Is 'g' Negative? Understanding the Confusion of Definitions

[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!

 

[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.

 

[Simon Bridge]

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!

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.

 

[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.

 

[technician]

I have never seen g = -GM/r^2 in any textbook ! 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 textbook?
It looks (to me !) like some confusion between the force of gravity and gravitational potential

 

[D H]

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.

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.

 

[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 textbooks until the topic of potential is raised.

 

[D H]

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/s<sup>2[/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.</sup>

 

[PeterO]

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!

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.

 

[Simon Bridge]

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 textbooks until the topic of potential is raised.

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 textbooks ... poor textbooks 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.

 

[darkxponent]

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

.

'G' a vector!. Sure?

 

[PeterO]

'G' a vector!. Sure?

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

 

[D H]

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.

 

[PeterO]

hen we have the weight Force F = mg, The Force, like all Forces, is a vector, but mass is a scalar. Where else does F get its vector reference, but from g?

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.

 

[darkxponent]

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

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 Earth's gravity.

 

[PeterO]

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 Earth's gravity.

So g is not a vector, yet acceleration is?

 

[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.

 

[darkxponent]

So g is not a vector, yet acceleration is?

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!

 

[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.

 

[darkxponent]

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.

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.

 

[Simon Bridge]

JIC: I edited my post #10 above to be clearer ... while I was doing that there were a lot of extra answers.

I think we all agree that the sign in the expression for the gravitational force is for direction and depends on the sign-convention for directions that are chosen and must reflect that gravity is attractive.

 

[Simon Bridge]

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.

But if it is a standard notation then it should be written down someplace - perhaps in some standard reference... and you will be able to provide a link to this?

I think this is a side issue to OPs question.
OPs confusion likely arises because of the common practice of working in 1D and using a sign convention to specify direction while all other variables are magnitudes.

 

[D H]

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.

That 9.81 m/s² (more precisely, 9.80665 m/s²) has a name. It's g (more precisely, it's g₀, but hardly anyone uses that subscript 0.). It's an SI defined constant. I personally prefer $\vec a_g$ as opposed to $\vec g$ to denote acceleration due to gravity to avoid confusion with that scalar.

 

[WannabeNewton]

http://en.wikipedia.org/wiki/Gravitational_field
http://astrowww.phys.uvic.ca/~tatum/celmechs/celm5.pdf
http://ocw.mit.edu/courses/earth-at...of-geophysics-fall-2004/lecture-notes/ch2.pdf
there's a start.

 

[WannabeNewton]

That 9.81 m/s² (more precisely, 9.80665 m/s²) has a name. It's g (more precisely, it's g₀, but hardly anyone uses that subscript 0.). It's an SI defined constant. I personally prefer $\vec a_g$ as opposed to $\vec g$ to denote acceleration due to gravity to avoid confusion with that scalar.

Sure but I was talking about technician's opposition to the statement that g = GM/r^2 is never used. The MIT OCW link above for example serves as a counter example.

 

[Simon Bridge]

http://en.wikipedia.org/wiki/Gravitational_field
http://astrowww.phys.uvic.ca/~tatum/celmechs/celm5.pdf
http://ocw.mit.edu/courses/earth-at...of-geophysics-fall-2004/lecture-notes/ch2.pdf
there's a start.

That's better :)
Otherwise it is just you vs him getting nowhere.
The next bit is to re-emphasize why this helps OP with the concepts, and you'll move forward (hopefully).
But there's water under the bridge since then - let's see what OP has to say?

 

[darkxponent]

Sure but I was talking about technician's opposition to the statement that g = GM/r^2 is never used. The MIT OCW link above for example serves as a counter example.

I was just going though a book by Bernstien, Fishbane. I think i just found the answer to the discussion.

They distinguish the Local gravity and Gravition. I would agree with with technician here that g= GM/r^2 is never used. The g technician is talking about is the local gravity constant which definitely is sclalar constant(a well defined one).

And the g you are talking about is gravitation field, which is vector of course.

The conventiin you are using to define g as gravitation field might be common in some texts but is not universally common. I haven't seen any of books.

 

[najm]

Big answers. By the way if you take a particular vector direction as positive then the opposite direction will be negetive. Here we can take downward direction accelaration due to gravity "g" as positive. But many a times it is just taken as negetive in order to coincide with the cartesian co ordinate system in which the vertically down direction is -y a negetive.

 

[voko]

In the end this all boils down to what is denoted by 'g'. There can be subtle differences, which may or may not be indicated by the type face or some other symbolism. It is not uncommon in physics and mathematics to denote closely related concepts by identical letters and varied shapes, accents, subscripts and superscripts. Which does get confusing at times, and texts are not always consistent.

If 'g' is the magnitude of the Earth's standard sea-level acceleration of gravity, then it is a positive scalar by definition. If it is anything else, and it can be, its definition must be carefully consulted.