Animated Solution for Physics - Gravitation: In the reported figure of Earth, the value of acceleration due to gravity is same at point $A$ and $C$ but it is smaller than that of its value at point $B$ (surface of the Earth). The value of $OA:AB$ will be $x:y$. The value of $x$ is .......... .

Imagine you are standing on the surface of the Earth at point $B$. The gravity here is at its absolute maximum. If you dig a deep tunnel down to point $A$, the gravity decreases. Similarly, if you fly up in a rocket to point $C$, the gravity also decreases.

The problem presents a fascinating scenario: the gravitational pull at point $A$ (deep underground) is exactly equal to the gravitational pull at point $C$ (high up in space). We are given the Earth's radius $R=6400\text{ km}$ and the height of point $C$ above the surface $h=3200\text{ km}$. Notice that $h$ is exactly half of the Earth's radius, so $h=R/2$.

Our goal is to find the ratio of the distance $OA$ to the distance $AB$.

To solve this, we need to recall how gravity behaves in different regions.

Inside the Earth ($r<R$): The acceleration due to gravity varies linearly with the distance from the center. The formula is:

Outside the Earth ($r>R$): The Earth behaves like a point mass concentrated at its center, and gravity follows the inverse square law:

Let's apply these formulas to our specific points. For point $A$, the distance from the center is simply $OA$. So, the gravity at $A$ is:

For point $C$, the distance from the center is the radius $R$ plus the height $h$. Since $h=R/2$, the total distance is $R+R/2=3R/2$. The gravity at $C$ is:

The problem states that $g_A=g_C$. Let's set our two expressions equal to each other:

First, let's simplify the right side of the equation. Squaring the denominator $(3R/2{)}^2$ gives $9R^2/4$. When we divide by a fraction, the denominator flips up to the numerator:

Now, we can cancel out the common terms. The gravitational constant $G$ and the mass of the Earth $M$ appear on both sides, so they vanish. We can also cancel $R^2$ from the denominator on both sides, leaving just one $R$ on the left side:

Cross-multiplying gives us the exact distance of point $A$ from the center:

We are almost there! We need the ratio $OA:AB$. We have $OA$, but we still need to find the length of segment $AB$.

Looking at the geometry of the problem, the total distance from the center to the surface is the radius $OB=R$. The segment $AB$ is just the total radius minus the inner segment $OA$:

Now, we simply divide $OA$ by $AB$ to find our ratio:

The $R$ and the $9$ in the denominators cancel out beautifully, leaving us with:

The problem states that this ratio is $x:y$, and we need to find the value of $x$. Comparing our result $4:5$ to $x:y$, it is clear that $x=4$.

This elegant problem perfectly demonstrates the symmetry of gravity inside and outside a spherical mass!