# Why is g negative?

1. Jun 23, 2013

### SecretSnow

Hi guys, I understand that g is by definition positive, as in the Gravitational field strength. However, my school's notes has defined g as
-(GM)/r^2. I don't understand what's that negative sign for, and in what scenario is the negative sign used? By the way, it also defined g as -(d phi/dr) where phi is the gravitational potential and d is the differentiation sign...in this case, it says F=-Gmm/r^2.

Thanks a lot!!

2. Jun 23, 2013

### darkxponent

g is the acceleration due gravity. It is attractive that is why taken as negative. It is just a reference that attractive forces are taken negative while repulsive forces are taken positive.

3. Jun 23, 2013

### Simon Bridge

The sign of the acceleration is to do with the direction. If "up" is taken to be positive, then the acceleration of gravity will be negative. Since F=ma, this means that the force is also negative.

That is all there is to it.

4. Jun 23, 2013

### voko

When you are given potential $\Phi$, then the equation for the strength of the field in direction $n$ will have the minus sign by definition: $g_n = - \frac {\partial \Phi} {\partial n}$.

In the case of (Newtonian) gravity, the potential for the field of a spherically symmetric mass is given by $\Phi = - \frac {GM} {r}$, where $r$ is the distance from the center of the mass, and, when the direction is $r$, the strength is given by $g_r = - \frac {\partial \Phi} {\partial r} = - \frac {\partial } {\partial r} \left( - \frac {GM} {r} \right) = - \frac {GM} {r^2}$. Because $r$ is the direction from the center, the minus sign in the resultant equation for the strength indicates that it is against the direction, that is, the force is toward the center. The latter is what we physically know about gravity, and that is why we have the minus sign in the original equation $\Phi = - \frac {GM} {r}$: we want that when we apply mathematics to find the force, we obtain physically sensible results.

Last edited: Jun 23, 2013
5. Jun 23, 2013

### technician

I have never seen g = -GM/r^2 in any text book !! If it is defined like this then the force of gravity on an object should always be written F=-mg.....have you ever seen this, It is unusual.
What exactly do you mean by 'school notes'....is it in a text book?
It looks (to me !!!) like some confusion between the force of gravity and gravitational potential

6. Jun 23, 2013

### D H

Staff Emeritus
That was exactly what I was about to write.

Some texts define down as positive for some problems, and then g is positive. Most introductory physics texts use height above the surface as the dependent variable, making g negative in this context. (Note that it's intro physics texts where this issue of g being a signed scalar arises; you'd be using vectors otherwise.)

Which of upwards or downwards is arbitrarily be positive dictates the sign of g. There's nothing physical going on. It's just a sign convention.

7. Jun 23, 2013

### technician

For a mass 'm' the force of gravity on the mass is 'F' so F = mg.
F and g are bound to be in the same direction (only attraction is possible in gravitation!) so we should write
F = mg ? Or......
-F = -mg ? ...purists may see a subtle physics relevance.
Never seen this referred to in standard text books until the topic of potential is raised.

8. Jun 23, 2013

### D H

Staff Emeritus
You are putting way too much emphasis on this, technician. It's just a sign convention on whether upward vertical displacement is positive or negative: Height versus depth. If height is the dependent variable, then acceleration due to gravity is -9.81 m/s2. Does it make one bit of difference if I use g = 9.81 m/s2, in which case acceleration due to gravity is a=-g, versus using g = -9.81 m/s2[/sup, in which case acceleration due to gravity is a=g?

To me its better to assign coordinates later rather than sooner (in which case g is unsigned 9.81 m/s2 rather than +9.81 m/s2), but that's personal preference. There's nothing wrong with the other approach.

9. Jun 23, 2013

### PeterO

g is a vector, directed towards the Centre of Mass.

r is the displacement out from from that C of M.

If we take the r direction as positive, then the g direction must be negative.

The negative sign is probably just to emphasise that g and r have opposite directions.

Last edited: Jun 23, 2013
10. Jun 23, 2013

### Simon Bridge

Yes you have - all the time.
It's in all that work on vectors.

Both the expressions above should read $\vec{F}=m\vec{a}$
If we choose our coordinate system so that the ground is the x-y plane, and the +z axis points upwards, then you get $\vec{a}=-g\hat{k}$ for a mass m in free fall. That is, an acceleration of magnitude $g$ in the $-\hat{k}$ direction.

Similarly, the force on mass $m_1$ due to mass $m_2$, is properly given as: $$\vec{F}=-\frac{Gm_1m_2}{r_{21}^3}\vec{r}\!_{21}=\frac{Gm_1m_2}{r_{12}^3}\vec{r} \! _{12}$$ ... where $\vec{r}\!_{ab}$ is a vector pointing from the position of $m_a$ to the position of $m_b$ ... notice how the minus sign depends which way the radial vector points?

These expressions are found in good text books ... poor text books may just expect you to join the dots between the work you do on vectors earlier in your education and the gravity stuff they do.

Last edited: Jun 23, 2013
11. Jun 23, 2013

### darkxponent

'G' a vector!. Sure?

12. Jun 23, 2013

### PeterO

Was trying to type g, but instinctively pressed shift at the start of the sentence.

13. Jun 23, 2013

### D H

Staff Emeritus
g certainly doesn't point the same direction in Rio as it does in New York. g is a scalar. Some intro level texts might use g as a signed scalar with a negative value. I'm not thrilled with that, but it's not invalid. Calling g a vector is invalid.

14. Jun 23, 2013

### PeterO

The vector form of Newton's law of gravitation is

\begin{aligned} \vec F &= -\,\frac {GMm}{||\vec r||^3}\vec r \\ &= -\,\frac {GMm}{||\vec r||^2}\frac{\vec r}{||\vec r||} \\ &= -\,\frac {GMm}{||\vec r||^2}\hat r \end{aligned}

The linearization of this for points near the surface of the Earth is
$$\vec F = -mg \frac{\vec r}{||\vec r||} = -mg\hat r$$

So where does F get its vector reference? From the radial vector. Not from g. g is a scalar, and it's best expressed as an unsigned scalar.

Last edited by a moderator: Jun 23, 2013
15. Jun 23, 2013

### darkxponent

Never seen g or G written as Vectors in any textbooks. What I know is both g and G are constants. By definition both g and G are different. The relation OP wrote between G and g is seldom used as g is well defined constant as the acceleration due to earths gravity.

16. Jun 23, 2013

### PeterO

So g is not a vector, yet acceleration is?

17. Jun 23, 2013

### WannabeNewton

I didn't want to post here because many people have already answered the main question with very simple answers. Regardless, $\mathbf{g}$ is certainly a vector, what else did you think it was? Acceleration is a vector by definition. The Newtonian field is given by $\mathbf{g} = -\frac{GM}{r^{2}}\hat{r}$. When you are standing on the surface of the earth, the gravitational field is approximately uniform and takes the form $\mathbf{g} = -9.81\frac{m}{s^{2}}\hat{z}$ where $\hat{z}$ is the upwards vertical direction. More generally, the gravitational field can be defined as the negative gradient of the gravitational potential $\mathbf{g} = -\nabla \Phi$ as voko pointed out; this is obviously a vector field.

18. Jun 23, 2013

### darkxponent

Yes because it always say that acceleration due to gravity is downwards. If g was vector they need not write that acceleration is downwards every now and then!

19. Jun 23, 2013

### WannabeNewton

It is very standard notation to use $\mathbf{g}$ to represent the gravitational field. Just because you yourself have never seen it doesn't mean it isn't standard.

20. Jun 23, 2013

### darkxponent

You are merging two different things here. g and Gravitational field are two very different things. Gravitational field is definitely a Vector. The values of acceleration and Field are same in a Gravition but they are very different quantities. Remember when we solve classical mechanics question, we never take g as a vector there.

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