Question

The magnitude of gravitational potential energy of a body at a distance ' \( R \) ' from the centre of the earth is ' \( E \) '. Its weight at a distance ' \( 1.5 R \) ' from the centre of the earth is

• A. \( \frac{2E}{9R} \)
• B. \( \frac{4E}{5R} \)
• C. \( \frac{4E}{9R} \)
• D. \( \frac{2E}{7R} \)

Hint:

Potential Energy $U \propto \frac{1}{r}$ and Weight $W \propto \frac{1}{r^2}$. Thus, $W = \frac{|U|}{r}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Gravitational potential energy is the energy possessed by a mass due to its position in a gravitational field.
Weight is the gravitational force acting on a body.
We need to relate the given potential energy at one distance to the weight at another distance.
Step 2: Key Formula or Approach:
The magnitude of gravitational potential energy \( U \) of a body of mass \( m \) at a distance \( r \) from the center of Earth (mass \( M \)) is given by: \[ |U| = \frac{GMm}{r} \] The weight \( W \) of the body at distance \( r \) is the gravitational force acting on it: \[ W = F_g = \frac{GMm}{r^2} \] Step 3: Detailed Explanation:
We are given that the magnitude of gravitational potential energy at distance \( R \) is \( E \).
Using the formula for potential energy magnitude: \[ E = \frac{GMm}{R} \quad \dots \text{(Equation 1)} \] We can rearrange this equation to express the constant term \( GMm \) in terms of \( E \) and \( R \): \[ GMm = E \cdot R \] We need to find the weight of the body at a new distance \( r' = 1.5R = \frac{3}{2}R \).
The formula for weight at distance \( r' \) is: \[ W = \frac{GMm}{(r')^2} \] Substitute \( r' = \frac{3}{2}R \) into the weight equation: \[ W = \frac{GMm}{(\frac{3}{2}R)^2} \] \[ W = \frac{GMm}{\frac{9}{4}R^2} \] \[ W = \frac{4}{9} \left( \frac{GMm}{R^2} \right) \] Now, substitute the expression \( GMm = E \cdot R \) into this equation: \[ W = \frac{4}{9} \left( \frac{E \cdot R}{R^2} \right) \] Simplify the expression by canceling one \( R \) from the numerator and denominator: \[ W = \frac{4E}{9R} \] This is the weight of the body at a distance of \( 1.5R \) from the center of the Earth.
Step 4: Final Answer:
Its weight at a distance \( 1.5 R \) is \( \frac{4E}{9R} \).