A minus sign in the gravitational potential

[MathematicalPhysicist]

So we have the Newtonian gravitation potential given by $\phi_M(r)=-GM/r$, and in class the teacher said that the Newtonian force is given by $F_m = -m\nabla \phi_M(r)$.

Now, I was thinking about what was taught in UG or high school, isn't the force should be $F_m = GmM/r^2$, if I plug the above potential I get a minus sign, i.e $-(GmM/r^2)\hat{r}$.

So I asked my current teacher in class about that, it seemed he didn't take notice of this remark.

Am I right?
Also in Wiki we have the potential with a minus sign, I guess someone would have noticed this by now, right? :-)

 

[Orodruin]

The force you quote without the minus is just the magnitude of the force. The direction of the force is towards the gravitating body, ie, towards smaller r. The unit vector $\hat r$ is directed away from the gravitating body and that expression is a vector expression. To point in the right direction, the minus sign is necessary, i.e., $-\hat r$ is a unit vector pointing towards the gravitating body.

 

[Dale]

Now, I was thinking about what was taught in UG or high school, isn't the force should be $F_m = GmM/r^2$, if I plug the above potential I get a minus sign, i.e $-(GmM/r^2)\hat{r}$.

The minus sign is correct. Remember, $\hat{r}$ points away from the center, but gravity points towards the center. So the minus sign is required to ensure that gravity is attractive rather than repulsive

 

[MathematicalPhysicist]

The minus sign is correct. Remember, $\hat{r}$ points away from the center, but gravity points towards the center. So the minus sign is required to ensure that gravity is attractive rather than repulsive

Thanks, makes sense!

 

[MathematicalPhysicist]

BTW, in class the teacher said something about the cosmological constant.

He said something along the lines that it's increasing with time, and as that for astronomical distances much greater than the Earth to sun's distance Newton's gravitational law breaks down, it doesn't work anymore;
Now if that's the case how exactly do you fix it do you still assume such a potential as in my OP, or something with $1/r^\alpha$ where $\alpha>1$, or some other function of the distance?

 

[Orodruin]

He said something along the lines that it's increasing with time, and as that for astronomical distances much greater than the Earth to sun's distance Newton's gravitational law breaks down, it doesn't work anymore;

Depending on how precise you are, Newton’s law of gravitation breaks down on much smaller scales. For example, the perhelion precession of Mercury needs general relativity to get a satisfactory explanation.

That the cosmological constant would increase over time would break the concept of it being a constant. It is difficult to interpret what your professor said without knowing exactly what he said.

 

[MathematicalPhysicist]

Depending on how precise you are, Newton’s law of gravitation breaks down on much smaller scales. For example, the perhelion precession of Mercury needs general relativity to get a satisfactory explanation.

That the cosmological constant would increase over time would break the concept of it being a constant. It is difficult to interpret what your professor said without knowing exactly what he said.

I agree that it would contradict it, but don't we have the oxymoron term: running coupling constants, which I must confess I don't know quite a lot of but it sounds by its name that the constants don't stay constant.

 

[Orodruin]

I agree that it would contradict it, but don't we have the oxymoron term: running coupling constants, which I must confess I don't know quite a lot of but it sounds by its name that the constants don't stay constant.

The Hubble constant is not a constant either (except for a universe dominated by - a cosmological constant). Anyway, that has little to do with the cosmological constant, which indeed is a constant. You could have other forms of dark energy that display a different evolution, but by definition the cosmological constant is constant.

Again, without knowing more of exactly what your professor said, it is impossible to judge if you misunderstood it, if he was "bending the truth", or if he was simply wrong.

 

[Dale]

Now if that's the case how exactly do you fix it do you still assume such a potential as in my OP, or something with 1/rα1/rα1/r^\alpha where α>1α>1\alpha>1, or some other function of the distance?

It winds up being a little more complicated than that. You fix it with General Relativity. In GR there is not always a gravitational potential, in particular the spacetime used to describe the universe on cosmological scales is one of the spacetimes that do not have an associated potential