# B Is gravity ever negative here?

1. Aug 7, 2015

### Alameen Damer

In the range equation, is the value of g ever taken to be negative?

dx=Vi^2/g x sin 2 (theta)

Basically my question is where the y displacement is 0, and we are able to use the range equation, will the value of g ever be negative, and if so, when? I know Vi is simply magnitude and not direction, and thus does that mean g is alway positive?

Last edited by a moderator: Aug 7, 2015
2. Aug 7, 2015

### Orodruin

Staff Emeritus
Negative g would mean that the object was accelerating upwards and not have a range. The directional information comes from the sine term.

It is extremely confusing when you write an x in your equation like that. A priori, we have no idea if it is a multiplication symbol or a distance unless we can deduce it from already knowing what the equation should say.

3. Aug 8, 2015

### Andrew Mason

g is essentially constant. $\vec{g} = -\frac{GM_{earth}}{r^2}\hat{r}$ where r is the distance from the projectile to the centre of the earth and $\hat{r}$ is the unit radius vector from the centre of the earth.

AM

4. Apr 4, 2016

### Greg Wallace

Gravitational repulsion does not exist but negative gravity does. Here is a simple example using Earth. If you take a smaller mass below the surface of the Earth, it starts to have Mass behind it. So the gravitaional attraction to the center is negated by the force of mass behind it. Take the smaller object to say 1 meter from the center of gravity of the Earth and you can no longer use the standard formula where F = (G * M * m)/R^2 . Gravity forces tend to cancel out towards the center. At the very center of the Earth there should be no net gravity force.

5. Apr 4, 2016

### Andrew Mason

Welcome to PF Greg.

I am not sure how you are defining negative gravity. There are gravitational forces at the centre of be earth, of course. It is just that the forces sum to zero. The effect of gravitational force however, is significant. It creates enormous pressure. You just have to go under water a short distance to feel some of that pressure!

AM

6. Apr 4, 2016

### houlahound

Is there a standard distance one should stop using mgh and start using 1/r^2 for GPE?

7. Apr 4, 2016

### Orodruin

Staff Emeritus
This depends on how accurate you want your result to be.

8. Apr 4, 2016

### Andrew Mason

No. It depends on. how accurate you want to be. mgh is a reasonable approximation for h<<r.

AM

9. Apr 5, 2016

### Greg Wallace

For Andrew

Sorry if my post was not sufficiently clear.

I am not trying to define negative gravity but just give a simple example where mathematically, gravity can be seen as negative.

I am trying to model and focus on the effects of a standard 1 Kg Mass as it conceptually moves toward the center of the Earth. I would be more happy to be pointed to a thread that is not about negative gravity. I just propose that if a standard 1 Kg mass is studied then there can be Mass in front (Mf) and Mass behind (Mb) and therefore positive and negative Newtons involved.

To take a very intuitive example, if the Moon is above (Mass behind) then it tends to negate Mass in front which is the Earth.

The accepted Newtons for a 1 Kg Mass at the surface Earth seems to be F = 9.74 N.

To take a different (not the Moon) very simple example, if the force resulting from Mass in front is say 9 N (where N is in Newtons) and the Mass behind is a force of 1 N then the net force in the direction of travel is 9 N - 1 N = 8 N.

So theoretically, and to satisfy a simple equation, -1 N is negative gravity in a mathematical sense only. It seems current science that Gravity only attracts and this does not contradict that.

My tentative formula for a standard 1Kg mass moving below the center of the Earth this is:
F = (G * Mf * m / Rf * 10^2) - (G * Mb * m / Rb * 10^2)
Where Mf is the Mass in front
Mb is the Mass behind
Rf is the radius to the center of Mass for Mf
Rb is the radius to the center of Mass for Mb
m can be a 1kg small mass. Using 1 makes it simpler.
G is the Gravitational constant.

It assumes that the Gravitational Constant takes into account angular forces of Mass and near and far Mass forces. I could say much more on this.

I have much more that could be said in studying a 1Kg standard mass moving below the center of the Earth and toward its center.

If there is no such thread on this theme, I would much like a suggestion for a new thread.

Regards
Greg

10. Apr 5, 2016

### jbriggs444

Physics is normally done with vectors in three (or four) dimensions. The notions of positive and negative which are simple and sufficient to describe direction in the one dimensional case are not adequate in the three dimensional case. Instead, we describe gravity from the moon as pointed "over there" (in the direction of the center of mass of the moon) and gravity from the earth as pointed "over here" (in the direction of the center of mass of the earth).

In the vector formulation, the direction of the force is determined by a unit direction vector between the objects. The magnitude of the force is determined by Newton's formula ($F=\frac{Gm_1m_2}{r^2}$). The sign of the magnitude is always positive as long as the masses are always positive.

There is no room in the formula for negative gravity unless one of the masses is negative. No known entities have negative mass. However...

The gravitational attraction from a uniformly dense sphere with a spherical hole can be calculated as the sum of positive gravity from the sphere and negative gravity from the hole.

11. Apr 5, 2016

### Greg Wallace

Hi JBriggs

I am dealing with just 3 dimensions.

I basically agree with what you say. I agree that gravity is always positive and there is no proof yet for negative mass.

However, my simple examples show that Newtons of force can be treated as positive and negative to satisfy an equation where 9 N - 1 N = 8 N.

Resultant force vectors are always in a resultant line.

When one delves into examples where a standard mass is taken at some point in a 3D Earth and below the surface, then it has Mass in front and Mass behind. The resultant force vector is toward the center. Mass behind has a negative resultant force vector.

This does not contradict that the Mass behind is still positive Mass but the sum of vectors have to allow for -ve Newtons.

Regards
Greg

12. Apr 5, 2016

### Andrew Mason

For a symmetrical solid sphere, the only mass that matters (as far as gravity is concerned) is the mass "below" you - that is the mass within a sphere of radius r where r is the distance from you to the centre of the sphere. The contributions from all the mass "behind" you and around you but above that radius r will sum to zero.

AM

Last edited: Apr 5, 2016