Question

A spherical hollow cavity is made in a lead sphere of radius R, such that its surface touches the outside surface of the lead sphere and passes through its centre. The mass of the sphere before hollowing was M. With what gravitational force will the hollowed-out lead sphere attract a small sphere of mass 'm', which lies at a distance d from the centre of the lead sphere on the straight line connecting the centres of the spheres and that of the hollow, if d = 2R :

• A. $\frac{7GMm}{18R^2}$
• B. $\frac{7GMm}{36R^2}$
• C. $\frac{7GMm}{9R^2}$
• D. $\frac{7GMm}{72R^2}$

Answer 1

Answer: (B)

$\frac{7GMm}{36R^2}$

Answer 2

The problem asks for the gravitational force exerted by a hollowed-out lead sphere on a small sphere of mass 'm'. We can solve this using the principle of superposition. The hollowed-out sphere can be considered as a complete solid sphere of mass M and radius R, from which a smaller sphere (the hollow) with a certain mass and radius has been removed.

1. Determine the properties of the hollow cavity:

• The original lead sphere has a radius R and its center is O.
• The hollow cavity's surface touches the outside surface of the lead sphere. Let this point be P_out. So, the distance OP_out = R.
• The hollow cavity's surface also passes through the center of the lead sphere (O).
• Since O and P_out are on the surface of the hollow and are separated by a distance R, this distance R must be the diameter of the hollow cavity.
• Therefore, the radius of the hollow cavity, let's call it $R_h$, is $R/2$.
• The center of the hollow cavity, let's call it $C_h$, is midway between O and P_out. So, the distance $OC_h = R/2$.

2. Calculate the mass of the hollowed-out part:

• Let $\rho$ be the uniform density of the lead.
• The mass of the original solid sphere is $M = \rho \times \frac{4}{3}\pi R^3$.
• The mass of the material removed to create the hollow, $M_h$, would be the mass of a solid sphere of radius $R_h = R/2$ and density $\rho$: $M_h = \rho \times \frac{4}{3}\pi R_h^3 = \rho \times \frac{4}{3}\pi \left(\frac{R}{2}\right)^3 = \rho \times \frac{4}{3}\pi \frac{R^3}{8}$.
• Comparing $M_h$ with M: $M_h = \frac{1}{8} \left(\rho \times \frac{4}{3}\pi R^3\right) = \frac{M}{8}$.

3. Apply the principle of superposition:

The net gravitational force on mass 'm' due to the hollowed-out sphere is the gravitational force due to the complete solid sphere minus the gravitational force due to the removed mass (the hollow). The small sphere 'm' is placed at a distance d = 2R from the center of the lead sphere (O), along the line connecting O and $C_h$. So, the arrangement is O - $C_h$ - 'm'.

• Gravitational force due to the complete solid sphere ($F_s$):

  • Mass = M
  • Distance from its center (O) to 'm' = d = 2R
  • $F_s = \frac{GMm}{d^2} = \frac{GMm}{(2R)^2} = \frac{GMm}{4R^2}$.
  • This force acts towards O.

• Gravitational force due to the removed mass ($F_h$):

  • Mass = $M_h = M/8$
  • The distance from the center of the hollow ($C_h$) to 'm' is $d_h = d - OC_h = 2R - R/2 = \frac{3R}{2}$.
  • $F_h = \frac{G M_h m}{d_h^2} = \frac{G (M/8) m}{(3R/2)^2} = \frac{GMm}{8 \times \frac{9R^2}{4}} = \frac{GMm}{18R^2}$.
  • This force acts towards $C_h$.

• Net gravitational force ($F_{net}$):

  Both forces act along the same line and are attractive. Since the hollow removes mass, its gravitational effect is subtracted from that of the complete sphere.

  $F_{net} = F_s - F_h$

  $F_{net} = \frac{GMm}{4R^2} - \frac{GMm}{18R^2}$

  To subtract, find a common denominator, which is 36$R^2$:

  $F_{net} = \frac{9GMm}{36R^2} - \frac{2GMm}{36R^2}$

  $F_{net} = \frac{(9-2)GMm}{36R^2}$

  $F_{net} = \frac{7GMm}{36R^2}$