# gravitation question

• Mar 23rd 2011, 12:28 PM
prasum
gravitation 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 3R from the centre of the sphere a spherical cavity of radius r/2 is now made in the sphere .the sphere with cavity now applies a gravitational force F2 on the same particle.the ratio F1/F2 is
• Mar 23rd 2011, 05:31 PM
topsquark
Quote:

Originally Posted by prasum
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 3R from the centre of the sphere a spherical cavity of radius r/2 is now made in the sphere .the sphere with cavity now applies a gravitational force F2 on the same particle.the ratio F1/F2 is

The problem is spherically symmetric so no problems with using the usual force law for gravitation:
$\displaystyle F_1 = \frac{GMm}{(3R)^2}$
where M is the mass of the sphere and m is the mass of the object.

To evaluate this we need the mass of the whole sphere. It has a constant mass density, call it $\rho$. Then we know that
$\displaystyle V = \frac{4}{3} \pi R^3 \implies M = \rho V = \frac{4}{3} \pi \rho R^3$

So
$\displaystyle F_1 = \frac{GMm}{(3R)^2} = \frac{G \left ( \frac{4}{3} \pi \rho R^3 \right ) m}{(3R)^2}$

Now your turn. Find the mass of the removed section. Notice that the problem is still spherically symmetric so that
$\displaystyle F_2 = \frac{G(M - \Delta M)m}{(3R)^2}$

where $\Delta M$ is the removed mass.

Now form the fraction and you will find that practically everything cancels out.

-Dan