/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 67 Infinite number of masses, each ... [FREE SOLUTION] | 91Ó°ÊÓ

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Infinite number of masses, each of \(1 \mathrm{~kg}\) are placed along the \(x\)-axis at \(x=+1 m, \pm 2 m_{1}, \pm 4 m\), \(\pm 8 m, \pm 16 m \ldots\) The magnitude of the resultant gravitational potential in terms of gravitational constant \(G\) at the origin \((x=0)\) is (a) \(\mathrm{G} / 2\) (b) \(\underline{G}\) (c) \(2 \mathrm{G}\) (d) \(4 G\)

Short Answer

Expert verified
The magnitude of the gravitational potential at the origin is \( 2G \).

Step by step solution

01

Understanding Gravitational Potential

The gravitational potential due to a point mass \( m \) at a distance \( r \) from it is given by \( V = -\frac{Gm}{r} \). For multiple masses, the potentials add up algebraically.
02

Identifying Mass Locations

Infinite masses are placed at \( x = 1 \), \( x = \pm 2 \), \( x = \pm 4 \), \( x = \pm 8 \), \( x = \pm 16 \), and so on. Note the pattern where distance doubles each step.
03

Calculating Potential for All Points

The potential at the origin due to the mass at \( x = 1 \) is \( -\frac{G}{1} \). For \( x = \pm 2 \), we have \( 2 \) masses, each contributing \( -\frac{G}{2} \), so total is \( -G \cdot \frac{1}{2} \cdot 2 = -G \). Similarly, for \( x = \pm 4 \), total potential is \( -G \cdot \frac{1}{4} \cdot 2 = -G/2 \). This pattern continues, halving each step.
04

Summing the Infinite Series

The total potential is the sum of all individual potentials: \[ V = -G - 2\left(\frac{G}{2}\right) - 2\left(\frac{G}{4}\right) - 2\left(\frac{G}{8}\right) - \ldots \] This is an infinite geometric series where the first term is \( -G \) and the common ratio is \( \frac{1}{2} \).
05

Evaluating the Geometric Series

The sum of an infinite geometric series \( S \) with first term \( a \) and common ratio \( r \) is \( S = \frac{a}{1-r} \). Here, \( a = -G \) and \( r = \frac{1}{2} \), so the sum is \[ V = \frac{-G}{1 - \frac{1}{2}} = -2G \].
06

Neglecting the Negative Sign

Gravitational potential is negative, but when asked for magnitude, we take the absolute value, giving \( 2G \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Point Mass
A point mass is a theoretical concept used in physics to simplify the analysis of an object with mass. It is considered to be a mass that is concentrated at a single point in space. This allows for simpler calculations because we can assume all gravitational effects are coming from that one point.
- Think of it like a single dot with mass but no size. - We often use this to model objects in problems involving gravity where the size of the object is negligible compared to other dimensions in the problem.
In the exercise, each mass placed along the x-axis is treated as a point mass. Using the formula for gravitational potential, the contribution of each mass to the potential at the origin is calculated based on its distance from the origin. Because point masses simplify the calculations, it is crucial to understand where each mass is located and how it affects the total potential.
Geometric Series
A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. The series continues infinitely when there's no endpoint.
- The general form of a geometric series is: \( a, ar, ar^2, ar^3, \, \ldots \), where \( a \) is the first term, and \( r \) is the common ratio.- Important features of a geometric series include the first term and the common ratio.
In the problem, the various masses along the x-axis form a sequence where the distance doubles each time, creating a geometric series in terms of their contributions to the gravitational potential. The series uses a first term of \( -G \) with a common ratio of \( \frac{1}{2} \). Understanding the geometric nature of these contributions helps in summing the potentials correctly.
Infinite Series
An infinite series is the sum of an infinite sequence of terms. When dealing with physics problems, infinite series often arise when you're adding up an unending set of contributions, as is frequently the case with gravitational potentials or electric fields.
- The series may converge to a specific value, meaning that as more terms are added, the sum approaches a certain limit.- Conditions for converging depend on the nature of the series, such as whether it is geometric, arithmetic, etc.
In the context of this exercise, once the gravitational contributions of each point mass are algebraically summed, the resulting infinite series needs evaluation. With a geometric series, like in our problem, convergence occurs if the absolute value of the common ratio is less than one. The convergence formula allows us to find the sum of the infinite series. Here, the convergence of the series gives us a neat value of \( -2G \). Taking the absolute value for interpretation gives the exercise's solution: \( 2G \). Understanding how to handle infinite series is vital for accurate evaluations of physical problems like this one.

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Most popular questions from this chapter

Moment of inertia of cylinder about an axis through the centre and perpendicular to its axis is $$ I_{c}=M\left(\frac{R^{2}}{4}+\frac{L^{2}}{12}\right) $$ Using theorem of parallel axes, moment of inertia of the cylinder about an axis through its edge would be $$ I=I_{c}+M\left(\frac{L}{2}\right)^{2}=M\left(\frac{R^{2}}{4}+\frac{L^{2}}{12}+\frac{L^{2}}{4}\right)=M\left(\frac{R^{2}}{4}+\frac{L^{2}}{3}\right) $$ When \(L=6 R, \quad I_{h}=\frac{49}{4} M R^{2}\)

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