For the gravity problem, the potential will be the scalar addition of each of the potentials (since they are in opposite directions, they are subtractive).
V=GM/x -GM/(ve-x)
Lets test that. IF x is large negative, they should be same sign.
V=GM(1/large- -1/large+ so the terms add in same direction.
If x is between ve and zero, they should subtract..
V=GM(1/small+ - 1/small) they do have opposing signs.
If x is large +, both should add
V=GM(1/large+ -1/large- _) and the add again.
(1) 2 particles of equal mass are fixed at X=0 and another +ve point on the x axis.
No other grav influences are in the system. (they can be ignored)
I have to derive an expression for Vgrav at a general position x, on the x axis.
I believe this is related to V=-GM/r, but can't see how to develop it for 2 masses.
(2) I have to say what the significance of the point where dV/dx=0 is, and what it's coords are in the system.
i believe the derivative is where there is no slope, so no V. Am I right?
I believe the point is on the x axis at infinity, and y=0
Any steers appreciated. Thanks
Reposted under 'Physics'
3 answers
AT the dV/dx =0 point, there is a potential, but it is not changing with x.
Isn't dpotential/dx equal to force?
Isn't dpotential/dx equal to force?
Thanks (I'm not much wiser!), but what's ve as in 'between ve and zero' please?
0 answers