Problem 72 - Master Resource Book in JEE Main Physics

Question

A body of mass \(m\) rises to a height \(h=R / 5\) from the surface of earth, where \(R\) is the radius of earth. If \(g\) is the acceleration due to gravity at the surface of earth, the increase in potential energy is (a) \((4 / 5) m g h\) (b) \((5 / 6) m g h\) (c) \((6 / 7) m g h\) (d) \(m g h\)

Step-by-Step Solution

Answer

The increase in potential energy is (b) \(\frac{5}{6} m g h\).

Step 1: Understanding Potential Energy Change

The change in gravitational potential energy, when a mass \( m \) is raised to a height \( h \) in the gravitational field of the Earth, is given by the difference in potential energy at height \( h \) and at the Earth's surface.

Step 2: Express Potential Energy at a Height and Surface

Potential energy at a distance \( r \) from the center of the Earth is given by \( U = -\frac{G M m}{r} \), where \( G \) is gravitational constant and \( M \) is the mass of the Earth. Thus,- At the surface, \( U_{surface} = -\frac{G M m}{R} \)- At height \( h = \frac{R}{5} \), the distance from center is \( R' = R + h = \frac{6R}{5} \), hence \( U_{height} = -\frac{G M m}{\frac{6R}{5}} = -\frac{5G M m}{6R} \).

Step 3: Calculate Change in Potential Energy

The increase in potential energy, \( \Delta U \), is:\[ \Delta U = U_{height} - U_{surface} = \left(-\frac{5G M m}{6R}\right) - \left(-\frac{G M m}{R}\right) = \frac{5G M m}{6R} - \frac{G M m}{R} \]\[ \Delta U = \frac{5G M m - 6G M m}{6R} = -\frac{G M m}{6R} \]

Step 4: Simplify for Increase in Potential Energy Using \( g \)

Using \( g = \frac{G M}{R^2} \), the change in potential energy can be simplified to a more familiar form:\[ \Delta U = \frac{G M m}{6R} = \frac{m g R}{6} \]Given that \( h = \frac{R}{5} \), the expression for \( \Delta U \):\[ \Delta U = \frac{m g \times \frac{R}{5}}{6} = \frac{m g \times R/5}{6} = \frac{m g h}{6} \]Rewriting the entire expression comparing to \( mgh \), it is clearly equal to:\[ \Delta U = \frac{5m g h}{6} \]

Step 5: Conclude the Solution

Upon simplification, the increase in potential energy when raised a height \( h = \frac{R}{5} \) yields \( \frac{5}{6} m g h \) to the first-order approximation.Thus, the correct answer is option (b): \( \frac{5}{6} m g h \).

Key Concepts

Gravitational Field

The gravitational field is a region in space where a mass experiences a force due to gravitational attraction. Think of it like the invisible arms of gravity reaching out to pull objects towards each other.
It is represented by the symbol \( g \) and its strength can be measured by how much force it exerts on a mass inside this field.

• On Earth, this is typically measured as the force per kilogram exerted on a mass, which is approximately \( 9.81 \, \text{m/s}^2 \).
• The gravitational field is stronger nearer to the Earth and weakens as you move further away.

This field is crucial for understanding how planets, stars, and satellites interact with each other. It helps us predict how objects like an apple will fall to the ground, or how the moon stays in orbit around the Earth.

Acceleration Due to Gravity

Acceleration due to gravity is the rate at which an object accelerates when falling freely in the gravitational field of a massive body, like Earth. Denoted by \( g \), it simplifies many calculations involving gravity.
On the surface of the Earth, \( g \) is approximately \( 9.81 \, \text{m/s}^2 \).

• This value can change slightly based on your location because the Earth is not a perfect sphere and its mass is not evenly distributed.
• For instance, gravity is slightly stronger at the poles compared to the equator.

Understanding acceleration due to gravity is essential for calculating how forces act on bodies in motion. For example, when a ball is dropped, it accelerates at \( g \) until it impacts the ground.

Potential Energy Change

Potential energy change refers to the difference in the energy stored in an object due to its position within a gravitational field when it is moved.
In the exercise, when a mass \( m \) is lifted to a height \( h \), its potential energy changes.

• At the surface, the potential energy is calculated using \( U = -\frac{G M m}{R} \).
• At a height \( h \) above the surface, it becomes \( U_{height} = -\frac{5G M m}{6R} \) when considering heights in terms of Earth’s radius \( R \).
• The change in potential energy is the energy at the new height minus the energy at the surface: \( \Delta U = \frac{m g h}{6} \).

By calculating this potential energy change, we can predict how much work would be needed to move an object within Earth's gravitational pull. Hence, it illustrates the work-energy principle and can help solve practical problems like determining how high a rocket can travel with a given amount of fuel.