Center of Mass between the Earth and Sun

[chaotixmonjuish]

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

 

[Vijay Bhatnagar]

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 relevant (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.]

 

[Doc Al]

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?

 

[chaotixmonjuish]

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

I'm guessing I made it too complicated?

 

[Doc Al]

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

That's all there is to it.

 

[chaotixmonjuish]

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

 

[Doc Al]

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?

 

[chaotixmonjuish]

84.5m?

 

[Doc Al]

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.

 

[Gokul43201]

(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)

 

[chaotixmonjuish]

so my equation looks like this

(6578.1*85M+385815*M)/(84.5M+M)

 

[Doc Al]

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.)

 

[chaotixmonjuish]

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

 

[Doc Al]

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?

 

[dynamicsolo]

(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...

 

[chaotixmonjuish]

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.

 

[Doc Al]

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.)

 

[chaotixmonjuish]

so the practice problems

(M(10)+M(0))/2M =5

So this is my new idea:

(385815*85M+M(0))/(84.5M+M)

 

[Doc Al]

so the practice problems

(M(10)+M(0))/2M =5

Exactly. One of the masses is at the zero point.

So this is my new idea:

(385815*85M+M(0))/(84.5M+M)

That would be right (almost) if we were measuring from the moon. But we're measuring from the earth.

The general equation (as you know) is:

$$(M_1 X_1 + M_2 X_2)/(M_1 + M_2)$$

Where M_1 is the mass of object 1 and X_1 is its coordinate. Since we want to measure things from the earth, we put the Earth at X = 0. And the moon will be at X = 385,815 km. So:

[(84.5M)(0) + (M)(385,815)]/(84.5M + M)

You finish it up.

 

[chaotixmonjuish]

A dog, with a mass of 9.0 kg, is standing on a flatboat so that he is 22.4 m from the shore. He walks 8.4 m on the boat toward the shore and then stops. The boat has a mass of 45.0 kg. Assuming there is no friction between the boat and the water, how far is the dog from the shore now?

Would I tackle this problem in a similar way? I noticed that since the surface is frictionless, would the boat move in the opposite direction to compensate.

 

[Dick]

The boat would move in the opposite way to compensate, yes.

 

[chaotixmonjuish]

Would it necessarily move 8.4?

I had this general equation in my head

9*22.4-8.4+45*(22.4+x)/45+9

 

[Doc Al]

Would I tackle this problem in a similar way? I noticed that since the surface is frictionless, would the boat move in the opposite direction to compensate.

Yes, the boat must move in the opposite direction. This problem is trickier, but it could be treated as a center of mass problem.

Would it necessarily move 8.4?

I had this general equation in my head

9*22.4-8.4+45*(22.4+x)/45+9

No. The first thing to realize is that the center of mass of "boat + dog" must remain the same distance to the shore no matter how the dog moves on the boat.

Now, if the dog moved 8.4 m in one direction, how much does that change the center of mass of the "dog + boat" with respect to the boat? (Pretend that the dog was in the middle of the boat before he moved. So the center of mass of "dog + boat" is in the center of the boat, say. After the dog moves, where's the new center of mass?)

Use that change in center of mass to figure out (1) how far the boat must have moved away from shore, and then (2) how far the dog ends up from the shore.

 

[chaotixmonjuish]

I have one question, I eventually solved the problem with the aid of an old thread:

https://www.physicsforums.com/showthread.php?t=97043

I don't understand the part where you have (D-y). I understand where all the others parts of the equation come from except that part.

 

[Doc Al]

I don't understand the part where you have (D-y). I understand where all the others parts of the equation come from except that part.

I'm not sure what equation you are referring to. The thread you linked to doesn't have it. Perhaps you saw it in a different thread.

 

[chaotixmonjuish]

oh well, an equation I found for a similar problem looked like this:

mass of dog(initial position-x)+mass of boat(distance from shore-y)/dog+boat

x+y=displacement caused by dog walking (d)

mass of dog(initial position-x)+mass of boat(distance from shore-d-x)/dog+boat