Problem 58 - Master Resource Book in JEE Main Physics

Question

There are two bodies of masses \(100000 \mathrm{~kg}\) and \(1000 \mathrm{~kg}\) separated by a distance of \(1 \mathrm{~m} .\) At what distance (in metre) from the smaller body, the intensity of gravitational field will be zero? (a) \(1 / 9\) (b) \(1 / 10\) (c) \(1 / 11\) (d) \(10 / 11\)

Step-by-Step Solution

Answer

(c) \(1 / 11\)

Step 1: Understanding the System

We have two masses, a larger mass, denoted as \( M = 100000 \, \text{kg} \), and a smaller mass, denoted as \( m = 1000 \, \text{kg} \), placed \( 1 \, \text{m} \) apart. We need to find a point along the line that connects these two masses where their gravitational fields cancel each other, resulting in zero gravitational field strength.

Step 2: Setting Up Gravitational Fields

The gravitational field intensity \( E \) due to a mass \( M \) at a distance \( r \) is given by \( E = \frac{GM}{r^2} \), where \( G \) is the gravitational constant. We find the point at which the gravitational field due to the mass \( M \) is equal in magnitude but opposite in direction to that due to the smaller mass \( m \).

Step 3: Determine the Position from Smaller Mass

Let the point be at a distance \( x \) from the smaller mass \( m \). Then, the distance from the larger mass \( M \) is \( (1 - x) \) since the total distance is \(1 \text{ m}\).

Step 4: Setting up the Equation for zero Gravitational Field

At the point where fields cancel, \( \frac{Gm}{x^2} = \frac{GM}{(1 - x)^2} \). We can cancel \( G \) from both sides to get \( \frac{m}{x^2} = \frac{M}{(1-x)^2} \).

Step 5: Simplifying the Equation

Substitute the values of \( m = 1000 \, \text{kg} \) and \( M = 100000 \, \text{kg} \) into the equation: \( \frac{1000}{x^2} = \frac{100000}{(1-x)^2} \). Cross-multiplying gives: \( 1000(1-x)^2 = 100000x^2 \).

Step 6: Solving the Equation

By expanding and simplifying, we have an equation: \( 1000 - 2000x + 1000x^2 = 100000x^2 \). Rearranging gives: \( 1000 = 100002000x - 99000x^2 \). Dividing through by 1000 for simplicity: \( (1-x)^2 = 100x^2 \). Solving this: \( (1 - x)^2 = 100x^2 \).

Step 7: Finding the Value of x

Take the square root of both sides: \( 1 - x = 10x \). Solving for \( x \) gives: \( 1 = 11x \) or \( x = \frac{1}{11} \).

Step 8: Conclusion

The position where the gravitational field intensity is zero is \( \frac{1}{11} \) meters away from the smaller body, \( m \).

Key Concepts

Gravitational Force

Gravitational force is one of the most fundamental forces of nature. It is the force that attracts two bodies with mass towards each other. This force is responsible for keeping planets in orbit around stars and governs many phenomena in the universe.
The formula to calculate the gravitational force between two masses is given as:

• \( F = \frac{Gm_1m_2}{r^2} \)

where \( F \) is the gravitational force, \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses of the two bodies, and \( r \) is the distance between the centers of the two masses.
This equation shows that gravitational force is directly proportional to the product of the two masses and inversely proportional to the square of the distance between them.

Newton's Law of Gravitation

Newton's law of gravitation is a universal principle that defines the gravitational attraction between two masses. This law states that every point mass attracts every other point mass in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
Mathematically, it is expressed as:

• \( F = \frac{Gm_1m_2}{r^2} \)

This law explains why objects fall towards Earth and why the Earth revolves around the Sun. The gravitational constant \( G \) is a crucial part of this equation and is a measure of the strength of gravity in our universe.
The significance of Newton's law is that it applies universally, meaning it can be used to analyze gravitational interactions from small scales, like apples falling from trees, to large cosmic interactions, like galaxy movements.

Equilibrium Point in Gravitational Field

An equilibrium point in a gravitational field is where the net gravitational force on a body is zero. This means if you place an object at this point, it would not feel any gravitational pull in either direction, essentially being stationary without any net force.
In the context of two masses, this concept is explored by finding a point where the gravitational field intensity is zero. At this point, the attractions from both masses cancel out, creating a balance.
The equation used to find this point considers the gravitational field due to each mass, such as:

• \( \frac{Gm}{x^2} = \frac{GM}{(d-x)^2} \)

where \( d \) is the total distance between the masses, \( x \) is the distance from one mass where the field is zero, \( m \) and \( M \) are the masses, and \( G \) is the gravitational constant.
By solving this equation, you can determine the exact location of the equilibrium point, important in understanding mass interactions and designing stable orbits in space missions.