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Newton’s Law of gravitation states that force of attraction between two point masses is proportional to the product of the masses and inversely proportional to the square of the distance between them. I know that the force of attraction between two spheres turns out to be of the same mathematical form as a consequence of Newton’s law. But I am not able to prove how the force between any two rigid masses is only proportional to the product of their masses (as my teacher says) and the rest depends upon the spatial distribution of the mass. So $F$ is ONLY proportional to $Mmf(r)$ where $f(r)$ maybe be some function based on the specifics of the situation.

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    $\begingroup$ Interesting question. I don't know if this will satisfy you, but the force must be proportional the product of the masses, otherwise the units will be wrong. To get Newtons from a combination fo $G$, the two masses, and the distance between them you have to multiply the first power of the masses. $\endgroup$
    – garyp
    May 8, 2020 at 16:39
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    $\begingroup$ Does this answer your question? Why in Newton's law of gravity, we do $M_1 \times M_2$ and not $M_1 + M_2$? $\endgroup$ May 8, 2020 at 18:43
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    $\begingroup$ It seems that a lot of people are not reading the question. $\endgroup$
    – garyp
    May 8, 2020 at 20:33
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    $\begingroup$ @Aditya Sharma To clarify: Do you want to know how come that if I have two objects and replace one of them with an identical one that has twice the mass (so twice the density),the attraction will be doubled, or do you want to know if replacing one of the two objects with another one having the same mass but another shape or shape of mass distribution, will create an identical force? $\endgroup$ May 8, 2020 at 21:02
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    $\begingroup$ I resonate with the second case. I have edited the question $\endgroup$ May 9, 2020 at 7:46

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The statement

...the force between Any two rigid masses is only proportional to the product of their masses

is not true in general, or at least it is misleading. The shapes of the mass distributions and their relative positions matter when computing the gravitational force.

It is true that once you hold constant the shapes of the mass distributions and their relative positions, then the force will be proportional to the product of the total masses of the bodies.

There are certain situations where treating two extended massive bodies as point sources can be exactly correct (in the context of Newtonian gravity). For a spherically symmetric mass distribution, the gravitational potential outside of it is the same as that arising from a point source of the same mass. This is an application of Gauss' law.

In general, one can build up an increasingly good approximation of the gravitational potential arising from a given mass distribution via a multi-pole expansion. . The leading-order term, which drops off least rapidly with distance (force $\propto r^{-2}$), is that of a monopole like what arises for a point mass or outside a spherically symmetric system. But a general mass distribution will have contributions from higher-order terms (dipole, quadropole, octopole...), all of which drop off increasingly rapidly with distance. As one considers two bodies at increasing separation, reducing them both to their monopole terms becomes increasingly more accurate.

Finally, the fact that the gravitational force of attraction on an extended body due to another body can vary with position is essential when considering phenomena such as tidal forces.

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I edited the answer to make it more readable.

It is true in general that the force between two separate and rigid bodies is equal to $ G M_1 M_2 f$ with $f$ depending on the details of the mass distributions. The main reason because this is true is that the mass is an extensive quantity. However, it is also true that the force is not always proportional to the inverse of the square of the distance. This means that in general, changing the mass distribution will change the mutual force, especially in the case where the two bodies are very close. If we also assume that the distance between the two bodies is much larger than their individual sizes, then one can also recover the inverse square law, because in this case one has that $F\approx G M_1 M_2/R^2$. In the following I demonstrate that

1) $F= G M_1 M_2 f$ with $f$ not depending on the masses but only on the details of the distributions

2) if the distance between the bodies is much larger than the individual sizes, then $F\approx G M_1 M_2/R^2$.

3) I show a counterexample where the mutual force is not $\propto 1/R^2$ but still proportional to the masses.

1) $F= G M_1 M_2 f$

For point masses, one has $$ \mathbf{F}= G M_1 M_2 \frac{\mathbf r_1- \mathbf r_2}{|\mathbf r_1- \mathbf r_2|^3} $$ So, assume that the mass distributions of the two bodies are rigid and separated in space. The total force between the two bodies can be written as an integral over the two mass distributions $$ \mathbf F_{12}= \int d r_1^3 d r_2^3 G \rho_1(\mathbf r_1) \rho_2(\mathbf r_2) \frac{\mathbf r_1- \mathbf r_2}{|\mathbf r_1- \mathbf r_2|^3} $$ The mass distributions satisfy $$ \int d r^3 \rho_{1,2}(\mathbf r) =M_{1,2} $$ In order to get rid of the mass dependences, one can factorize the masses in the density distributions, by defining $$ \rho_{1,2}(\mathbf r) = M_{1,2} \psi_{1,2}(\mathbf r) $$ The crucial point here is that the functions $\psi_1(\mathbf r)$ and $\psi_2(\mathbf r)$ do not depend on the masses $M_1$ and $M_2$, but only on the geometrical distribution of their densities. They satisfy the property $$ \int d r^3 \psi_{1,2}(\mathbf r) =1 $$ One can imagine them as ''normalized'' and adimensional density distributions. So, why these functions do not depend on the mass? Imagine a rigid body or a mass distribution in space with mass $M$ and volume V. Imagine that this body is made of, for example, wood. Then imagine one could change every infinitesimal portion of this body with, for example, iron. The final result is a body with the same ''normalized'' mass distribution $\psi$ but with a mass much greater than the initial one $M'\gg M$. Then, the function $\psi$ do not depend on the mass but only on the geometrical distribution. The same argument can be applied for rigid bodies but also for more complicated mass distributions, for example nonuniform or deformable objects, fluids, etc.

Using the functions $\psi$ one has $$ \mathbf F_{12}= G M_1 M_2 \int d r_1^3 d r_2^3 \psi_1(\mathbf r_1) \psi_2(\mathbf r_2) \frac{\mathbf r_1- \mathbf r_2}{|\mathbf r_1- \mathbf r_2|^3}=$$$$= G M_1 M_2 \cdot \mathbf f_{12}(\psi_1,\psi_2) $$ where the function $$ \mathbf f_{12}(\psi_1,\psi_2)= \int d r_1^3 d r_2^3 \psi_1(\mathbf r_1) \psi_2(\mathbf r_2) \frac{\mathbf r_1- \mathbf r_2}{|\mathbf r_1- \mathbf r_2|^3} $$ depends only in the geometrical details of the distributions and not on the masses $M_!$ and $M_2$.

2) $F\approx G M_1 M_2/R^2$ at large distances

One can go further and redefine the spatial coordinates with respect to the center of mass of the two bodies 1 and 2. That means that $\mathbf r_1={\mathbf r}_1'+\mathbf R_{1}$ and $\mathbf r_2={\mathbf r}_2'+\mathbf R_{2}$ where $\mathbf R_{1}$ and $\mathbf R_{2}$ are the positions of the center of mass of the two bodies. If $ \mathbf R_{12} = \mathbf R_{1}-\mathbf R_{2}$ is their distance, then one has

$$ \mathbf f_{12}(\psi_1,\psi_2)= \int d {r}_1^{\prime 3} d {r}_2^{\prime 3} \psi_1({\mathbf r}_1') \psi_2({\mathbf r}_2') \frac{{\mathbf r}_1'- {\mathbf r}_2'+\mathbf R_{12}}{|{\mathbf r}_1'- {\mathbf r}_2' +\mathbf R_{12}|^3} $$ depends on the distance of the two masses and on the geometrical details of their distributions, but not on the value of the masses $M_1$ and $M_2$.

The dependence on the distance of the center of masses does not necessarily goes as the inverse of the square of the distance. However, if the distance between the two bodies is much larger than the sizes of the two bodies, one has that $|\mathbf R_{12}|\gg |\mathbf r_1- \mathbf r_2| $ then one has, as a first approximation $$ \mathbf f_{12}(\mathbf R_{12}) \approx \int d {r_1}^{\prime 3} d {r}_2^{\prime 3} \psi_1({\mathbf r}_1') \psi_2({\mathbf r}_2') \frac{\mathbf R_{12}}{|\mathbf R_{12}|^3} =\frac{\mathbf R_{12}}{|\mathbf R_{12}|^3} $$ The last equality follows from the fact that the functions $\psi$ are normalized to 1, which gives $$ \int d {r_1}^{\prime 3} d {r}_2^{\prime 3} \psi_1({\mathbf r}_1') \psi_2({\mathbf r}_2')= \int d {r}^{\prime 3} \psi_1({\mathbf r}') \times \int d {r}^{\prime 3} \psi_2({\mathbf r}') =1$$ Therefore in this case one has $$ |\mathbf F_{12}|\approx \frac{G M_1 M_2}{|\mathbf R_{12}|^2} $$

2) $F\neq G M_1 M_2/R^2$ but still $F= G M_1 M_2 f$

The obvious violation to the case 2) is where the distance $R$ between the center of masses of the bodies is not large. This condition is obviously violated if the bodies are one inside the other. This is realized, for example, in the famous case where one has a spherical shell containing a sphere. In this case (see wikipedia) all contributions to the mutual gravitational force cancel each other, and one has $$ \mathbf f_{12}(\psi_1,\psi_2)= \int d r_1^3 d r_2^3 \psi_1(\mathbf r_1) \psi_2(\mathbf r_2) \frac{\mathbf r_1- \mathbf r_2}{|\mathbf r_1- \mathbf r_2|^3}=0 $$ This still satisfy $F=G M_1 M_2 f$ with $f=0$.

Now consider the case of a sphere and imagine to dig a tunnel from one pole to the other and consider the gravitational force of an object inside the tunnel. In this case one has that $ \mathbf f_{12}(\psi_1,\psi_2)=1/R$ which gives $$ {F}_{12}= G M_1 M_2 \frac{1}{|\mathbf r_1- \mathbf r_2|} =\frac{G M_1 M_2}{R_{12}} $$ The dependence on the distance is now $\propto1/R$ but the force is still proportional to $M_1 M_2$ and satisfy the equation derived in 1).

Example of mass distribution

To make things more intuitive, consider the easiest example of mass distribution: A rigid sphere of radius $R$. In this case one has $$ \psi(\mathbf r)= \begin{cases} 1 & \text{for} |\mathbf r|<R\\ 0 & \text{for} |\mathbf r|>R \end{cases} $$

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    $\begingroup$ I didn't go through this in detail. But what if the mass distributions are the same, varying only in mass? It's still not clear to me what exactly the OP is asking. For example, I think that he is not asking about distance dependence. And although he mentions spatial distribution it's not clear to me whether or not he wants an answer for fixed distributions. Anyway, if you can summarize in words what your analysis yields for fixed distributions you might nail what the OP is after. $\endgroup$
    – garyp
    May 27, 2020 at 17:07
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    $\begingroup$ I think you still need to prove that the $\psi_i$ do not depend on the masses. How can I see that? Also you should probably write $\rho_i(r,M_i)$, as they must depend on the mass such that the $\psi$ doesn't. $\endgroup$
    – Azzinoth
    May 27, 2020 at 17:47
  • $\begingroup$ @garyp about your first questions, what do you mean if the mass distributions are the same? Do you mean the case, for example, of two spheres at a finite distance, or the case where the two mass distributions coexist in the same volume? About your last sentence, what do you mean with fixed distributions? $\endgroup$
    – sintetico
    May 28, 2020 at 8:43
  • $\begingroup$ @ Azzinoth Actually, the mass is defined as the integral of the density distribution. If I divide the density distribution by the mass, I obtain a normalized quantity which is adimensional and does not depend on the total mass. $\endgroup$
    – sintetico
    May 28, 2020 at 9:12
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    $\begingroup$ I wonder if there is a way to convert the function $\mathbf f_{12}(\psi_1,\psi_2) = \int \ldots$ into a PDE? $\endgroup$ May 30, 2020 at 17:47
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How does this sound?

Let's say you have two extended bodies, A and B, each made up of a number of particles. Let's consider the force on a particle in A, call it P, due to body B. Each particle in body B exerts a force on P that is proportional to that B particle's mass. The sum of such forces gives the net force on P due to B. Now let's suppose we double the mass of each B particle. That would result in doubling the force on P. But doubling the mass of each B particle is doubling the total mass of body B. So doubling the total mass of body B doubles the force on P. Hence, the force due to body B is proportional to B's total mass.

To get the total force due to body B on the body A, we have to add up the forces on all of A's particles. Again, if we doubled each particle's mass, the total force on A would double, as would its total mass. Hence that total force is proportional to A's total mass.

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  • $\begingroup$ Yes, you can combine it with the symmetry and dimensional argument given by @Dvij D.C. to prove that it is only proportional to the product of the masses. And the rest shall depend upon the specifics of the situation. $\endgroup$ May 10, 2020 at 5:26
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    $\begingroup$ I was starting to have doubts about my own answer - suppose the masses of the B particles are not all doubled but increased randomly so that the total B mass doubles. Then the force on the A particle will not double as B particles are at different distances from it. But then that's a different body B with a different f(r), to use your notation. I suppose that for a given f(r) then it's true the force of attraction between A and B is proportional to the product of their masses. But what about @Dvij D.C.'s example of a particle in a spherical shell? $\endgroup$ May 10, 2020 at 13:42
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As is stated in the previous answer huge non-spherical objects are rarely (never) seen in Nature so let's consider this as a purely theoretical enterprise. Let's go!
Consider two bodies of mass (let's assume they are rigid so they can't be deformed and tidal effects omitted). The gravitational force vector a massive body exerts on another body originates in its center of gravity (CG). Also, the point where it gets a grip on the other body is the other body's CG. For most celestial bodies (which are spherical symmetric) as well as for a number of other bodies where symmetry is involved (think of an ellipsoid), the center of mass (CM) and CG coincide, but in general, this is not the case. In fact, there is for these bodies not one CG, but they lie on a line piece on which the CM is situated. Where the CG is situated depends on where the other body is to be found. For example on a cube Earth:

enter image description here

The direction of the gravity you feel when walking on it doesn't point most of the time towards the CM. In the article is written:

“… Gravity on the surface wouldn’t generally point toward the exact center of the [cube] Earth anymore.”

And the center of this cube Earth is where the CM is situated.

The same holds for a barbel formed mass. Although some symmetry is involved the CG lies on a line piece on the principal axis of rotation with the smallest moment of inertia (in the barbel's length) going through the CM. The endpoints of the line piece lie a bit before the center of the two balls, as seen from the CM, and on the same distance from it (unless the balls are different in size, in which case the CM shifts from the middle and doesn't lay in the middle of the line piece anymore).
And also in this case the position of the CG depends on the position of the other body.

Now, what does this all mean? It means that if two rotating bodies of mass (with masses $M$ and $m$ gravitate towards each other in a straight line, the forces of gravity lay on the line connecting the two CGs that lay on one of the endpoints of the line of CGs (dependent on their relative orientations). This means they make a torque (a pseudovector produced by a cross product) come into existence on both bodies:

$$\vec{{\tau}_{1net}}=\vec{F_{gnet}}\times \frac{1}{2}(\overrightarrow{CM_1}-\overrightarrow{CG_{1max}})=\lVert{\vec{F_{gnet}}}\rVert \lVert{\frac{1}{2}(\overrightarrow{CM_1}-\overrightarrow{CG_{1max}}})\rVert \sin{{\theta}_1}$$ $$\vec{{\tau}_{2net}}=\vec{F_{gnet}}\times \frac{1}{2}(\overrightarrow{CM_2}-\overrightarrow{CG_{2max}})=\lVert{\vec{F_{gnet}}}\rVert \lVert{\frac{1}{2}(\overrightarrow{CM_2}-\overrightarrow{CG_{2max}}})\rVert \sin{{\theta}_2}$$

The factor $\frac{1}{2}$ appears in front of the vectors because their magnitudes vary between zero (when the CMs and CGs coincide) and the maximum values (when the CMs and CGs are the most far apart). This is the case on both sides of the CM, but on one side the $\vec{F_g} s$ have bigger magnitudes, which is why I write $\vec{\tau_{net}}$ and $F_{gnet}$, which I will call just $\vec{\tau}$ and $\vec{F_g}$ in what follows.
When the bodies have an initial minimum angular rotation the bodies make a whole rotation, while they rotate back and forth when they rotate below this minimum angular rotation, and no rotation at all results when the initial angular momentum is zero and the line piece between the two CGs is perpendicular to the line between the two CMs of the bodies.
The angles ${\theta}_1$ and ${\theta}_2$ are the angles between the concerning vectors. Their maximum values increase when the bodies get closer. The torque vectors are perpendicular to the plane containing $\vec{F_g}$ and $(\overrightarrow{CM_1}-\overrightarrow{CG_{1max}})$ or $(\overrightarrow{CM_2}-\overrightarrow{CG_{2max}})$ (if these are parallel, no torques are present, because in that case $\sin{\theta}_1$ and $\sin{\theta}_2$ are zero) and rotate around the axis connecting ${CM}_1$ and ${CM}_2$.
Now a torque makes a body rotate in the plane perpendicular (or rotating back and forth as we saw) to the plane just mentioned and are also given by:

$$\vec{{\tau}}_M=I_M\frac{\vec{{d\omega}}_M}{dt}$$ $$\vec{{\tau}}_m=I_m\frac{\vec{{d\omega}}_m}{dt},$$

where $I_M$ is the moment of inertia of the body of mass $M$ and $I_m$ the moment of inertia of the body of mass $m$ (different momenta of inertia $I$, depending on the form of the mass, can be calculated or looked up) and $\frac{\vec{d\omega}}{dt}$ the time derivative of the angular velocity. It might be clear that the time derivative of the angular velocity (pseudo)vector ($\vec{\omega}=\vec{v}\times \vec{l}$) lies on the same line as the torque vector because $I$ is a scalar, i.e. a positive number. Let's assume the torques makes the bodies rotate only around the principal axes with the highest moment of inertia (the rotation can be around any axis, but the principle is the same).
When the bodies initially don't rotate, and the CMs and the CGs of both bodies coincide or all lie on one line (which is the case if the lines of CGS are parallel or orthogonal, so no torque is present) then the bodies just accelerate linearly to each other with a force $F=G\frac{Mm}{r^2}$, where $r$ is the distance between both CMs (or CGs).
If this is not the case the bodies make each other rotate. After each full rotation of each body, they have the same angular velocity.

So we have the mutual linear acceleration caused by the force $F=G\frac{Mm}{r^2}$, in which $F$ is the force component of the force pulling on the CMs and $r$ the distance between the two CMs. This linear acceleration is minimal when the total rotational acceleration is maximal, and vice-versa.

This linear acceleration is attenuated periodically by the variable (but periodic) angular rotation of both bodies. The linear acceleration grows, diminishes, grows, diminishes, etc. (because the rotational energy of the bodies varies periodically).
When the bodies are very far apart there will (approximately) only be linear acceleration, because the torques go to zero. But the linear acceleration also goes to zero when they are far apart so $f(r)$ reduces to one, so you can reduce the force formula to $G\frac{Mm}{r^2}$, with $M$ and $m$ considered as point particles. The torques and linear acceleration will not have the same ratio when the distance between the bodies increases. The torques of the bodies (making their rotation vary between a maximum and minimum value) are just as the force $F_g$ (producing the linear acceleration) dependent on a squared distance ($F\propto{\frac{1}{{r_F}^2}}$ and $\tau\propto{\frac{r_{\tau}^2}{r_F}^2}$) so initially, the linear acceleration is very small, just as the torques. The angles involved in the torques go to zero though when the bodies approach infinity and so does $\sin{\theta}$ for both bodies.
So the attenuating function $f(r)\rightarrow 1$ when $r\rightarrow \infty$ so $F_g$ approaches the $G\frac{Mm}{r^2}$ form. When the bodies approach each other the ratio of the torque of the bodies and linear accelerating force is not equal at every distance between the bodies (see the previous alinea). So $f(r)$ is a periodic function (depending on the initial rotations of the bodies, the momenta of inertia of them both, and the varying torques) and it gives the linear acceleration a periodic component. This periodic variation is small when the linear acceleration is small (when they are far apart) and gets bigger when the linear acceleration increases (when they are getting closer). But because the ratio of the torque and F grows when the distance between $CM_1$ and $CM_2$ diminishes (e.g. when the distance gets half as big, the force becomes $\frac{1}{4}$ big, while the torque becomes more than $\frac{1}{4}$ as big, because in the two cross products definitions I gave above, a factor $\sin{\theta}$ is involved, which grows when the distance of the bodies gets smaller), the period of rotation in time get. Nevertheless, $f(r)$ still attenuates the linear acceleration periodically.
Given the necessary information, $f(r)$ can be calculated. Of course, we also have to include the stretching of the bodies because they are not truly rigid. This stretching is due to the rotations and the tidal forces (the last grows when the distance gets smaller), which can be calculated too. When the bodies stretch potential energy is given to the bodies and this diminishes the linear acceleration, but this effect I neglected (though it makes a small contribution to $f(r)$ and also approaches one when the distance approaches infinity) which is why I assumed them to be rigid.

In most cases, there is no constant periodicity (i.e. after a certain number of rotations the initial relative position of the bodies arises again) which is the case when the ratio of the momenta of inertia is a non-rational (real) number, but nevertheless a periodicity is present.

Pffff....I think this is more than enough.

Just one more thing. I just realized that the CGs don't have to lay on a straight line piece (this is only the case when symmetry is involved), but in general, they lay on a curved line piece. As a first approximation, it will do though (like a first approximation in a multipole expansion).

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  • $\begingroup$ In your example, the mutual force is still proportional to $m M$, although the dependence on the distance is not $\propto 1/r^2$. This is obviously a consequence of the fact that the two objects are not separate in space by a large distance. In fact, they are one inside the other. It is not a consequence of the non spherical shape. $\endgroup$
    – sintetico
    May 28, 2020 at 11:18
  • $\begingroup$ No matter how far apart, because the center of gravity and the center of mass don't coincide, the gravitational force that accelerates the two masses linearly (the force acting on the center of mass) varies, no matter how far they are apart. So $mM$ is preceded by a factor $f(r)$. What do you mean when you write they are one inside the other? $\endgroup$ May 28, 2020 at 11:43
  • $\begingroup$ I think I was misled by the picture :D Ok, forget my last comment. In any case, in your example, if I understand it now correctly, you have $F=G M m f$ where $f$ depends on the distance. If the body are far apart, one has that $F\approx G M m/R^2$ asymptotically for $R\to\infty$. $\endgroup$
    – sintetico
    May 28, 2020 at 11:57
  • $\begingroup$ Ah, you mean the picture of the cube Earth. I understand ;D But suppose we have, for example, two equal dumbbells very far apart ($R\rightarrow \infty$). $F$ goes to zero (so also the force which gives rise to the linear acceleration), but also the torque. The torque though is dependent on the angle between the force taking hold on the CG and the line CM-CM, which is zero when both dumbbells are at an infinite distance to each other. So what you say is indeed true. The function f(r) goes asymptotically to zero when infinity is approached. $\endgroup$ May 28, 2020 at 13:11
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    $\begingroup$ Sorry for reacting a bit late! Anyhow, yes. When $R\rightarrow \infty$ then $f(r)\rightarrow 1$ so $F\cdot{R^2}\rightarrow GMm$. I realized just now though that the lines on which the $CG$s lie are in general not straight lines. But as a first approximation, it will do. $\endgroup$ May 29, 2020 at 12:26
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This is not true in the general case if the two masses are close to each other. That is, if you have 2 different objects of the same mass but with different shapes, the gravitational attraction between these 2 objects and a third mass that is close enough to the objects will depend on the specifics of the situation. For example, imagine a large dumbbell. In the middle between the two big masses at the end, the gravitational force of these masses is equal and opposite, so with a zero resultant. Only the small bar in between the two "counts". Compare that to a large plate of the exact same mass. In this case, all the mass "counts". Of course, if the objects have the same shape, but one has a mass double the other, then the forces will be doubled.

If the objects between which you calculate attraction are far enough from each other however, this is a good approximation, irrespective of shape. To get the attraction you integrate the attraction between each small element of object 1 and each element of object 2. If you are not at the calculus level, you can say that you divide both objects in small pieces and add the contribution of attraction between all the pieces with each other. In that case, the 1/r^2 factor is approximatively a constant for all "force couples" and the inverse square law is a good approximation.

To address your comment in the question, changing the mass distribution but keeping shape the same can be equivalent to changing the distance, so this can change attraction. For example, in my dumbbell, make one of the dumbbell much more massive than the other one, then the attraction at the center changes direction.

As a last note, even if you use the center of mass of the objects to get the distance "r", changing mass distribution will affect attraction. The center of mass moves as a linear function of distance in the mass distribution, but the attraction moves as a function of 1/r^2. These cannot compensate each other exactly all the time. In my dumbbell example, you are at the center of mass in the middle between the two balls, yet attraction is zero. Now, transfer half of the mass of one sphere to the other. The center of mass moves towards the mass that gets heavier. Go to the new center of mass. The attraction is not zero, as you are now close to a larger mass and far from a small mass.

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    $\begingroup$ Why do add a third mass? You are answering a question that was not asked. Regarding the force changing when shapes change: sure, the force changes, but that's not the question. The question is, is it still proportional to the product of the masses? $\endgroup$
    – garyp
    May 8, 2020 at 20:31
  • $\begingroup$ The third mass is a "test mass". It is used to compare the attraction with two different objects having the same shape. I don't agree with you regarding shape. I think part of the question is change in shape. See the second comment on the question: "Thanks , but I am still not completely satisfied. Is it not possible that the force be some function of their masses that still conserves the units?( maybe a function that depends on the mass distribution in space or something?) I" $\endgroup$ May 8, 2020 at 20:42
  • $\begingroup$ The person asked: "But I am not able to prove how the force between Any two rigid masses is proportional to the product of their masses( As my teacher says..)." I think he meant. "Only depended on the product of their masses". However, there is a "form factor" if you wish. Yes, doubling one of the masses doubles the attraction, but I don't think that's what the OP wants to know. (as per his comment about mass disctribution. $\endgroup$ May 8, 2020 at 20:45
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    $\begingroup$ So basically, I am not answering the literal question, I am answering what I think the OP wants to know and which seems to be not well defined in the question. $\endgroup$ May 8, 2020 at 20:47
  • $\begingroup$ I asked for clarifications in the comments. I can always be wrong. $\endgroup$ May 8, 2020 at 21:03
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It is not true that the gravitational force between extended mass distributions depends only on the product of the total masses. It is true that the time averaged total force integrated over each body is $$\vec F = Gm_1m_2 \frac{\vec r_{12}}{r_{12}^3} ~.$$ However, unless both mass distributions are spherical, the attraction has higher moments. These higher moment forces cause the bodies to be stressed and to nonuniformly rotate or wiggle. Only for certain relative orientations these higher moment forces exactly cancel the internal stresses.

An example is the Earth-Moon system. The moon is deformed but it is almost at rest in the corotating frame. It only wiggles a little. Weirder is the rotation of Mercury. It has a slight permanent dipole deformation causing it to rotate in a tidal 3:2 resonance. See https://en.wikipedia.org/wiki/Mercury_(planet)#Spin-orbit_resonance.

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If to be summarized in short - you need to apply and solve second Newton law equation for two-body problem :

$$ \vec F_G = \mu \, \vec r^{\,\prime \prime} $$

Where $\mu$ is two-body system reduced mass : $$ \mu ={\cfrac {m_{1}m_{2}}{m_{1}+m_{2}}} $$

Btw, it's interesting to note that reduced mass has reciprocal additive property :

$$ {\frac {1}{\mu }}={\frac {1}{m_{1}}}+{\frac {1}{m_{2}}} $$

Reduced mass helps to analize two-body problem as it were just 1 single body. And $\vec r$ is displacement between bodies.

That's why gravity force is proportional to mass product of both bodies. (I.e. product increases faster than sum of masses). Another way which is helpful to physical intuition is to check the moment of inertia of binary system :

enter image description here

Which is :

$$ I={\frac {m_{1}m_{2}}{m_{1}\!+\!m_{2}}}x^{2}=\mu x^{2} $$

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It is not true in general that the gravitational force of attraction between extended bodies is proportional to their masses. It happens that we usually deal with gravitational attraction between celestial bodies, and that celestial bodies above a certain size are almost invariably close to spherical (in consequence of the self gravity of the body). In the particular case of spherical bodies, the result is true as a consequence of Newton's shell theorem.

In the general case, simply note that the inverse square law of gravity is basically the same (up to the sign of charge) as the Coulomb law of electrostatics, and apply the argument of any number of text book examples, such as the electrostatic attraction/repulsion for a charge uniformly distributed on a long rod, or a large plate. Clearly the force does depend on the distribution of charge/mass.

OTOH, with regard to gravity, because gravity is such a weak force, most of the practical examples with rigid bodies in celestial mechanics do involve spherical bodies. One important exception is to treat the gravitational field of a spiral galaxy (it is not rigid, but its mass distribution can be treated as constant). This is not the same as the gravitational field of a central mass. I have shown how it can be treated in The effects of turbulence generated in Big Bang nucleosynthesis

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The simple explanation is that any finite body (i.e. occupying a bounded region of space) looks like a point from sufficiently far away. This observation also tells you what is the range of validity of this "law". The distance between the bodies needs to be much larger than the linear size of each body.

Using math and calculus it is possible to turn this intuition into precise and predictive equations. This approach goes under the name of multipole expansion

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It is a fundamental "law" of nature that just happens to be true.

You can prove laws to a very high degree of accuracy, but from theory alone it is impossible to predict something fundamental by definition. Maths can't tell you the universe's axoims!

To give an example, Coulomb's law for the force between two charged masses is very similar to the gravitational law.

If this law is correct, it can be shown that the electric field inside any conductor is zero. Thus, instead of trying to test the law directly (separating to charges and measuring their forces very precisely), which is not very accurate, one can check the inverse square law for electrostatics to a very high precision by testing that the electric field really is zero inside a conductor by seeing if any charge flows onto a test rod.

Thus, although some laws may be hard to verify experimentally directly (such as by moving stars), we can check that the laws match up with our measurement of elliptical orbits and their time periods, amongst other things.

In fact, it was the orbit of Mercury that led scientists (Einstein) to show that Newton's Law of Gravitation was in fact incorrect as it did not match the experimental data.

So my answer to your question is that no, this law cannot be proven from some other law since it is fundamental, however, through experimental data, we can show that it is true and thus "good enough for the time being" until we find a better one! Can you really ever say that it is really true everywhere in the Universe without visiting every location in space and time? I'd say that's philosophy not Physics.

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  • $\begingroup$ I see your point but taking the law for point masses as fundamental, I believe that there should be some way to deduce mathematically that the statement in question is true. The way Newton precisely proved the statement for spheres as a consequence of the law for point masses. So as long as the statement for point masses is true, The other statements deduced mathematically from it are equally true..That deduction is what I require $\endgroup$ May 8, 2020 at 17:15
  • $\begingroup$ You treat the body as a combination of particles. The vector sum of the forces from each particle is the same as a force acting through the centre of mass. $\endgroup$
    – Joe Iddon
    May 8, 2020 at 19:06
  • $\begingroup$ It should also be noted that experimental physics had a large hand in helping to nail it down and continues to do so $\endgroup$
    – Triatticus
    May 9, 2020 at 5:18

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