Problem 29 - Master Resource Book in JEE Main Physics

Question

A suitable unit for gravitational constant is (a) \(\mathrm{kg}-\mathrm{m} \mathrm{s}^{-1}\) (b) \(\mathrm{Nm}^{-1} \mathrm{~s}\) (c) \(\mathrm{Nm}^{2} \mathrm{~kg}^{-2}\) (d) \(\mathrm{kg} \mathrm{ms}^{-1}\)

Step-by-Step Solution

Answer

The correct unit for the gravitational constant is (c) \( \mathrm{Nm}^{2} \mathrm{~kg}^{-2} \).

Step 1: Understand the Problem

We need to determine the correct units for the gravitational constant, which appears in Newton's law of universal gravitation: \( F = G \frac{m_1 m_2}{r^2} \). The formula relates the force \( F \), masses \( m_1 \) and \( m_2 \), and the distance \( r \).

Step 2: Recall the Formula for Gravitational Force

The gravitational force formula is \( F = G \frac{m_1 m_2}{r^2} \), where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses, and \( r \) is the distance between the centers of the masses.

Step 3: Isolate the Gravitational Constant Units

Rearrange the formula to solve for \( G \): \( G = \frac{F \cdot r^2}{m_1 \cdot m_2} \).

Step 4: Identify Units for Each Quantity

Identify the SI units: \( F \) is in Newtons (N), \( r \) is in meters (m), and \( m_1 \) and \( m_2 \) are in kilograms (kg). Hence, \( G = \frac{N \cdot m^2}{kg^2} \).

Step 5: Match with Provided Options

Check the given options to see which matches with \( \mathrm{Nm}^{2} \mathrm{~kg}^{-2} \). Option (c) is \( \mathrm{Nm}^{2} \mathrm{~kg}^{-2} \), which is the correct representation.

Key Concepts

Newton's Law of Universal Gravitation

Newton's Law of Universal Gravitation is a fundamental principle that explains the attractive force between two masses. This law states that every particle in the universe attracts every other particle 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 mathematically as:\[F = G \frac{m_1 m_2}{r^2}\]Here, \( F \) is the gravitational force, \( m_1 \) and \( m_2 \) are the two masses, \( r \) is the distance between the centers of these masses, and \( G \) is the gravitational constant. Newton formulated this law in the 17th century, and it fundamentally changed the way we understand gravity and motion. The value of \( G \) allows us to calculate the magnitude of the force, which is why identifying its units is crucial.

SI Units for Force and Mass

SI units are the standard units of measurement in science, allowing us to express quantities in a uniform system that is universally understood. For forces and masses involved in Newton's Law of Universal Gravitation:

• Force \( F \) is measured in Newtons (N). 1 Newton is defined as the force required to accelerate a 1 kg mass by 1 m/s².
• The basic unit of mass in the International System of Units (SI) is the kilogram (kg).
• Distance, \( r \), is measured in meters (m).

These units provide consistency in our calculations, allowing us to understand and compare forces across different scales and situations. When using these units to calculate the gravitational constant \( G \), it simplifies the expression as \( G = \frac{N \cdot m^2}{kg^2} \).

Dimensional Analysis in Physics

Dimensional analysis is an essential tool in physics that involves checking the consistency of dimensions in equations. It can also be used to deduce relationships between different physical quantities. This technique allows scientists and students to understand if an equation makes dimensional sense without doing detailed calculations.To apply dimensional analysis in Newton's Law of Universal Gravitation and find the units for \( G \), consider the dimensions in the equation:- Force \( F \) has dimensions of mass \( M \) times length \( L \) divided by time squared \( T^{-2} \), i.e., \([F] = [M][L][T^{-2}]\).- The distance \( r \) has dimensions of length \( L \), i.e., \([r] = [L]\).- Masses \( m_1 \) and \( m_2 \) both have dimensions of mass \( M \), i.e., \([m_1] = [M]\) and \([m_2] = [M]\).Rearranging the equation \( G = \frac{F \cdot r^2}{m_1 \cdot m_2} \) helps us determine the dimensions for \( G \) as \([G] = \frac{[M][L][T^{-2}][L]^2}{[M][M]} = [L^3][T^{-2}][M^{-1}]\). This confirms that the SI units for \( G \) are indeed \( \mathrm{Nm}^{2}\mathrm{kg}^{-2} \), which matches the correct option provided in the exercise.