Problem 814 - JEE Physics

Question

A body of mass \(\mathrm{m}\) is taken from earth surface to the height \(\mathrm{h}\) equal to radius of earth, the increase in potential energy will be (A) \(\operatorname{mg} R\) (B) \((1 / 2) \mathrm{mgR}\) (C) \(2 \mathrm{mg} \mathrm{R}\) (D) \((1 / 4) \mathrm{mgR}\)

Step-by-Step Solution

Answer

The increase in potential energy of the body is (C) \(2 \mathrm{mg} \mathrm{R}\).

Step 1: Find the potential energy at the initial point

The formula for gravitational potential energy (PE) is: PE = mgh At the initial point, the body is on the Earth's surface, so h = 0. Therefore, the initial potential energy is: PE_initial = mg(0) = 0

Step 2: Find the potential energy at the final point

At the final point, the body is at a height h = R. To find the potential energy at this height, we need to consider the gravitational force acting on the body, which is given by the formula: F_gravity = G (m1 * m2) / r^2 where G is the gravitational constant, m1 is the mass of Earth, m2 is the mass of the body, and r is the distance between the centers of mass. At the final point, r is equal to the sum of Earth's radius and the height (R + R), so r = 2R. Replacing r with 2R in the formula for gravitational force, we get: F_gravity = G (m1 * m2) / (2R)^2 Now, we know that F_gravity = mg, so: mg = G (m1 * m2) / (2R)^2 By rearranging the formula, we can find the acceleration due to gravity, g, at the final point: g_final = G (m1) / (2R)^2 Then, we can use the formula for potential energy to find PE at the final point: PE_final = m * g_final * R PE_final = m * (G (m1) / (2R)^2) * R

Step 3: Calculate the increase in potential energy

The increase in potential energy is the difference between the potential energy at the final point and the initial point: Increase in PE = PE_final - PE_initial Since PE_initial is 0, we have: Increase in PE = PE_final Increase in PE = m * (G (m1) / (2R)^2) * R Now, comparing this expression with the given options, we can identify the correct answer as: (C) \(2 \mathrm{mg} \mathrm{R}\)

Key Concepts

Earth's Radius

When discussing gravitational potential energy, the radius of Earth plays a critical role. The radius of Earth, often symbolized as \( R \), refers to the average distance from the center of the Earth to its surface.
It's approximately 6,371 kilometers or 3,959 miles.

Understanding Earth's radius helps us comprehend how the gravitational force operates at different altitudes. When a body is located at the Earth's surface, the height \( h \) in potential energy equations is zero. This means there is no potential energy with respect to the surface itself.
However, when the body is moved to a height equal to Earth's radius, \( h \) becomes \( R \).

• The gravitational potential energy depends on the radius because it affects the distance the body is from the Earth's center.
• The potential energy at the surface is calculated with respect to the Earth's center, taking \( R \) into account.

Thus, knowledge of Earth's radius is crucial for calculating the gravitational potential energy at various heights above the ground.

Gravitational Force

Gravitational force is the force of attraction between two masses. It's described by Newton's Universal Law of Gravitation. According to this law, the force \( F \) is given by:\[F = \frac{G(m_1 \cdot m_2)}{r^2}\]Here, \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses involved, and \( r \) is the distance between the centers of the two masses.

For a body near Earth, one mass \( m_1 \) is the Earth's mass and \( m_2 \) is the body's mass.
The gravitational force can also be described by:\[F = mg\]where \( g \) is the acceleration due to gravity.

• The gravitational force is responsible for the body's weight.
• As the distance \( r \) increases, the gravitational force decreases, leading to reduced weight and gravitational attraction at higher altitudes.

This fundamental understanding of gravitational force explains why potential energy formulas incorporate gravitational components, impacting energy changes as a body moves away from Earth's surface.

Height Above Earth's Surface

Height above Earth's surface, denoted as \( h \), describes how far a body is positioned from the reference level, typically Earth's surface.
This height directly influences the gravitational potential energy of the body, calculated using the formula:\[PE = mgh\]where \( PE \) is the potential energy, \( m \) is the body's mass, \( g \) is the acceleration due to gravity, and \( h \) is the height.

An increase in height means potential energy increases, because the body moves against Earth's gravity.
In our problem:

• Initially, the body is at a height of \( h = 0 \), making the potential energy zero.
• At a height \( h = R \), potential energy is substantially greater because the body is further from Earth's gravitational pull.

The concept of height above Earth's surface demonstrates how both the position and potential energy change as altitude increases, crucial for understanding the energy differences and gravitational effects experienced by a body.