Simplified LaGrange Point Calculation

AI Thread Summary
The discussion centers on finding the gravitationally neutral point between the Earth and the Moon, specifically the L1 point, without considering centripetal forces or other gravitational influences. The user attempts to derive the relationship between the gravitational forces exerted by the Earth and the Moon using the formula for gravitational force. They express confusion over their derivation, particularly regarding the distances involved, mistakenly equating the distance from the Earth's center to the Lagrangian point with the Earth's radius. A response clarifies that the distance should be measured from the Earth's center to the Lagrangian point, which is closer to the Moon. The conversation highlights the importance of correctly understanding the distances in gravitational calculations.
seanhogge
Messages
1
Reaction score
0
I am attempting for my own curiosity to find out at what point during a geodesic path from the Earth to the Moon one would reach a gravitationally neutral point.

This is essentially the L1, but without adjustments for centripetal force of a moving system, and ignoring all other gravitational bodies (i.e. the sun).

It's one of those back-of-envelope things that I've run out of envelope for. My stumbling block seems to be the math, but I figured this was a better place to ask for guidance since the subject may have something to do with it.

So basically, I set

Fearth = G (mme * mearth) / r2earth and Fmoon = G (mme * mmoon) / r2moon to be equal.

So when I remove G and move everything so that distances are on one side, and masses are on another, I get

sqrt( mearth / mmoon ) = rearth / rmoon.

This can't be right. It's supposed to be an inverse square. However, I cannot for the life of me find where this derivation goes wrong. I suspect there is something fundamental with my original concept, or I'm missing some very obvious mathematical issue in my complete ignorance. Which is caused by my never having taken anything higher than high school geometry/trig and no physics. Please feel free to correct me on any and all points, including my terminology and minutiae.

Thanks in advance!
 
Physics news on Phys.org
seanhogge said:
I am attempting for my own curiosity to find out at what point during a geodesic path from the Earth to the Moon one would reach a gravitationally neutral point.

This is essentially the L1, but without adjustments for centripetal force of a moving system, and ignoring all other gravitational bodies (i.e. the sun).

It's one of those back-of-envelope things that I've run out of envelope for. My stumbling block seems to be the math, but I figured this was a better place to ask for guidance since the subject may have something to do with it.

So basically, I set

Fearth = G (mme * mearth) / r2earth and Fmoon = G (mme * mmoon) / r2moon to be equal.

So when I remove G and move everything so that distances are on one side, and masses are on another, I get

sqrt( mearth / mmoon ) = rearth / rmoon.

This can't be right. It's supposed to be an inverse square. However, I cannot for the life of me find where this derivation goes wrong. I suspect there is something fundamental with my original concept, or I'm missing some very obvious mathematical issue in my complete ignorance. Which is caused by my never having taken anything higher than high school geometry/trig and no physics. Please feel free to correct me on any and all points, including my terminology and minutiae.

Thanks in advance!

Your derivation looks good to me. Perhaps you're confusing rearth with the Earth's radius? rearth is the distance from the Earth's center to the Lagrangian point. At the Lagrangian point the object will be closer to the moon than to the earth.
 
Thread 'Gauss' law seems to imply instantaneous electric field propagation'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Thread 'A scenario of non-uniform circular motion'
(All the needed diagrams are posted below) My friend came up with the following scenario. Imagine a fixed point and a perfectly rigid rod of a certain length extending radially outwards from this fixed point(it is attached to the fixed point). To the free end of the fixed rod, an object is present and it is capable of changing it's speed(by thruster say or any convenient method. And ignore any resistance). It starts with a certain speed but say it's speed continuously increases as it goes...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Back
Top