Problem 684 - JEE Physics

Question

If \(R\) is the radius of the earth and \(g\) the acceleration due to gravity on the earth's surface, the mean density of the earth is \(=\ldots \ldots \ldots\) (A) \([(4 \pi \mathrm{G}) /(3 \mathrm{~g} \mathrm{R})]\) (B) \([(3 \pi R) /(4 \mathrm{gG})]\) (C) \([(3 \mathrm{~g}) /(4 \pi \mathrm{RG})]\) (D) \([(\pi R G) /(12 g)]\)

Step-by-Step Solution

Answer

The short answer is: The mean density of the Earth is given by the formula \(\rho = \frac{3g}{4 \pi RG}\).

Step 1: Gravitational Force Formula

Recall Newton's law of gravitation, which states that the gravitational force F between two objects with masses m1 and m2, separated by a distance r, is given by the equation: \[F = G\frac{m1 \cdot m2}{r^2}\] Here, G is the universal gravitational constant.

Step 2: Define Variables and Find Mass of Earth

Let m be the mass of an object on the surface of the Earth and M be the mass of the Earth. According to Newton's law of gravitation, the gravitational force acting on the object at the surface of Earth is: \[F = G\frac{m \cdot M}{R^2}\] We know that F = m * g (gravitational force is equal to the mass of the object times the acceleration due to gravity) So, \(G\frac{mM}{R^2} = mg\) Now, we can solve for the mass of the Earth (M): \[M = \frac{g \cdot R^2}{G}\]

Step 3: Find Mean Density of the Earth

The mean density (ρ) of the Earth can be calculated using the formula for the mass and volume of a sphere: \[ρ = \frac{M}{(4/3) \pi R^3}\] Substitute the value of M from Step 2 into the mean density formula: \[ρ = \frac{\frac{g \cdot R^2}{G}}{(4/3) \pi R^3}\] Now, simplify the formula: \[ρ = \frac{3g}{4 \pi RG}\] So, the correct answer is: (C) \([(3 \mathrm{~g}) /(4 \pi \mathrm{RG})]\)

Key Concepts

Universal Gravitational Constant

The universal gravitational constant, denoted as \(G\), is a fundamental constant in physics that plays a key role in the law of universal gravitation. It characterizes the strength of the gravitational force between two bodies. The value of \(G\) is approximately \(6.674 \, \times 10^{-11} \, \text{Nm}^2/\text{kg}^2\). This constant is essential across various calculations involving gravitational forces and aids in comparing gravitational interactions across different contexts.

• \(G\) is constant and universal, meaning it does not change regardless of the location in the universe.
• It beautifully bridges gravitational force to other physical quantities like mass and distance.
• The precision with which \(G\) is measured underpins the accuracy of many gravitational physics problems.

Understanding \(G\) helps to better grasp the forces that govern the motions and interactions of celestial bodies.

Newton's Law of Gravitation

Newton's Law of Gravitation provides a comprehensive explanation for the gravitational attraction existing between two masses. According to this law, 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. The law can be mathematically expressed as:\[ F = G \frac{m_1 \cdot m_2}{r^2} \]Here, \(F\) is the gravitational force, \(m_1\) and \(m_2\) are the masses of the objects, \(r\) is the distance between the centroids of these objects, and \(G\) is the universal gravitational constant.

• The law reveals how mass and distance govern the strength of gravitational attraction.
• It also highlights that gravitational force is always attractive and acts along the line joining the center of the two masses.
• This principle is foundational in understanding both terrestrial and celestial moving bodies, such as how the moon orbits the Earth.

Newton’s law was revolutionary, enabling many advancements in predicting celestial events and exploring space dynamics.

Acceleration due to Gravity

The acceleration due to gravity, commonly denoted as \(g\), is the rate at which objects accelerate downwards towards the Earth. This acceleration is due to the planet’s gravitational pull and is approximately \(9.8 \, \text{m/s}^2\) on Earth's surface. This value can vary slightly depending on geographical locations due to differences in Earth's shape and composition.

• \(g\) plays a crucial role in the dynamics of objects under free-fall near the Earth's surface.
• It impacts how fast an object speeds up when dropped from a height without any initial velocity.
• Understanding \(g\) is essential for solving various physics problems related to projectile motion and forces.

Recognizing \(g\) helps to evaluate phenomena such as why rain falls, how tides are influenced, and how planets maintain their orbits.

Gravitational Force Formula

The gravitational force formula describes the interaction between two masses, as detailed in Newton's Law of Gravitation. This can be specifically applied to calculate the force of gravity between an object and the Earth’s surface by considering the mass of the Earth and the effects of gravitational acceleration.
The gravitational force \(F\) can be given by:\[ F = G \frac{m \cdot M}{R^2} = m \cdot g \]where:

• \(m\) is the mass of an object on the Earth's surface.
• \(M\) is the mass of the Earth.
• \(R\) is the radius of the Earth.
• \(g\) is the acceleration due to gravity.

By setting \( F = m \cdot g \), we effectively relate this force with universal constants and the Earth’s properties. This relationship is vital when deriving expressions like mean density or understanding concepts like weight. It explains how forces act on everyday objects to enormous celestial bodies.