Inverse square law of gravitation and force between two spheres

[ShaunPereira]

Homework Statement

two identical spheres are placed in contact with each other. The force of gravitation between the spheres will be proportional to ( R = radius of each sphere)

Relevant Equations

$$F= \frac{Gm_1m_2} {r^2} $$
$$ m=\frac{4}{3}\pi R^3\rho $$

I recently encountered this problem on a test where the solution for the above problem was given as follows:
$$F= \frac{Gm_1m_2} {r^2} $$ (1)
but
$$ m=\frac{4}{3}\pi R^3 $$
substituting in equation (1)
$$F= \frac{{G(\frac{4}{3}\pi R^3\rho})^2 }{2R^2} $$
where r=radii of the two spheres
m=mass of the two spheres
$\rho$=density of the two spheres

solving further we get
$$F \alpha \frac{R^6}{R^2}$$

Therefore,
$$F \alpha R^4$$

But isn't gravity an inverse square force. An explanation I read in a book previously said that
$$F \alpha R^4$$
cannot be correct because it is against the basic concept which is more important than a numerical value and that an inverse square law is fundamental and gravitational force is for masses and not for densities.

This makes more sense to me than the solution provided for the test but I am still really confused as to which explanation is right

 

[Orodruin]

The inverse square law $1/r^2$ proportionality holds when the masses are fixed. In your case, keeping density of the spheres fixed and increasing the size also increases the masses of the spheres.

 

[jbriggs444]

Centripetal acceleration in uniform circular motion is given by both $F = \frac{v^2}{r}$ and by $F=\frac{4\pi^2r}{t^2}$. So is it inversely proportional to $r$ or directly proportional to $r$?

It depends on what you are holding fixed. If you hold velocity fixed, it is inversely proportional to $r$. If you hold the rotational period fixed, it is directly proportional to $r$.

 

[PeroK]

... the gravitational force between two adjacent marbles is less than that of two adjacent planets. Even though the marbles centres are closer together.

 

[ShaunPereira]

The inverse square law $1/r^2$ proportionality holds when the masses are fixed. In your case, keeping density of the spheres fixed and increasing the size also increases the masses of the spheres.

Right. So does that mean that the force between the two spheres obeys the inverse square law when the masses are fixed and the distances vary. However when the sizes of the two spheres vary keeping density constant the force is no more proportional to $R^2$ but rather proportional to $R^4$.

Newton's law of Gravitation still holds but we don't get the familiar equation as the answer. Its just a matter of what you take as constant and what you take as variable.

Right?

 

[PeroK]

Right. So does that mean that the force between the two spheres obeys the inverse square law when the masses are fixed and the distances vary. However when the sizes of the two spheres vary keeping density constant the force is no more proportional to $R^2$ but rather proportional to $R^4$.

Newton's law of Gravitation still holds but we don't get the familiar equation as the answer. Its just a matter of what you take as constant and what you take as variable.

Right?

Newton's law of gravitation says:$$F = \frac{Gm_1m_2}{r^2}$$where $m_1$ and $m_2$ are the masses of two point particles, and $r$ is the distance between them. That is an inverse square law. Note that the masses of the particles are relevant, not only the distance between them. There is no law that says $F = \frac{G}{r^2}$.

 

[ShaunPereira]

Newton's law of gravitation says:$$F = \frac{Gm_1m_2}{r^2}$$where $m_1$ and $m_2$ are the masses of two point particles, and $r$ is the distance between them. That is an inverse square law. Note that the masses of the particles are relevant, not only the distance between them. There is no law that says $F = \frac{G}{r^2}$.

Yes. and the mass here is proportional to $R^3$ which when we put in our equation gives us the answer.
The inverse square law takes into account both mass and distances. Since both change we have to take both into account

Right?

 

[PeroK]

Yes. and the mass here is proportional to $R^3$ which when we put in our equation gives us the answer.
The inverse square law takes into account both mass and distances. Since both change we have to take both into account

Right?

Yes, the force between any two objects is a specific calculation - in this case using the inverse square law and shell theorem. If the objects were cubes you'd have a different calculation.

 

[ShaunPereira]

Yes, the force between any two objects is a specific calculation - in this case using the inverse square law and shell theorem. If the objects were cubes you'd have a different calculation.

Ok thanks.