The thread discusses the location of the center of gravity (CG) between the Earth and the Moon, exploring the implications of their mass ratio and gravitational interactions. Participants engage in calculations and theoretical reasoning related to the CG's position, considering both algebraic and gravitational approaches.
Participants express differing views on the calculations and interpretations of the CG's location, with no consensus reached on the correct approach or result.
Some calculations depend on the definitions of the CG and zero gravity point, and there are unresolved mathematical steps in the reasoning presented by participants.
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 said: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.)
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 said: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 ...
Ah, ok. Misunderstood it.Charles Link said: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... :-)
Yes.thetexan said:Is it that simple?
(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.Bandersnatch said:Yes.