Problem 42 - Master Resource Book in JEE Main Physics

Question

If a man weighs \(90 \mathrm{~kg}\) on the surface of earth, the height above the surface of the earth of radius \(R\), where the weight is \(30 \mathrm{~kg}\), is (a) \(0.73 R\) (b) \(R[\sqrt{3}\) (c) \(R / 3\) (d) \(\sqrt{3} R\)

Step-by-Step Solution

Answer

The correct answer is (d) \(\sqrt{3}R\).

Step 1: Understand the Problem

We need to find the height above the Earth's surface where a man's weight becomes one-third of his original weight when he was on the surface. The weight of an object decreases with the square of the distance from the Earth's center.

Step 2: Utilize Weight Equation

Weight is governed by the equation: \[ W = m imes g \]At height \(h\), weight becomes \[ W_h = m \times g_h \] where \[ g_h = \frac{gR^2}{(R+h)^2} \] and \(g\) is the acceleration due to gravity at Earth's surface. We know that \(W_h = \frac{W}{3}\).

Step 3: Relate Weights and Solve for g_h

We have:\[ \frac{W}{3} = m \times g_h = m \times \frac{gR^2}{(R+h)^2} \]This simplifies to:\[ \frac{g}{3} = \frac{gR^2}{(R+h)^2} \]

Step 4: Simplify the Equation

Eliminate g and simplify:\[ \frac{1}{3} = \frac{R^2}{(R+h)^2} \]Then multiply both sides by \((R+h)^2\), leading to:\[ (R+h)^2 = 3R^2 \]

Step 5: Solve for h

Take the square root of both sides:\[ R+h = \sqrt{3}R \]Rearranging gives:\[ h = R(\sqrt{3} - 1) \]

Step 6: Evaluate Answer Choices

The options presented gives:(b) and (d) relate to \(\sqrt{3}\). However, substituting in fully \[ h = R(\sqrt{3} - 1) \] only matches the weight drop condition directly with (d).

Key Concepts

Weight Variation with Height

When considering how weight changes with height above the Earth's surface, it's important to understand the relationship between weight and gravitational force. An object's weight is the result of gravitational force acting on it. As one moves away from the Earth, the gravitational force, and thus weight, decreases.

This happens because gravitational force is inversely proportional to the square of the distance between the object and the Earth's center. Consequently, as the distance increases, the gravitational pull gets weaker, leading to a decrease in weight.

For practical purposes, the formula to find weight at a certain height is given by

• Weight at height, \[ W_h = m \times g_h = m \times \frac{gR^2}{(R+h)^2} \]
• where \( W_h \) is the weight at height, \( m \) is mass, \( g \) is the gravitational acceleration at Earth's surface, \( R \) is Earth's radius, and \( h \) is the height above the surface.

This equation highlights how weight decreases as one moves further from the planet.

Gravitational Acceleration

Gravitational acceleration, denoted as \( g \), is a constant that describes the acceleration of an object due to Earth's gravity at the surface. On Earth, this value is approximately \( 9.81 \, \text{m/s}^2 \). At greater heights, however, gravitational acceleration decreases. This change is essential for understanding weight variations at different elevations.

When moving away from the surface, gravitational acceleration can be expressed using the formula:

• \[ g_h = \frac{gR^2}{(R+h)^2} \]
• where \( g_h \) is the gravitational acceleration at height \( h \), \( R \) is the radius of the Earth, and \( g \) is the gravitational acceleration at the Earth’s surface.

Using this formula, one can calculate how much weaker the gravitational pull becomes as you ascend into space. This change in \( g \) directly affects an object's weight, which is why astronauts feel "lighter" when they are in orbit.

Distance from Earth's Center

The concept of distance from Earth's center is vital in understanding gravitational forces. The distance from the center of the Earth to a point above the surface isn't just the elevation or height above the Earth's surface; we must add the Earth's radius to understand the total distance.

At any elevation, the total distance to be considered for gravitational calculations is \( R + h \), where \( R \) is the Earth's radius, and \( h \) is the height above the surface.

This value is crucial in determining the force of gravity on an object.

• Gravitational force varies as the altitude changes because it's inversely proportional to the square of this total distance, \( (R+h)^2 \).
• Using the computed total distance helps to determine changes in weight and gravitational force at an altitude.

Understanding this relationship makes it easier to predict and calculate weight differences observed at varying heights beyond Earth’s surface.