Problem 70 - 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 earth's surface is \((R=\) radius of the earth) (a) \(m g R \frac{n}{(n-1)}\) (b) \(m g R\) (c) \(m g R \frac{n}{(n+1)}\) (d) \(m g R \frac{n^{2}}{\left(n^{2}+1\right)}\)

Step-by-Step Solution

Answer

The correct answer is (c): \(m g R \frac{n}{(n+1)}\).

Step 1: Understand the Problem

We need to find the change in gravitational potential energy when an object of mass \(m\) is raised to a height \(nR\) above the Earth's surface. The potential energy is related to the gravitational force between the Earth and the object.

Step 2: Recall the Gravitational Potential Energy Formula

The gravitational potential energy \(U\) at a distance \(r\) from the Earth's center is given by \(U = -\frac{G M m}{r}\), where \(G\) is the gravitational constant, \(M\) is the mass of the Earth and \(r\) is the distance from the center of the Earth.

Step 3: Calculate Initial Potential Energy

Initially, the object is on the Earth's surface, so \(r = R\). The initial potential energy is \(U_1 = -\frac{G M m}{R}\).

Step 4: Calculate Final Potential Energy

When the object is raised to a height \(nR\), the final distance from the Earth's center \(r\) becomes \(R + nR = (n+1)R\). Thus, the final potential energy is \(U_2 = -\frac{G M m}{(n+1)R}\).

Step 5: Determine Change in Potential Energy

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

Step 6: Simplify the Expression

Simplify \(\Delta U = \frac{G M m}{R} \left(1 - \frac{1}{n+1}\right) = \frac{G M m}{R} \cdot \frac{n}{n+1}\). Since the gravitational force near the Earth's surface is \(mg = \frac{G M m}{R^2}\), we multiply our expression by \(R\) to convert into standard potential energy units: \(\Delta U = m g R \cdot \frac{n}{n+1}\).

Step 7: Compare with Options

The derived expression corresponds to option (c): \(m g R \frac{n}{(n+1)}\). Thus, the correct answer is option (c).

Key Concepts

Gravitational Force

Gravitational force is the attractive force that exists between any two masses. It is the force that keeps planets in orbit, causes objects to fall to the ground, and influences the motion of all celestial bodies.
In the context of the Earth's gravity, it is estimated using Newton's law of universal gravitation, which is given by the formula:

• \( F = \frac{G M m}{r^2} \)

Here, \( F \) is the gravitational force, \( G \) is the gravitational constant \( (6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2) \), \( M \) is the mass of the Earth, \( m \) is the mass of the object, and \( r \) is the distance from the center of the Earth to the object.
This force decreases with the square of the distance between the masses, making it a long-range force that gets weaker as distance increases. However, near the Earth's surface, we often approximate \( F \) as \( mg \), where \( g \) is the acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \). This simplification is useful for solving many problems related to gravitational potential energy.

Potential Energy Formula

Potential energy due to gravity is the energy an object possesses due to its position in a gravitational field. The formula for gravitational potential energy \( U \) when an object is at a distance \( r \) from the center of the Earth is:

• \( U = -\frac{G M m}{r} \)

The negative sign indicates that work must be done against the gravitational force to increase the distance \( r \), thus increasing the potential energy. The closer you are to the Earth, the stronger the gravitational force and the lower the potential energy.
In problems involving a change in height, such as raising an object from the Earth's surface to a height \( nR \), we use this formula to calculate the initial and final potential energies and find their difference. The formula must be adapted to reflect the change in distance and then simplified, sometimes requiring conversion to a form involving the gravitational acceleration \( g \), such as \( \Delta U = mgR \cdot \frac{n}{n+1} \), as shown in the solution.

JEE Physics Problem Solving

Solving physics problems like those encountered in the JEE exams requires a clear understanding of the underlying concepts as well as systematic problem-solving skills.
Here's a typical strategy to tackle such problems:

• Understand the Problem: Carefully read the problem to identify what's asked and which physical principles are relevant.
• Recall Relevant Formulas: Use well-established formulas to relate known quantities to the unknowns.
• Calculate Step by Step: Break down the problem into smaller, manageable parts. Here, calculating initial and final potential energies separately was crucial before finding the change.
• Simplify and Solve: Use algebraic techniques to simplify expressions. Conversion to known units (like \( mg \) for gravitational force) makes recognizing patterns or comparing with options easier.
• Verify and Compare: Double-check calculations and see if the derived solution matches one of the given choices.
  

By following these steps, even complex problems become approachable, and you can confidently solve them by leveraging both conceptual understanding and methodical solving techniques.