Animated Solution for Physics - Gravitation: A solid sphere of radius $R$ gravitationally attracts a particle placed at $3R$ from its centre with a force $F_1$. Now, a spherical cavity of radius $(R/2)$ is made in the sphere (as shown in figure) and the force becomes $F_2$. The value of $F_1:F_2$ is

Imagine a massive solid sphere of radius $R$ and mass $M$. Now, place a tiny particle of mass $m$ at a point $A$, which is at a distance of $3R$ from the center $O$. This is a classic setup, but it gets incredibly interesting when we start removing pieces of the sphere.

According to Newton's law of gravitation, the entire mass of a uniform solid sphere can be assumed to be concentrated at its center for any point outside it. So, the initial force $F_1$ on our particle is simply the gravitational constant $G$ times the product of the masses, divided by the square of the distance.

Squaring the denominator, we get $9R^2$. So, our initial force is:

Let's keep this equation safe; it is the baseline we will compare our final result against.

Now comes the most interesting part of this question. We scoop out a smaller sphere of radius $R/2$ to create a cavity. How do we find the gravitational force of a sphere with a hole in it? The geometry is no longer symmetric, so we cannot just assume the mass is at the center.

This is where we use the Principle of Superposition! We can treat the sphere with a cavity as a complete solid sphere plus a smaller sphere of negative mass located exactly where the cavity is.

To find the effect of this missing mass, we first need to find out how much mass was removed. The volume of a sphere is proportional to the cube of its radius. Since the radius of the cavity is half of the original sphere, its volume will be $(1/2{)}^3=1/8$ of the total volume. Because the density is uniform, its mass will also be one-eighth of the total mass $M$.

Let's calculate the force due to this imaginary removed mass. The center of this cavity, let's call it $B$, is at a distance of $R/2$ from the main center $O$. Since the particle is at $3R$ from $O$, the distance from the cavity's center to the particle is $3R-R/2=2.5R$ or $\frac{5R}{2}$.

We substitute $M^{\prime }$ as $M/8$, and the distance as $5R/2$ into the gravitational force formula:

Squaring $5R/2$ gives $25R^2/4$. The $4$ goes to the numerator. Simplifying the fraction, we get:

Now, let's find the net force $F_2$. We subtract the cavity's force from the full sphere's force.

Taking $\frac{GMm}{R^2}$ common, we are left with $\frac{1}{9}-\frac{1}{50}$. Taking the LCM as $450$, the numerator becomes $50-9$, which is $41$.

Finally, we need the ratio of $F_1$ to $F_2$. We divide the two expressions. The $\frac{GMm}{R^2}$ terms cancel out beautifully.

Solving this, $450$ divided by $9$ is $50$. So, the final ratio is:

Always remember, treating a missing mass as a negative mass is a superpower in physics! It turns complex, asymmetric integration problems into simple algebraic subtractions.