Question

An object is placed at a distance of \(\mathrm{R} / 2\) from the centre of earth. Knowing mass is distributed uniformly, acceleration of that object due to gravity at that point is: (g = acceleration due to gravity on the surface of earth and \(\mathrm{R}\) is the radius of earth) (1) \(\mathrm{g}\) (2) \(2 \mathrm{~g}\) (3) \(\mathrm{g} / 2\) (4) none of these

Short answer

The acceleration due to gravity at \( \frac{R}{2} \) is \( \frac{g}{2} \).

Step-by-step solution

Understanding the problem

Given an object placed at a distance of \(\frac{R}{2}\) from the center of the Earth. The aim is to find the acceleration due to gravity at that point. We know the acceleration due to gravity on the surface of the earth is \(g\) and \(R\) is the radius of the Earth.

Formula for gravitational force inside the Earth

When the object is inside the Earth, the gravitational force varies linearly with the distance from the center. The formula to use is: \[ g' = g \times \frac{d}{R} \] where \(d\) is the distance from the center of the earth.

Substitute the given values

Here, \(d = \frac{R}{2}\). Substitute this into the equation: \[ g' = g \times \frac{\frac{R}{2}}{R} \]

Simplify the expression

Simplifying the equation: \[ g' = g \times \frac{1}{2} = \frac{g}{2} \]

Conclusion

The acceleration due to gravity at a distance of \(\frac{R}{2}\) from the center of the earth is \( \frac{g}{2} \).

Key concepts

gravitational force variation

Gravitational force varies depending on where you are within the Earth. Particularly, when you move closer to the center, the force does not remain constant. Instead, it decreases. Why does this happen? Here's a simple explanation:

• **Outside Earth**: Gravitational force is stronger because the entire mass of the Earth pulls you towards the center.
• **Inside Earth**: The force varies linearly with the distance from the center, meaning the further inside you go, the less force you feel because the mass effect becomes divided. This is because only the mass within the radius you are at contributes to the gravitational pull.

It's crucial to understand this uniform linear variation as described by the formula: \[ g' = g \times \frac{d}{R} \] where \(g\) is the gravitational force at the surface, \(g'\) is the force at a distance \(d\), and \(R\) is the total radius of the Earth.

distance from the center of Earth

The distance from the center of the Earth plays a significant role in determining the gravitational force you experience. Here's why this matters:

• **Surface Level**: When you are on the surface, you feel the maximum gravitational pull because you are at the maximum radius \(R\).
• **Inside Earth**: Moving inside, the effective radius decreases. For example, if at half the radius (\(\frac{R}{2}\)), the gravitational acceleration will be half of what it is on the surface.

Mathematically, you use the distance in the formula \(g' = g \times \frac{d}{R}\) to determine the gravity at any point inside. This provides a clear and organized way to find gravitational variations based on your position relative to the Earth's center.

uniform mass distribution

To determine gravitational forces and variations inside the Earth, we need to assume the mass distribution of the Earth is uniform. Here's a deeper look:

• **Uniform Mass**: This means that the Earth’s density is the same throughout. Such an assumption simplifies calculations as every layer of Earth contributes equally to gravity.
• **Gravity Calculations**: When mass is uniformly distributed, gravity inside decreases linearly with distance from the center. With non-uniform mass, calculations would require complex integration.

This assumption is crucial in exercises because it allows us to use simpler formulas and establish clear relationships between distance, mass distribution, and gravitational force.