What is the distance R between centers of mass in a barbell-shaped object?

[Maurogaetano]

Homework Statement

Not a homework, just a curious chap
I think I learned that the distance R in the force of gravity equation is the distance between the two "center of mass" of the masses. May not be true but this is the assumption I have to work. So two balls of equal mass (m) separated by a distance 2x should be attracted by a distant body of mass M with the same force regardless of the relative orientation. In other word a bar bell body should weight the same regardless of relative orientation.

Relevant Equations

F = k2mM/R^2

Normalizing respect m, M, R and k then the normalized force for the case of the bar bell body aligned with the direction of R become Fn = 1/(1+x)^2 + (1-x)^2 is a vectorial sum but in this case the vectors are aligned
We can assume x<<1 but is not really material to the problem

For x = 0 the force is 2 as it should be if the ball were to fuse into one

No amount of algebra manipulation will make Fn invariant from x which is the crux of my question

What is the distance R? Is the distance between centers of mass of each body? Or something else?

 

[BvU]

Hello @Maurogaetano, !

I think I learned that the distance R in the force of gravity equation is the distance between the two "center of mass" of the masses.

Thinking is always good. Do you also remember under what assumptions this claim was derived ?

 

[Maurogaetano]

You answered while I was typing Thanks
Do not remember exactly, but there was some mention of "spherical symmetry".

Now a bar bell O======O does not seem to have spherical symmetry but anything missing to a full sphere could (not really sure) be seen as a collection of bar bell each behaving the same to the effect of the problem. But I am beginning to see that a single bar bell inclined respect the axis presents the two balls at slightly different angles so the vectors have a lateral component beside the straight pull and can not be simply added like scalars.

 

[PeroK]

Do not remember exactly, but there was some mention of "spherical symmetry".

The $R$ in that equation is really the distance between point masses (or masses whose size is negligible compared to the distance between them). It does, however, also work when the objects are spheres: this is known as Newton's Shell Theorem. That's why $R$ is the radius of the Earth when calculating the surface gravity.

In addition, close to the Earth the force per unit mass is approximately constant, so the shape of an object and its orientation do not matter.

 

[etotheipi]

As @PeroK says the result holds exactly for point masses and uniform spheres (use Gauss' Law!), but there is another useful principle that I will just quote from Classical Mechanics, D. Gregory:

Let $\mathcal{S}$ be any bounded system of masses with total mass $M$. Then the force $\mathbf{F}$ exerted by $\mathcal{S}$ on a particle $P$, of mass $m$ and position vector $\mathbf{r}$, has the asymptotic form$$\mathbf{F} \sim - \frac{mMG}{r^2} \hat{\mathbf{r}}$$as $r \rightarrow \infty$

where the origin of coordinates $\mathcal{O}$ is presumably at the centre of mass of the system $\mathcal{S}$.

 

[Maurogaetano]

Got it. At least one of the masses must be a point mass and then only when considering the force exercised on it by a collection of dispersed masses. Then the collection of dispersed masses can be lumped around the center of mass and only for great distances.
Thanks

 

[etotheipi]

Yeah, I think what you said is right. If you want to find the attractive force between two extended bodies that are not both spherical and uniform, then you might first consider the two bodies as systems of particles $\mathcal{S}$ and $\mathcal{S}'$. The field due to $\mathcal{S}$ is then going to be$$\mathbf{g}(\mathbf{r}) = \sum_{i \in \mathcal{S}} -\frac{Gm_i (\mathbf{r} - \mathbf{r}_i)}{|\mathbf{r} - \mathbf{r}_i|^3} $$Then the force acting on any particle in $\mathcal{S}'$ due to only the particles in $\mathcal{S}$ $$\mathbf{F}_j = m_j \mathbf{g}(\mathbf{r}_j)$$Then you take continuum limit $\sum m_i \rightarrow \int d^3 \mathbf{r} \rho(\mathbf{r})$.

 

[haruspex]

Yeah, I think what you said is right. If you want to find the attractive force between two extended bodies that are not both spherical and uniform, then you might first consider the two bodies as systems of particles $\mathcal{S}$ and $\mathcal{S}'$. The field due to $\mathcal{S}$ is then going to be$$\mathbf{g}(\mathbf{r}) = \sum_{i \in \mathcal{S}} -\frac{Gm_i (\mathbf{r} - \mathbf{r}_i)}{|\mathbf{r} - \mathbf{r}_i|^3} $$Then the force acting on any particle in $\mathcal{S}'$ due to only the particles in $\mathcal{S}$ $$\mathbf{F}_j = m_j \mathbf{g}(\mathbf{r}_j)$$Then you take continuum limit $\sum m_i \rightarrow \int d^3 \mathbf{r} \rho(\mathbf{r})$.

If either one is spherically symmetric (doesn't have to be uniform radially) then that one can be reduced to a point, no?
@Maurogaetano, your question leads into another interesting aspect, that of tidal forces heating a satellite and tidal locking.

 

[etotheipi]

If either one is spherically symmetric (doesn't have to be uniform radially) then that one can be reduced to a point, no?

Yeah sorry, you're right, I meant to say each concentric shell is uniform, i.e. $\rho = \rho(r)$.

 

[Maurogaetano]

your question leads into another interesting aspect, that of tidal forces heating a satellite and tidal locking.

That is where it started, by observing some satellite structures. A barbell shaped object in space will orient radially to a massive body.