Problem 726 - JEE Physics

Question

The Gravitational P.E. of a body of mass \(\mathrm{m}\) at the earth's surface is \(-\mathrm{mgRe}\). Its gravitational potential energy at a height \(\operatorname{Re}\) from the earth's surface will be \(=\ldots \ldots \ldots\) here (Re is the radius of the earth) (A) \(-2 \mathrm{mgRe}\) (B) \(2 \mathrm{mgRe}\) (C) \((1 / 2) \mathrm{mg} \mathrm{Re}\) (D) \(-(1 / 2) \mathrm{mg} \operatorname{Re}\)

Step-by-Step Solution

Answer

The gravitational potential energy of the body at a height Re from the earth's surface is \(-\frac{1}{2}mgRe\). The correct answer is (D) \(-\frac{1}{2}mgRe\).

Step 1: Convert the given potential energy formula to GmM/r form

First, let's convert -mgRe to the standard P.E. formula: \(-mgRe = -\frac{GmM}{Re}\) Here, g is the acceleration due to gravity near the surface of the earth, which can be written as: \(g = \frac{GM}{Re^2}\) Combine these two equations to get: \(-mgRe = -\frac{GmM}{Re}\)

Step 2: Find the potential energy at height Re

Now we find the potential energy at a height Re above the earth's surface. In this case, the distance from the center of the earth will be the sum of the radius of the earth and the height: \(r = Re + Re = 2Re\) The potential energy at this height is given by: \(P.E. = -\frac{GmM}{2Re}\)

Step 3: Convert the potential energy to the given Re form

Now let's convert this potential energy formula back to the given Re form. By using the earlier mentioned relationship: \(g = \frac{GM}{Re^2}\) We get: \(P.E. = -\frac{mgRe^2}{2Re} = -\frac{1}{2}mgRe\)

Step 4: Write the final answer

The gravitational potential energy of the body at a height Re from the earth's surface is: \(-\frac{1}{2}mgRe\) Comparing this with the given options, we find that the correct answer is: (D) \(-\frac{1}{2}mgRe\)

Key Concepts

Gravitational Force

Gravitational force is a fundamental natural force that pulls objects toward each other. It's responsible for keeping planets in orbit around stars, like the Earth around the Sun, and for holding moons around planets. Gravity acts on all objects with mass. The larger the mass, the stronger the gravitational pull.

The equation for gravitational force is given by Newton's Law of Universal Gravitation:

• \( F = \frac{G \cdot m_1 \cdot m_2}{r^2} \)

Here:

• \( F \) is the gravitational force between two objects,
• \( G \) is the gravitational constant \( (6.674 \times 10^{-11} \ \text{Nm}^2/\text{kg}^2) \),
• \( m_1 \) and \( m_2 \) are the masses of the objects,
• \( r \) is the distance between the centers of the two masses.

When calculating gravitational force, the distance \( r \) greatly influences the result. If you double the distance, the force becomes a quarter of its original value. This concept of distance also extends to calculating gravitational potential energy, as seen in the problem solution when considering the effect of height above Earth's surface on potential energy.

Earth's Radius

Earth's radius is a crucial parameter in gravitational calculations because it helps determine distances relevant to gravitational force and potential energy. The Earth's average radius \( R_e \) is approximately 6,371 kilometers (or about 3,959 miles). It's important to know that this value is an average, as Earth is not a perfect sphere but rather an oblate spheroid—wider at the equator than at the poles.

When solving problems involving gravitational potential energy at heights above Earth's surface, the concept of Earth's radius becomes essential. In the given exercise, you saw the need to double the Earth's radius when calculating potential energy at a distance \( r = 2R_e \) from Earth's center of mass.

Understanding Earth's radius allows for calculations that take into account the change in distance, an important aspect when examining how gravitational pull and potential energy change with altitude.

Acceleration due to Gravity

Acceleration due to gravity, denoted by \( g \), is a measure of how quickly an object accelerates when falling freely under the influence of Earth's gravitational field. Near Earth's surface, \( g \) is approximately \( 9.8 \, \text{m/s}^2 \). This value decreases slightly with elevation above sea level or when moving towards the poles due to Earth's shape and rotation.

The equation relating \( g \) to the gravitational constant \( G \), Earth's mass \( M \), and the Earth's radius \( R_e \) is:

• \( g = \frac{G \cdot M}{R_e^2} \)

This equation shows that \( g \) is determined by the constant \( G \), the mass of the Earth, and inversely by the square of Earth's radius. This inverse-square relationship indicates that as you move away from Earth (increase \( r \)), \( g \) decreases.

In gravitational potential energy calculations, knowing \( g \) is vital. It helps convert gravitational potential energy from the standard formula \( -\frac{GmM}{r} \) into forms involving \( g \), allowing us to express energy in terms of more familiar parameters, as handled in the exercise.