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
P, distance 2R from the centre O of the sphere. A spherical cavity of radius R/2 is now made in the sphere as shown in figure. The sphere with cavity now applies an gravitational force F₂ on same particle placed at P. The ratio F₂/ F₁ will be

Moderate

A

1 /2
B

7/9
C

3
D

7

Solution

Gravitational force due to solid sphere, $F_{1} = \frac{GMm}{(2 R)^{2}}$ where M and mare masses of the solid sphere and particle respectively
and R is the radius of the sphere. The gravitational force on particle due to sphere with cavity = force due to solid sphere creating cavity, assumed to be present above at that position.
i, e, $F_{2} = \frac{GMm}{4 R^{2}} - \frac{G (M / 8) m}{(3 R / 2)^{2}} = \frac{7}{36} \frac{GMm}{R^{2}}$
So, $\frac{F_{2}}{F_{1}} = \frac{7 GMm}{36 R^{2}} / (\frac{GMm}{4 R^{2}}) = \frac{7}{9}$

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