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In the equation for universal gravitation $(1)$ between two objects, where from is $r$ calculated? From the surface, from the center? Also, are the objects assumed to be particles in this equation or could the be multiple particles like molecules?

$$F=G \frac{m_{1} m_{2}}{r^{2}}\tag{1}$$

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Strictly speaking, this equation for the force due to gravity only holds between point-like objects. In the case of a point-like object, the notion of "distance" between them is simply the distance between their positions.

For extended objects, things get a little more complicated. Really, if you had an extended object (a sphere, cylinder, amorphous blob) you would need to imagine breaking the object into many many tiny pieces of mass, compute the force due to gravity caused by each small piece, and add them all up (as vectors! not all the forces will point in the same direction, see for example tidal force).

The only exception to this is the gravity due to a sphere...in this case we can use the distance from the center of the sphere to compute the force due to gravity. But this is a very special result that comes about from the "breaking apart" that I described above...it just so happens that things work out the same as if we had just used the center of the sphere. For example, see here.

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  • $\begingroup$ Thank you! So am I right that point-particles do not actually exist in that every particle has some spatial extension, but the spatial extension of point particles is so small that we can neglect it? If I am right, how do we calculate gravitation between atoms or even quarks, where we can't neglect the spatial extension? $\endgroup$
    – timtam
    Commented Mar 16, 2021 at 20:12
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    $\begingroup$ @timtam Saying something is a point particle is necessarily an approximation, sure. We do not calculate the gravitation between atoms and such things. Such a calculation would be poorly defined at best since, though these objects are often depicted as point particles, the truth is more complicated. In the case of such small objects, however, the gravitational forces between them are, in essentially all cases, entirely negligible. $\endgroup$
    – Richard Myers
    Commented Mar 16, 2021 at 22:22
  • $\begingroup$ @timtam In any case, my answer was entirely about classical Newtonian gravity which makes no comment about atoms or anything of the sort. $\endgroup$
    – Richard Myers
    Commented Mar 16, 2021 at 22:22
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The object in this law assumed to be points (so the distance is just distance between points). Any two object that are very small compare to the distance between them (no matter from which points on the objects you measure it) can be considered as points. So, for example if you have a two rigid molecule that are small compare to the distance between them, then you can use this equation to calculate the force between them.

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  • $\begingroup$ Thanks, so otherwise we cannot use this equation? What is "small" compared to the distance? Would $10^{-9}$ be small? $\endgroup$
    – timtam
    Commented Mar 16, 2021 at 20:24
  • $\begingroup$ It depends how accurate you want to be. In reality, 10^(-9)*(the object distance) will be always be small enough. Please note that you can still find the accurate force, using this equation multiple time for each particles and adding together the forces (using vector addition) $\endgroup$
    – ziv
    Commented Mar 16, 2021 at 20:29
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The distance between two objects is the distance between centre of gravity of this two objects.

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For the two-body problem, assuming the bodies have respective masses $m_0$ and $m_1$, with respective positions $𝐫_0$ and $𝐫_1$, and assuming there are no forces at play other than their mutual gravity, the equations of motion are: $$\begin{align} 𝐩_0 &= m_0\frac{d𝐫_0}{dt}, &\quad \frac{d𝐩_0}{dt} &= \frac{Gm_0m_1}{|𝐫_1 - 𝐫_0|^2}\frac{𝐫_1 - 𝐫_0}{|𝐫_1 - 𝐫_0|},\\ 𝐩_1 &= m_1\frac{d𝐫_1}{dt}, &\quad \frac{d𝐩_1}{dt} &= \frac{Gm_1m_0}{|𝐫_0 - 𝐫_1|^2}\frac{𝐫_0 - 𝐫_1}{|𝐫_0 - 𝐫_1|}, \end{align}$$ where $𝐩_0$ and $𝐩_1$ are their respective momenta.

Under the conversion: $$\begin{align} 𝐏 &= 𝐩_0 + 𝐩_1, &\quad M &= m_0 + m_1, &\quad 𝐑 &= \frac{m_0𝐫_0 + m_1𝐫_1}{m_0 + m_1},\\ 𝐩 &= \frac{m_0𝐩_1 - m_1𝐩_0}{m_0 + m_1}, &\quad m &= \frac{m_0m_1}{m_0 + m_1}, &\quad 𝐫 &= 𝐫_1 - 𝐫_0, \end{align}$$ the system reduces equivalently to: $$\begin{align} 𝐏 &= M\frac{d𝐑}{dt}, &\quad \frac{d𝐏}{dt} &= 𝟬, \\ 𝐩 &= m\frac{d𝐫}{dt}, &\quad \frac{d𝐩}{dt} &= -\frac{GmM}{|𝐫|^2}\frac{𝐫}{|𝐫|}. \end{align}$$

It's the full distance $𝐫 = 𝐫_1 - 𝐫_0$, but with the reduced mass $m = m_0m_1/(m_0 + m_1)$ for the orbiting body, and the total mass $M = m_0 + m_1$ for the body being orbited.

As a set of second-order equations, this can also be rewritten as: $$\frac{d^2𝐑}{dt^2} = 𝟬, \quad \frac{d^2𝐫}{dt^2} = -\frac{GM}{|𝐫|^2}\frac{𝐫}{|𝐫|},$$ which underscores that it is with the full distance $𝐫$ for the orbiting body and the total mass $M$ for the body being orbited. When written in this form, the reduced mass $m$ drops out of the picture.

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