A solid sphere of constant mass density 𝜌 applies force F1 on mass m when a spherical cavity of radi

A solid sphere of constant mass density 𝜌 applies force F1 on mass ‘m’ when a spherical cavity of radius R/ 3 is made, the force acting on ‘m’ is F2 . Find F1/ F2 ?
*Understanding the Problem: Force in Gravitational Field* 🌍
We are given a solid sphere with a constant mass density \(
ho\), and the mass \(m\) is subjected to a force due to the gravitational field of the sphere. When a spherical cavity of radius \(R/3\) is made within the sphere, the gravitational force acting on the mass changes. We are tasked with finding the ratio of the gravitational forces, \(F_1/F_2\), before and after the cavity is made.
Let’s break down the concepts and solve this step by step with minimal calculations and more focus on the theory behind it. 💡
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*Key Concepts Involved* 🌟
1. **Gravitational Force in a Solid Sphere**:
A solid sphere creates a gravitational force on an object placed inside it. According to the law of gravity, the force at a distance \(r\) from the center of the sphere (inside the sphere) depends on the mass enclosed within a sphere of radius \(r\). The force at this point can be derived from the gravitational field due to the sphere’s mass.
2. **Effect of a Cavity on Gravitational Force**:
When a cavity is created inside the sphere, the region of the cavity no longer contributes to the gravitational force. Essentially, the sphere's gravitational field is altered by removing mass, which reduces the overall gravitational force acting on any object inside or near the sphere.
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*Theory: Gravitational Force Before and After the Cavity* 🔍
1. **Before the Cavity (Force \(F_1\))**:
For a solid sphere of uniform density \(
ho\), the gravitational force \(F_1\) acting on a mass \(m\) placed inside the sphere can be determined by the mass enclosed within a radius \(r\) (assuming the object is at distance \(r\) from the center). For a solid sphere of total radius \(R\), the gravitational force at a distance \(r\) inside the sphere is proportional to \(r\) (directly proportional to the mass enclosed).
2. **After the Cavity (Force \(F_2\))**:
When a spherical cavity of radius \(R/3\) is made, the mass that would have been present in this region is now removed. This reduces the gravitational force, as the object no longer feels the gravitational pull from the mass that would have been in the cavity. The key point here is that the mass removed from the sphere causes a decrease in the gravitational force, but it’s not just a simple subtraction because of the spherical symmetry.
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*Finding the Ratio \(F_1/F_2\)* 🧠
The gravitational force inside a solid sphere (without a cavity) is given by:
\[
F_1 \propto \text{Enclosed mass} \quad \text{(Directly proportional to } r\text{)}
\]
When a spherical cavity is created, the mass within the cavity is removed, and this affects the overall gravitational force. The removed mass can be considered as creating a "negative" contribution to the gravitational force. The spherical symmetry ensures that the force reduction is proportional to the volume of the cavity.
Since the cavity’s radius is \(R/3\), the volume of the cavity is proportional to \((R/3)^3 = R^3/27\), and thus the mass removed is also proportional to this volume. The resulting force \(F_2\) will therefore be reduced by a factor related to this volume ratio.
From this, we can conclude that:
\[
\frac{F_1}{F_2} = 27
\]
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*Conclusion* 💡
The ratio of the gravitational forces \(F_1/F_2\) is **27**. 🎯
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*Key Insights* 🔑
The creation of a cavity reduces the gravitational force due to the removal of mass.
The gravitational force inside a sphere is proportional to the mass enclosed within a given radius.
The ratio \(F_1/F_2\) reflects how much the gravitational force changes when a cavity is created.
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