Question

Consider two solid uniform spherical objects of the same density $\rho $. One has a radius R and the other has a radius 2R. They are in outer space where the gravitational fields from other objects are negligible. If they are arranged with their surface touching, what is the contact force between the objects due to their traditional attraction?
A) \[G{\pi ^2}{R^4}\]
B) $\dfrac{{128}}{{81}}G{\pi ^2}{R^4}{\rho ^2}$
C) \[\dfrac{{128}}{{81}}G{\pi ^2}\]
D) $\dfrac{{128}}{{87}}G{\pi ^2}{R^2}$

Answer 1

Hint: When there is no other kind of force acting on two bodies only attraction due to gravity will work between them. This force is proportional to both their masses and inversely proportional to the square of the distance between the objects’ center of mass.

Formula used:
Gravitational attraction between two point objects are given as:
$F = \dfrac{{G{M_1}{M_2}}}{{{r^2}}}$ …………………….(1)
Where,
$F$ is the attractive force,
$G$ is gravitational constant,
${M_1}$and ${M_2}$are the masses of two objects,
$r$ is the distance between the center of masses of the two objects.

The volume of a sphere of radius R is given by:
$V = \dfrac{4}{3}\pi {R^3}$ ……………. (2)
Where,
$V$ is the volume of the sphere,
$R$ is the radius of the sphere.

Mass of an object with given density and volume:
$M = \rho .V$ ………………. (3)
Where,
$M$ is the mass of the object,
$\rho $ is the density of the object,

Complete step by step solution:
Given:
The radius of the smaller sphere is $R$.
The radius of a larger sphere is $2R$.
The density of both spheres is $\rho $.
The spheres are kept with their surface touching each other.
To find: Contact force between the spheres.
Step 1:
Use eq.(2) in eq.(3) to get the mass of the first sphere of $R$ as:
\[
  {M_1} = \rho \times \left( {\dfrac{4}{3}\pi {R^3}} \right) \\
  \therefore {M_1} = \dfrac{4}{3}\pi \rho {R^3} \\
 \] ……………….(4)
Step 2:
Similarly, Use eq.(2) in eq.(3) to get the mass of the first sphere of $2R$ as:
$
  {M_2} = \rho \times \left( {\dfrac{4}{3}\pi {{\left( {2R} \right)}^3}} \right) \\
  \therefore {M_2} = \dfrac{{32}}{3}\pi \rho {R^3} \\
 $ …………….(5)
Step 3:
For a uniform sphere, its center of mass always stays at its center. As they are kept just in touch so the distance between their center is $r = R + 2R = 3R$. Now, substitute r and the values of \[{M_1}\] and \[{M_2}\] obtained from eq.(4) and eq.(5) in eq.(1) to get the attractive force value as:
$
  F = \dfrac{{G \times \left( {\dfrac{4}{3}\pi \rho {R^3}} \right) \times \left( {\dfrac{{32}}{3}\pi \rho {R^3}} \right)}}{{{{\left( {3R} \right)}^2}}} \\
  \therefore F = \dfrac{{128}}{{81}}G{\pi ^2}{R^4}{\rho ^2} \\
 $

The contact force between the objects will be: (b) $\dfrac{{128}}{{81}}G{\pi ^2}{R^4}{\rho ^2}$.

Note:
This problem can be solved in a tricky manner without any calculation. Notice that, each of the two masses is proportional to the volume which is proportional to \[{R^3}\]. Now, the force term itself is proportional to \[{R^{ - 2}}\]. Hence, the total power of R in the expression will be $(3+3-2)=4$. Again each mass term is proportional to density \[\rho \]. So, \[\rho \] will have power $(1+1)=2$. Hence, ${R^4}{\rho ^2}$ term must be there in the expression. And an only possible option with it is option (B).