# Finding the maximum value of a gravitional force field

1. Mar 28, 2015

### Calpalned

1. The problem statement, all variables and given/known data
Two identical particles, each of mass m are located on the x-axis at $x=+x_0$ and $x = -x_0$. At what point (or points) on the y-axis is the magnitude of $\vec g$ a maximum value and what is its value there?

2. Relevant equations
Formula for gravitational field $-2Gm \frac {y}{(x_0^2 + y^2)^{3/2}}$

3. The attempt at a solution
The textbook's first step is to keep y a positive quantity. " If we keep y as a positive quantity, then the magnitude of the field is $2Gm \frac {y}{(x_0^2 + y^2)^{3/2}}$

How is this a legal move? We want to find the maximum value, so shouldn't we be considering the entire y-axis and not just the part above the x-axis? How can we be so sure that the maximum is actually above the x-axis?

2. Mar 28, 2015

### jbriggs444

By symmetry it is obvious that the pattern of gravitational force below the x axis will be a mirror image of the pattern above. If you find a maximum for positive y, there will be an identical maximum for negative y. There is no point looking for both maxima when finding the one will reveal the other.

3. Mar 28, 2015

### Calpalned

The textbook also gives me a hint "Take the derivative of $\frac {d \vec g }{dy}$ However, there are more than one variable in the formula for gravitational field. Therefore, why don't we take a partial derivative?

Never mind. I just realized that $x_0$ is a constant. Therefore, a partial derivative is not needed.

4. Mar 28, 2015

### jbriggs444

There is only one variable. The other values are constants. You could call them variables and take a partial derivative with respect to y. Or you can call them constants and take an ordinary derivative with respect to y. Both amount to the same thing.

5. Mar 28, 2015

### Calpalned

We find locations of the maximum magnitude by setting the first derivative equal to 0. Since the expression is never negative, any extrema will be maxima.
Expression: $g = 2Gm \frac {y}{(x_0^2 + y^2)^{3/2}}$.

What makes this statement true? For example, the equation $y = x^2 + 1$ is never negative, but its extrema is a minimum.

6. Mar 28, 2015

### jbriggs444

You are right. As written, that reasoning is incorrect. More correctly, one could reason that since the expression is zero at zero, zero at infinity, bounded, strictly positive and continuously differentiable between that it must have at least one extremum which is a global maximum and at which the first derivative is zero somewhere in that range.

7. Mar 28, 2015

thank you