Q.A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?

The key idea here is the Variation of Gravity with Height. The gravitational force exerted by the Earth on a body decreases as its distance from the Earth's center increases.

1. The weight of the body on the Earth's surface is given by $W_s = mg$, where $g$ is the acceleration due to gravity on the surface. We are given $W_s = 63 \text{ N}$.
2. The acceleration due to gravity at a height $h$ above the Earth's surface, $g_h$, is related to $g$ by the formula:

   $g_h = g \left( \frac{R}{R+h} \right)^2$

   where $R$ is the radius of the Earth.

3. We are given that the height $h$ is half the radius of the Earth, so $h = R/2$. Substituting this into the formula for $g_h$: $g_h = g \left( \frac{R}{R + R/2} \right)^2 = g \left( \frac{R}{3R/2} \right)^2 = g \left( \frac{2}{3} \right)^2 = \frac{4}{9}g$ …

Understanding Gravitational Force and its Variation with Height

The weight of a body is the gravitational force exerted on it by the Earth. This force is given by $W = mg$, where $m$ is the mass of the body and $g$ is the acceleration due to gravity at that location.

The acceleration due to gravity, $g$, is not constant. It depends on the distance from the center of the Earth. According to Newton's Law of Universal Gravitation, the force between two masses is inversely proportional to the square of the distance between their centers. This means that as a body moves further away from the Earth's surface, its distance from the Earth's center increases, and thus the gravitational force, and consequently the acceleration due to gravity, decreases.

At the Earth's surface, the acceleration due to gravity is $g_s$. At a height $h$ above the Earth's surface, the distance from the center of the Earth becomes $R+h$, where $R$ is the radius of the Earth. The acceleration due to gravity at this height, $g_h$, is given by the formula:

This formula is exact and should be used when the height $h$ is comparable to the Earth's radius $R$. There is an approximation $g_h \approx g_s (1 - 2h/R)$ which is valid only for $h \ll R$. Since the problem specifies $h = R/2$, which is not much smaller than $R$, we must use the exact formula.

Now, let's apply this understanding to solve the problem.

Step-by-Step Solution

1. Identify the initial condition:

   The body weighs $63 \text{ N}$ on the surface of the Earth. This means the gravitational force on the body at the surface is $W_s = 63 \text{ N}$.

   We know that weight is $W = mg$. So, $W_s = m g_s = 63 \text{ N}$. Here, $m$ is the mass of the body, which remains constant regardless of its location, and $g_s$ is the acceleration due to gravity at the Earth's surface.

2. Identify the target condition:

   We need to find the gravitational force (weight) on the body at a height $h$ equal to half the radius of the Earth.

   So, $h = \frac{R}{2}$.

   Let the gravitational force at this height be $W_h$. We can write $W_h = m g_h$, where $g_h$ is the acceleration due to gravity at height $h$.

3. Apply the formula for $g_h$:

   Using the formula for acceleration due to gravity at height $h$:

   $g_h = g_s \left( \frac{R}{R+h} \right)^2$

4. Substitute the given height into the formula:

   We are given $h = \frac{R}{2}$. Substitute this into the equation for $g_h$: …

Step 1: Weight on the surface $W_s = mg = 63$ N.

Step 2: Use the height-variation formula $g_h = g\left(\dfrac{R}{R+h}\right)^2$. With $h = R/2$: $\dfrac{R}{R+h} = \dfrac{R}{3R/2} = \dfrac23$. …