(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'

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.

AT the dV/dx =0 point, there is a potential, but it is not changing with x.

Isn't dpotential/dx equal to force?

Thanks (I'm not much wiser!), but what's ve as in 'between ve and zero' please?

To derive an expression for the gravitational potential, Vgrav, at a general position x on the x-axis for two particles of equal mass fixed at X=0 and another positive point on the x-axis, you can use the formula for the gravitational potential due to a point mass.

The gravitational potential due to a point mass M at a distance r is given by:

V = -GM/r

In this case, we have two point masses, so we need to consider the gravitational potential due to both masses. Let's call the distance between the two masses d and the position of the first mass X=0. The position of the second mass can be represented as x. The distance between the second mass and the general position x is then (d - x).

The gravitational potential at position x due to the first mass (at X=0) is given by:

V1 = -GM / |x|

Note that we used the absolute value of x since the first mass is located at X=0.

Similarly, the gravitational potential at position x due to the second mass is given by:

V2 = -GM / |d - x|

Now, to find the total gravitational potential at position x, we need to add the potentials due to both masses:

Vgrav = V1 + V2
= -GM / |x| - GM / |d - x|

This expression gives the gravitational potential at a general position x on the x-axis when two equal masses are fixed at X=0 and another positive point on the x-axis.

Moving on to the second part of your question, the significance of the point where dV/dx = 0 can be understood in terms of the slope of the gravitational potential curve. The derivative, dV/dx, represents the rate of change of the gravitational potential with respect to position x.

When dV/dx = 0, it means the slope of the potential curve is zero. In other words, at this point, the potential is not changing as we move along the x-axis. This point is known as an equilibrium point. In the context of gravitational potential, it signifies a point of stable equilibrium.

In this particular case, the point where dV/dx = 0 corresponds to the point where the two masses are located, i.e., at x = 0 (X=0) and x = d. At these positions, the gravitational potential has reached its maximum or minimum value, resulting in an equilibrium point. The y-coordinate of these equilibrium points would depend on the specific problem, but they are not necessarily at infinity with y = 0.

I hope this helps! Let me know if you have any further questions.