# Gravitational field strength equals zero?

1. Jun 27, 2012

### pussimatti

There is a point between earth and moon where the gravitational field strength is zero.
What about the same thing with three or more objects?

With three of more objects, is there always at least one point where the gravitational field strength due to the objects is zero?
Why / Why not?

2. Jun 27, 2012

### math man

For the Moon/Earth part of your question, simply set the gravitational forces of the Earth and Moon equal to each other and solve for r.
$\vec{F} =$$\frac{GMm}{r^{2}}$$\hat{r}$

$\frac{GM_{E}m}{(R_{E}+d)^{2}}$$=$$\frac{GM_{M}m}{(D_{M}-R_{E}-d)^{2}}$

where
$R_{E} =$ the radius of the Earth
$D_{M} =$ the distance from the center of the Earth to the center of the Moon
$d =$ the distance above the surface of the Earth
$M_{E} =$ the mass of the Earth
$M_{M} =$ the mass of the Moon

3. Jun 27, 2012

### Simon Bridge

Welcome to PF.
You can test this yourself ... in general you will be able to find a place where the gravitational fields balance out for any number of bodies. Usually several. These places will generally wander around as the bodies move ... if you sit in one it moves away, leaving you behind.

More interesting are the points where an object can be set to orbit the Earth at the same rate as the Moon (or the Sun at the same Rate as the Earth etc).

4. Jun 28, 2012

### pussimatti

I know about Lagrangian points and I understand that in real world planets (or other objects) move unlike in my problem.
So I'm interested in only a math problem where the radiuses of the objects are zero and the objects don't move.
And I've tested this with my graphing calculator by drawing a 3D-plot where z(x,y)=magnitude of gravitational force (or accerlation) in point (x,y), and there seems to be always at least one zero-gravity-point.
So I've ended up in the same conclusion that there is always such point(s). But why?
I want some mathematical proof about it.

Last edited: Jun 28, 2012
5. Jun 28, 2012

### Simon Bridge

The field is zero where the slope of the potential is zero. Since the potential everywhere else is always higher than the potential close to a point, there must be some place where it "turns around".

To construct a mathematical proof, you'd have to start from the potentials. Most people are happy with just the logic - there is no way the potential can go from low to high to low again without going through a level spot. It's like proving that hills go down as well as up.

6. Jun 28, 2012

### pussimatti

So the zero-gravity-point is the point where both the partial derivatives (in 2D space, in 3D space all three partial derivatives) are zero. In one dimension I understand that there is a point where df/dx=0 but why should there be a point in a xy-plane where both ∂f/∂x=0 and ∂f/∂y=0 in the SAME point (f(x,y)=potential). I understand that somewhere ∂f/∂x must be zero and somewhere ∂f/∂y=0 but why they should be both zero in the same point. I don't understand that. (OK, I'm only in high school and I haven't studied vector calculus. Is it obvious that ∇f is zero at some point like it's obvious in one dimension that df/dx=0 at some point.)

7. Jun 28, 2012

### Simon Bridge

Yeah - you should be able to see it from the rubber sheet analogy :)
That gives you two dimensions, which will let you see what a mathematical proof will involve. The point masses would be like pinning points on the sheet to the floor. The shape of the sheet between these pinned points will be complicated... but you'll find places to balance a ball. The sheet has to go down as well as up.

This must happen because all the potentials are negative and the field has to be continuous everywhere.

In the case of an electric filed, where potentials can be positive, it's a different kettle of piranhas.

The subject that deals with this sort of problem is called "topology" and it is a senior and grad-school course at college.

8. Jun 28, 2012

### Staff: Mentor

An argument, not a proof:
In 2D: If you take all points with ∂f/∂x=0, you get a line (or multiple lines). ∂f/∂y=0 gives another line. Now you just have to show that the lines have to intersect.
In 3D, you just get two-dimensional objects for all partial derivatives, and 3 of them will intersect in single points.