Where is the CG between the Earth and the Moon

[thetexan]

When two bodies orbit each other the entire system rotates about a center of gravity point in the space between them causing a wobble effect of the entire system.

Approximately at what distance from Earth is that CG point between the Earth and moon please.

tex

 

[Charles Link]

The moon has a mass approximately 1/81 the mass of the earth. A little algebra gives $(M_e)x=(M_e/81)(R-x)$ where $R$ is the Earth moon distance.(from center of Earth to center of moon) ==>> $x=R/82 =240000$ miles $/82=3000$ miles approximately. With the radius of the Earth $r_e=4000$ miles this puts this about $1000$ miles below to the surface of the Earth on the side of the Earth that faces the moon. Please double check my calculations to verify. (I didn't google this, but I believe I have it correct.)

 

[fresh_42]

The moon has a mass approximately 1/81 the mass of the earth. A little algebra gives $(M_e)x=(M_e/81)(R-x)$ where $R$ is the Earth moon distance.(from center of Earth to center of moon) ==>> $x=R/82 =240000$ miles $/82=3000$ miles approximately. With the radius of the Earth $r_e=4000$ miles this puts this about $1000$ miles below to the surface of the Earth on the side of the Earth that faces the moon. Please double check my calculations to verify. (I didn't google this, but I believe I have it correct.)

This cannot be true. I used Newton's gravity formula $\frac{G \cdot M_E \cdot M_0}{x^2} = \frac{G \cdot M_M \cdot M_0}{(R-x)^2}$ which isn't linear and got $x=0.9 \, R$ which looks as if there were an easier argument since $9^2 = 81.$ And it violates a quote from "Armageddon", not that a movie would be a reliable source ...

 

[Charles Link]

This cannot be true. I used Newton's gravity formula $\frac{G \cdot M_E \cdot M_0}{x^2} = \frac{G \cdot M_M \cdot M_0}{(R-x)^2}$ which isn't linear and got $x=0.9 \, R$ which looks as if there were an easier argument since $9^2 = 81.$ And it violates a quote from "Armageddon", not that a movie would be a reliable source ...

I think you computed the zero gravity point. If I remember correctly (I computed it a few years ago), it lies 9/10 of the way to the moon. editing... yes, your calculation is for the zero gravity location... :-)

 

[fresh_42]

I think you computed the zero gravity point. If I remember correctly (I computed it a few years ago), it lies 9/10 of the way to the moon. editing... yes, your calculation is for the zero gravity location... :-)

Ah, ok. Misunderstood it.

 

[thetexan]

Im an instructor pilot.. Is it as simple as doing a weight and balance of the airplane using moments and arms as if the earth/moon combination was one big balanced teeter toter?

If that is the case then the 1/81 difference indicates to me that the arm would be a 1:81 ratio of the distance between the two bodies...one unit on one side of the balance point and 81 units on the other. Using 243,000 miles and dividing by 82 that puts the balance point 2963 miles from the center of the Earth or inside the Earth itself. Is it that simple?

tex

 

[Bandersnatch]

Is it that simple?

Yes.

 

[Charles Link]

Yes.

(For the OP) When the moon revolves around the Earth in 29 days or thereabouts, it is actually this dumbbell system that is rotating. Because the CG lies so near to the center of the earth, basically the moon revolves around the earth.

 

[tony873004]

Here is a simulation of the Moon and Earth orbiting the EMB (Earth/Moon barycenter)
http://orbitsimulator.com/gravitySimulatorCloud/simulations/1468674152290_emb.html