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Center of Mass between the Earth and Sun 
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#1
Oct1807, 11:11 AM

P: 287

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 earthmoon 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 


#2
Oct1807, 12:11 PM

P: 48

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.]



#3
Oct1807, 12:16 PM

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What's the generic definition of center of mass between two masses? 


#4
Oct1807, 02:36 PM

P: 287

Center of Mass between the Earth and Sun
(M1X1+M2X2....MnXn)/(M1+M2)
I'm guessing I made it too complicated? 


#6
Oct1807, 05:28 PM

P: 287

So are all the Ms unknown, or do I use the 84.5 as a mass



#7
Oct1807, 05:42 PM

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P: 41,560

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?



#8
Oct1807, 06:29 PM

P: 287

84.5m?



#9
Oct1807, 06:31 PM

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P: 41,560

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. 


#10
Oct1807, 11:14 PM

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P: 11,155

If you have a system of n masses, then the X(CoM) = (M1X1+M2X2+...+MnXn)/(M1+M2+...+MnXn) For a 2body system, this reduces to X(CoM) = (M1X1+M2X2)/(M1+M2) 


#11
Oct1907, 01:32 AM

P: 287

so my equation looks like this
(6578.1*85M+385815*M)/(84.5M+M) 


#12
Oct1907, 04:28 AM

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#13
Oct1907, 01:42 PM

P: 287

Oh, now I'm really confused as to how to write this equation.



#14
Oct1907, 01:48 PM

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#15
Oct1907, 02:06 PM

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#16
Oct1907, 02:11 PM

P: 287

The distance from the earth's center to the moon?
(385815*85M+3858156578*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 seesaw. 


#17
Oct1907, 02:29 PM

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P: 41,560

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.) 


#18
Oct1907, 02:36 PM

P: 287

so the practice problems
(M(10)+M(0))/2M =5 So this is my new idea: (385815*85M+M(0))/(84.5M+M) 


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