Problem 68 - Master Resource Book in JEE Main Physics

Question

The change in potential energy when a body of mass \(m\) is raised to a height \(n R\) from the centre of earth ( \(R=\) radius of earth) (a) \(m g R \frac{(n-1)}{n}\) (b) \(n m g R\) (c) \(m g R \frac{n^{2}}{n^{2}+1}\) (d) \(m g R \frac{n}{n+1}\)

Step-by-Step Solution

Answer

The change in potential energy is \( m g R \frac{(n-1)}{n} \).

Step 1: Understand the problem

We need to calculate the change in potential energy of a body when it is raised from the center of the Earth to a height of \( nR \) above the center, where \( R \) is the radius of the Earth.

Step 2: Calculate potential energy at the Earth's surface

The formula for gravitational potential energy (GPE) at a distance \( r \) from the center of the Earth is \(-\frac{G M m}{r}\), where \( G \) is the gravitational constant and \( M \) is the mass of the Earth. At the Earth's surface, \( r = R \), so \( U_1 = -\frac{G M m}{R} \).

Step 3: Calculate potential energy at height \( nR \)

At height \( nR \), the distance from the center of the Earth is \( nR \). The potential energy \( U_2 = -\frac{G M m}{nR} \).

Step 4: Find the change in potential energy

The change in potential energy is given by \( \Delta U = U_2 - U_1 = \left(-\frac{G M m}{nR}\right) - \left(-\frac{G M m}{R}\right) = \frac{G M m}{R} \left(1 - \frac{1}{n}\right) \).

Step 5: Relate gravitational force with acceleration due to gravity

We know that \( g = \frac{G M}{R^2} \). Thus, \( G M = g R^2 \). Substitute it into \( \Delta U \): \( \Delta U = \frac{g R^2 m}{R} \left(1 - \frac{1}{n}\right) = m g R \left(1 - \frac{1}{n}\right) \).

Step 6: Simplify further

Simplifying, we have \( \Delta U = m g R \frac{n - 1}{n} \). This matches option (a).

Key Concepts

gravitational potential energy

Gravitational potential energy (GPE) is the energy a body possesses due to its position within a gravitational field. When you lift an object against gravity, you're doing work and storing potential energy in that object. This energy depends on the mass of the object, the height it's raised to, and the gravitational field strength.
In general, the formula for GPE is given by: \[ U = -\frac{G M m}{r} \] where:

• \( U \): gravitational potential energy
• \( G \): gravitational constant
• \( M \): mass of the Earth
• \( m \): mass of the object
• \( r \): distance from the center of the Earth

This formula shows that GPE becomes more negative as you move closer to the Earth's core, meaning the energy required to escape gravity's hold increases.
When considering changes in GPE, such as when an object is raised from the surface to a height, we look at the difference in potential energy at two points. Understanding GPE is crucial for solving problems related to changes in an object's position in a gravitational field.

acceleration due to gravity

Acceleration due to gravity, denoted as \( g \), is the rate at which objects accelerate towards the Earth due to its gravitational pull. On Earth's surface, it is approximately \( 9.8 \, m/s^2 \). This constant plays a vital role in determining the gravitational field strength.The relationship between gravitational force \( F \) and acceleration due to gravity \( g \) is expressed as:
\[ F = m \, g \] where:

• \( F \): gravitational force
• \( m \): mass of the object
• \( g \): acceleration due to gravity

Furthermore, we find \( g \) using the formula:\[ g = \frac{G M}{R^2} \] This equation indicates that \( g \) is directly dependent on the Earth's mass \( M \) and inversely proportional to the square of the Earth's radius \( R \). When you understand \( g \), you can interpret how it affects the gravitational potential energy of an object. This concept helps in relating how potential energy changes as an object is further or nearer from the Earth’s surface.

radius of the Earth

The radius of the Earth, symbolized as \( R \), is a critical factor in many gravitational calculations. The Earth is not a perfect sphere, but a simplification can be made using an average radius of approximately 6,371 km. This average is useful for calculations that involve Earth's curvature and its gravitational interactions. In problems related to gravitational potential energy, the radius of the Earth helps determine the gravitational pull at various distances from the Earth's center. For example, the potential energy at the Earth's surface uses \( R \) as the distance from the center. As the elevation moves to a height \( nR \), the new distance becomes crucial for calculating the change in potential energy.In addition, \( R \) is an essential factor in the formula for acceleration due to gravity, \( g = \frac{G M}{R^2} \). It shows that even minor variations in \( R \) can significantly impact the value of \( g \), affecting all potential energy calculations based upon this constant. Understanding \( R \)'s role helps in grasping the full scope of gravitational interactions on Earth.