Problem 57 - Master Resource Book in JEE Main Physics

Question

Two bodies of masses \(100 \mathrm{~kg}\) and \(1000 \mathrm{~kg}\) are separated by a distance of \(1 \mathrm{~m}\). What is the intensity of gravitational field at the mid point of the line joining them? (a) \(6.6 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}\) (b) \(2.4 \times 10^{-8} \mathrm{Nkg}^{-1}\) (c) \(2.4 \times 10^{-7} \mathrm{Nkg}^{-1}\) (d) \(2.4 \times 10^{-6} \mathrm{Nkg}^{-1}\)

Step-by-Step Solution

Answer

The intensity of gravitational field at the midpoint is approximately \(2.4 \times 10^{-7} \mathrm{~N/kg}\). Option (c) is correct.

Step 1: Understand the Problem

We are asked to find the gravitational field intensity at the midpoint of the line joining two bodies with different masses. The formula for gravitational field intensity due to a body is given by \( g = \frac{Gm}{r^2} \), where \( G \) is the gravitational constant, \( m \) is the mass of the body, and \( r \) is the distance from the body to the point where the field intensity is being calculated.

Step 2: Identify Masses and Distance

The bodies have masses \( m_1 = 100 \text{ kg} \) and \( m_2 = 1000 \text{ kg} \). The distance between the two bodies is \( 1 \text{ m} \). Since we're looking for the field at the midpoint, the distance from each body to the midpoint \( r \) is \( \frac{1}{2} \text{ m} = 0.5 \text{ m} \).

Step 3: Calculate Gravitational Field Intensity Due to First Mass

Using the formula, the gravitational field intensity at the midpoint due to the first mass is \( g_1 = \frac{G \cdot 100}{(0.5)^2} \). Given \( G = 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \), plug in the values: \[ g_1 = \frac{6.674 \times 10^{-11} \times 100}{0.25} = 2.6696 \times 10^{-8} \, \text{N/kg} \]

Step 4: Calculate Gravitational Field Intensity Due to Second Mass

Similarly, calculate the gravitational field intensity at the midpoint due to the second mass: \[ g_2 = \frac{G \cdot 1000}{(0.5)^2} = \frac{6.674 \times 10^{-11} \times 1000}{0.25} = 2.6696 \times 10^{-7} \, \text{N/kg} \]

Step 5: Determine the Direction of the Field Intensities

Since gravitational field intensity is a vector, it is directed towards the mass. At the midpoint, \( g_1 \) is directed towards the 100 kg mass and \( g_2 \) is directed towards the 1000 kg mass. These directions are opposite to each other.

Step 6: Calculate Net Gravitational Field Intensity

To find the net gravitational field intensity at the midpoint, subtract \( g_1 \) from \( g_2 \) because they are in opposite directions: \[ g = g_2 - g_1 = 2.6696 \times 10^{-7} - 2.6696 \times 10^{-8} = 2.40264 \times 10^{-7} \, \text{N/kg} \] Rounding to two significant figures gives approximately \( 2.4 \times 10^{-7} \, \text{N/kg} \).

Key Concepts

Gravitational Force

Gravitational force is a fundamental force of nature that arises because of the mass of objects. Every object with mass attracts every other object with mass in the universe, albeit with a very small magnitude unless the masses are significantly large, such as planets or stars. This force is always attractive and acts along the line joining the centers of two bodies.

The gravitational force (\( F \)) between two masses (\( m_1 \) and\( m_2 \)) separated by a distance (\( r \)), can be calculated using the formula:

• \[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]

Where \( G \) is the gravitational constant,\( 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \). The concept of gravitational force is crucial in understanding interactions at both astronomical levels and everyday phenomena like falling objects.

Unlike other forces, gravity is unique because it acts over long distances and does not require direct contact. This makes it essential in governing the motion and orbits of celestial bodies, as well as influencing tides and even the structure of galaxies.

Newton's Law of Universal Gravitation

Newton's law of universal gravitation is a cornerstone of classical physics, introduced by Sir Isaac Newton in the 17th century. It 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 their centers.

This can be expressed as:

• \[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]

Where:

• \( F \) is the gravitational force between two objects,
• \( m_1 \) and\( m_2 \) are the masses of the objects,
• \( r \) is the distance between the centers of the two masses,
• \( G \) is the gravitational constant.

Newton's law was revolutionary because it provided a unified framework for describing both the motion of objects on Earth and celestial bodies. It applies universally, making it a crucial element in understanding the gravitational pull that not only affects planets but also smaller objects like moons and artificial satellites.

It's important to note, however, that while Newton's law is practically applicable to most scenarios in physics, it has limitations at quantum scales or where relativistic physics becomes necessary, such as near a black hole or at the speed of light.

Vector Addition in Physics

In physics, vector addition is a fundamental process used to sum two or more vectors, considering both their magnitude and direction. Vectors represent quantities that have both size and direction, like forces, velocity, and acceleration. This is different from scalar quantities that only have magnitude.

For any vector addition, it’s crucial to consider:

• The direction of each vector
• The point of application or start point

To perform vector addition graphically, one might use the head-to-tail method, where vectors are drawn as arrows and positioned so the tail of one vector connects to the head of the previous one. Alternatively, when vectors are in the same or opposite directions, simple algebraic addition or subtraction can be used, respectively.

In the context of gravitational forces, vector addition is especially significant. When calculating the net gravitational field intensity at a point from multiple masses, each field is a vector that can either cancel out or amplify another based on their directions. As such, the net gravitational field is found by summing these vector quantities, just as demonstrated in the example where gravitational fields due to two masses were subtracted because they acted in opposite directions.