Question

I'm working on a problem that involves calculating the gravitational potential energy at a certain altitude between the surface of the earth and the Lagrange point between the earth and the moon (where the gravitational forces cancel each other out). I've verified my expression, and gotten these 2 plots for GPE vs. altitude (first response). The peak in GPE is at an altitude of 6137 km, which is the same distance as the earth's radius. Can anyone explain why those two numbers might be the same?

Answer 1

Here are the plots. The one on the left is on a normal scale; the one on the right has a logarithmic scale for altitude.

Answer 2

My current thought is that it is perhaps related to the derivative of the expression for potential energy:
\[PE(y) = -G \cdot y\Bigg[\frac{M_m}{(D -(R_e +y))^2} - \frac{M_e}{(R_e +y)^2} \Bigg]\]

Answer 3

@mathmale any insights?

Answer 4

i think you are way off beam on this

i just typed out a long response and totally lost it for reasons that only OS can ever manage

so in short expressions

-you're not converting between energy and force properly
-you are omitting motion
-you are neglecting the barycentre

there's also a bunch of assumptions, maybe all well grounded.

the wonderment in all of this is that it contradicts Kepler's Law. that too is missing though alluded to in my suggest that you consider motion


i'll post now before i lose this much shorter post!! good luck!!!

Answer 5

not converting between energy and force - this is true. I am not multiplying by mass, so this is potential energy per unit mass.

you are omitting motion - where should I be accounting for motion? Potential energy is based solely on position, not on any motion. I'm treating the earth and moon like two stationary bodies in space, without any orbits.

you are neglecting the barycentre - I don't know what this is...could you explain it to me?

Answer 6

your equation is

\(PE(y) = -G \cdot y\Bigg[\frac{M_m}{(D -(R_e +y))^2} - \frac{M_e}{(R_e +y)^2} \Bigg]\)

and if we take this bit , ie
\(G \bigg [\frac{M_m}{(D -(R_e +y))^2} - \frac{M_e}{(R_e +y)^2} \Bigg]\)

that's the force on a unit mass spaced y above the earth's surface

that's not energy

but if we want to find the point at which a unit mass experiences no force we have

\(\bigg [\frac{M_m}{(D -(R_e +y))^2} - \frac{M_e}{(R_e +y)^2} \Bigg] = 0\)

but we have frozen all of the objects, the earth, the moon, the unit mass, and we are ignoring the sun.

Answer 7

so if you solve that, you get the point at which an object can be left that is net-net neither attracted to the earth or the moon

Answer 8

but if we freeze the rotation of the earth and the moon, they crash into each other!! it is the rotational motion that balances out the forces and so on!

Answer 9

I think I understand your points, but it still doesn't answer my question.

We are assuming that the earth and moon are 2 large bodies, fixed in space. They won't come crashing down (we are operating in a rotating reference frame).
There is a Lagrange point, where the gravity from the earth and moon cancel each other out. That point is at \(3.5 \cdot 10^8\) meters above the surface of the earth, which is much higher than the 6371 km we are looking at here.

Gravitational potential energy is \(mgh\), where \(h\) is the altitude above the reference (here, the surface of the earth). Is that much correct?

Answer 10

mgh is the linear approximation over a very small distance. it does not work here.

but thank you (!), that bit i now get. and it remains erroneous.

Answer 11

Thanks for pointing out the problem with the linear approximation. I had to go back and find the general formula for GPE and now it makes a lot more sense.

Answer 12

What is it you actually want to verify?

Answer 13

Your expression for grav pe \[\large PE(y) = -G \cdot y\Bigg[\frac{M_m}{(D -(R_e +y))^2} - \frac{M_e}{(R_e +y)^2} \Bigg]\]did you derive it yourself? If not, where'd you get it?

Answer 14

I based that off of \(PE = mgh \), which I now realize is only a linear approximation for low altitudes. The real expression is \(PE = G \frac{Mm}{R}\)

Answer 15

Yes, use that one. I was gonna derive the expression but I figured you'd found the correct formula. The main thing that stood out was the fact you had the denominators squared.

Answer 16

\[\Large PE(y) =-G m \left[\frac{ M_M }{ D-(R_E+y) }- \frac{ M_E }{ R_E+y } \right]\]I think that'd be it.

m is mass of the object, D is the Earth-Moon center to center distance

Answer 17

Might have mixed up negatives but meh.