2)/(M1+M2)
Oct19-07, 01:32 AM   #11
 
so my equation looks like this

(6578.1*85M+385815*M)/(84.5M+M)
Oct19-07, 04:28 AM   #12
 
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Quote by chaotixmonjuish View Post
so my equation looks like this

(6578.1*85M+385815*M)/(84.5M+M)
Two problems: (1) The mass of the earth is 84.5M, not 85M. (2) The mass of the earth is centered at the earth's center, not at its radius. (I have no idea why you are given the earth's radius in this problem.)
Oct19-07, 01:42 PM   #13
 
Oh, now I'm really confused as to how to write this equation.
Oct19-07, 01:48 PM   #14
 
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Quote by chaotixmonjuish View Post
Oh, now I'm really confused as to how to write this equation.
It's easier than you think. Since we are measuring distance from the center of the earth, what distance should the earth's mass have in your equation?
Oct19-07, 02:06 PM   #15
 
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Quote by Doc Al View Post
(I have no idea why you are given the earth's radius in this problem.)
You don't need it to solve the problem. However, it is of interest to compare the Earth's radius to your answer...
Oct19-07, 02:11 PM   #16
 
The distance from the earth's center to the moon?

(385815*85M+385815-6578*M)/(84.5M+M)

I'm running a bit low on ideas. The only example we were given on how to solve the five problems he gave us was a see-saw.
Oct19-07, 02:29 PM   #17
 
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Do this practice problem: Find the center of mass of two equal masses that are 10 m apart. Measure distances from the center of one of the masses. Use the same equation.

(This might be easier to see, since I presume you know what the answer must be. Once you see how easy it is, you'll probably smack yourself.)
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