First slide

Gravitational force and laws

Question

A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F₁ on a particle placed at a distance 2R from the centre of the sphere. A spherical cavity of radius R/2 is now made in the sphere as shown in the figure. The sphere with the cavity now applies a gravitational force F₂ on the same particle. The ratio F₁/F₂ is

Moderate

A

$\frac{1}{2}$
B

$\frac{3}{4}$
C

$\frac{7}{8}$
D

$\frac{9}{7}$

Solution

Mass of the cavity = M/8 if mass of the sphere = M as volume of the cavity is 1/8th of the sphere.
$\begin{array}{l} F_{2} = \frac{G M m}{4 R^{2}} - \frac{G M m}{8 {(\frac{3}{2} R)}^{2}} \\ = \frac{G M m}{R^{2}} [\frac{1}{4} - \frac{1}{18}] = \frac{G M m}{R^{2}} [\frac{9 - 2}{36}] \\ = \frac{G M m}{R^{2}} \frac{7}{36} \\ ∴ \frac{F_{1}}{F_{2}} = \frac{36}{4 \times 7} = \frac{9}{7} \end{array}$

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