Problem 60 - Master Resource Book in JEE Main Physics

Question

If \(g\) is the acceleration due to gravity on earth's surface, the gain of the potential energy of an object of mass \(m\) raised from the surface of the earth to a height equal to the radius \(R\) of the earth is [UP SEE 2008] (a) \(2 \mathrm{mgR}\) (b) \(m g R\) (c) \(\frac{1}{2} m g R\) (d) \(\frac{1}{4} m g R\)

Step-by-Step Solution

Answer

The gain in potential energy is \( \frac{1}{2} mgR \); so the answer is (c) \( \frac{1}{2} mgR \).

Step 1: Understanding Potential Energy

Potential energy at height is given by the formula: \( U = mgh \), where \( m \) is mass, \( g \) is acceleration due to gravity, and \( h \) is height.

Step 2: Calculating Initial Potential Energy

When the object is on the surface of the earth, the potential energy with respect to infinity is \( U_1 = -\frac{GMm}{R} \), where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

Step 3: Calculating Potential Energy at Height R

At height \( h = R \), the potential energy of the object becomes \( U_2 = -\frac{GMm}{2R} \).

Step 4: Finding the Gain in Potential Energy

The gain in potential energy is given by the change in potential energy: \( \Delta U = U_2 - U_1 \). Substituting the values, \( \Delta U = -\frac{GMm}{2R} - \left(-\frac{GMm}{R}\right) = \frac{GMm}{2R} \).

Step 5: Relating with g

The formula for gravity on the surface is \( g = \frac{GM}{R^2} \). Therefore, the gain in potential energy can be rewritten using \( g \) as \( \Delta U = \frac{GMm}{2R} = \frac{1}{2} mgR \).

Key Concepts

Acceleration Due to Gravity

The concept of acceleration due to gravity, denoted as \( g \), is vital in understanding gravitational potential energy. It represents the acceleration with which an object moves towards Earth due to Earth's gravitational pull. This value is approximately \( 9.81 \text{ m/s}^2 \), a universally accepted average, calculated at sea level. However, note that \( g \) can slightly vary depending on geographical location and altitude. Understanding \( g \) is crucial because it affects how potential energy is calculated when an object is elevated. When you lift an object off the ground, the change in height and hence potential energy depends directly on \( g \). For situations like raising an object to Earth's radius height, knowing \( g \) helps ascertain the increase in potential energy using the potential energy formula.

Potential Energy Formula

The potential energy formula is a fundamental tool for calculating the energy an object possesses due to its position. This formula is given by: \[ U = mgh \] Where:

• \( U \) is the potential energy
• \( m \) is the mass of the object
• \( g \) is the acceleration due to gravity
• \( h \) is the height above a reference point (usually the ground)

This equation tells us that potential energy is directly proportional to the mass of the object, the gravitational pull it feels, and how high it is lifted. When an object is elevated to a significant height, like one equal to Earth's radius, the height \( h \) becomes a central value, drastically affecting the overall potential energy. By substituting these values into the formula, we can determine how much energy has been stored in the object due to its heightened position.

Gravitational Constant

The gravitational constant, symbolized as \( G \), is a key figure in gravitational physics, known to be approximately \( 6.674 \times 10^{-11} \text{ m}^3\text{/kg s}^2 \). This constant plays a crucial role when calculating gravitational forces and potential energy for cosmic distances, like the radius of Earth. In the exercise solution, \( G \) is indispensable for determining the initial and final potential energy when an object is raised to a height equal to Earth's radius. The relationship \( F = \frac{GMm}{R^2} \) is core to these calculations, reflecting the force of gravity between two masses \( M \) and \( m \) separated by a distance \( R \). When addressing potential energy change in raising an object, the formula ties \( G \) with Earth's mass and radius, helping compute the energy transition in a comprehensive manner. Understanding \( G \) allows one to bridge the gap between simplified gravitational concepts and complex real-world applications.