Question

Mass \(M\) is divided into two parts \(\mathrm{xM}\) and \((1-\mathrm{x}) \mathrm{M}\). For a given separation, the value of \(\mathrm{x}\) for which the gravitational force between the two pieces becomes maximum is (A) 1 (B) 2 (C) \(1 / 2\) (D) \(4 / 5\)

Short answer

The value of x for which the gravitational force between the two pieces becomes maximum is 1/2. Therefore, the correct answer is (C) 1/2.

Step-by-step solution

Set up the gravitational force equation using the given masses

We are given two masses, xM and (1-x)M. We will plug these into the gravitational force equation: \(F = G \frac{(xM)((1-x)M)}{r^2}\)

Simplify the equation

We will now simplify the equation by expanding the terms in the numerator: \(F = G \frac{M^2(x-x^2)}{r^2}\)

Differentiate the equation with respect to x

In order to find the maximum value of F with respect to x, we will differentiate F with respect to x: \(\frac{dF}{dx} = G M^2 \frac{d(x-x^2)}{dx}\) \(\frac{dF}{dx} = G M^2 (1-2x)\)

Set the derivative equal to zero and solve for x

To find the maximum value, we will set the derivative equal to zero and solve for x: \(0 = G M^2 (1-2x)\) \(2x = 1\) \(x = \frac{1}{2}\)

Select the correct answer

The value of x for which the gravitational force between the two pieces becomes maximum is: x = 1/2 Therefore, the correct answer is (C) 1/2.

Key concepts

Newton's Law of Gravitation

Newton's Law of Gravitation is a key concept in physics that explains how two objects are attracted to each other by a force known as gravity. The gravitational force between two masses is calculated by the formula: \[ F = G \frac{m_1 m_2}{r^2} \]Where:- \(F\) is the gravitational force between the objects,- \(G\) is the gravitational constant, approximately \(6.674 \times 10^{-11} \) Nm²/kg²,- \(m_1\) and \(m_2\) are the masses of the two objects,- \(r\) is the distance between the centers of the two masses.This law shows that the force increases with the multiplication of the two masses and decreases as the distance between them increases. Hence, the separation or distance \(r\) inversely affects the gravitational force.
Newton's Law not only forms the basis for understanding planetary motions but also helps in calculating forces acting between any two massive objects in the universe.

Differentiation

Differentiation is a fundamental tool in calculus that helps us find the rate at which a quantity changes with respect to another quantity. In our exercise, we aim to determine the point where the gravitational force \(F\) is at a maximum between two parts of a mass. To find this maximum, we first set up the equation for gravitational force and then take its derivative with respect to the variable \(x\).
The derivative of a function gives us the slope of the function at any point. For our problem, differentiating \( F = G \frac{M^2(x-x^2)}{r^2} \) with respect to \(x\) gives us the change in force as \(x\) changes. This is mathematically expressed as: \[ \frac{dF}{dx} = G M^2 (1-2x) \]By setting this derivative equal to zero, we find critical points where the slope is zero, indicating potential maxima or minima. Finding where this derivation is zero helps us locate the value of \(x\) where the force between the pieces is maximized.

Optimization in Physics

Optimization in physics involves finding the conditions that yield the best possible outcome according to a particular criterion. For our problem, the goal is to determine the value of \(x\) that maximizes the gravitational force \(F\) between two portions of a divided mass. By using differentiation, particularly setting the derivative equal to zero \(0 = GM^2(1-2x)\), we solve for the values that optimize this force.
In optimization problems, the process typically involves the following steps:- Setting up an equation that describes the system.- Differentiating the equation to find how a variable affects the quantity we seek to optimize.- Solving the derivative equation for critical points to locate maxima or minima.For our exercise, solving \(2x = 1\) leads to \(x = \frac{1}{2}\), indicating that equal mass division maximizes the gravitational force. This shows how theoretical tools like calculus can be employed to find practical solutions in physical systems.