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The force, \(F,\) on a satellite in Earth's upper atmosphere depends on the mean path length of molecules, \(\lambda\), the density, \(\rho\), the diameter of the body, \(D\), and the molecular speed, \(c\). Express this functional relationship in terms of dimensionless groups.

Short Answer

Expert verified
The force can be expressed as a function of dimensionless groups: \(F \lambda^{-3} \rho^{-1} c^{-2} = f(D \lambda^{-1})\).

Step by step solution

01

Understanding Dimensions

First, identify the dimensions of each variable. Force, \(F\), has dimensions of mass \(M\), length \(L\), and time \(T\) as \([F] = MLT^{-2}\). Mean path length \(\lambda\), density \(\rho\), diameter \(D\), and molecular speed \(c\) have dimensions of \([\lambda] = L\), \([\rho] = ML^{-3}\), \([D] = L\), and \([c] = LT^{-1}\) respectively.
02

Select Repeating Variables

Choose three repeating variables that incorporate all the fundamental dimensions. Here, select \(\lambda\), \(\rho\), and \(c\) as the repeating variables, as they cover all necessary dimensions \((L, M, T)\). Note that \(\lambda\) is indicative of length, \(\rho\) indicates mass per volume, and \(c\) represents speed.
03

Constructing Pi Groups

Using the repeating variables, construct dimensionless groups (or Pi groups). Focus on the Force \(F\) initially and represent it as a product of powers of the repeating variables: \(\Pi_1 = F \lambda^a \rho^b c^c\). Balance the dimensional equation: \([MLT^{-2}] [L]^a [ML^{-3}]^b [LT^{-1}]^c = [1]\), giving three equations: \(M: 1 + b = 0\), \(L: 1 + a -3b + c = 0\), \(T: -2 - c = 0\). Solve these to find \(a, b, c\).
04

Solving Dimensional Equations

From the equations: \(b = -1\), \(c = -2\), and substituting into \(1 + a - 3(-1) + 2 = 0\) gives \(a = -3\). Thus, \(\Pi_1 = F \lambda^{-3} \rho^{-1} c^{-2}\).
05

Consider Remaining Variables

Now, consider the relationship involving the diameter \(D\). Construct another dimensionless group with diameter: \(\Pi_2 = D \lambda^{-1}\). Since \(\Pi_2\) is dimensionless, this relation is straightforward.
06

Formulate Functional Relationship

Combine these dimensionless groups into an equation: \(\Pi_1 = f(\Pi_2)\), or \(F \lambda^{-3} \rho^{-1} c^{-2} = f(D \lambda^{-1})\), indicating that the relationship between these variables can be expressed functionally using these non-dimensional groups.

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Key Concepts

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

Dimensionless Groups
In many scientific and engineering problems, **dimensionless groups** are used. These groups are combinations of variables from a problem, designed in such a way that they do not have any physical dimensions. This means they are pure numbers. The wizards behind these groups help simplify complex equations and make them easier to handle.

By transforming all involved quantities into dimensionless groups, we can often reduce the number of variables in a problem. This is done without losing physical meaning. For instance, when expressing a force like the one on a satellite, we create dimensionless groupings using the variables such as - mean path length, - density, - speed, and - diameter.
These groups capture the essence of the forces in play, without bringing in the specific units such as meters or seconds.
Pi Theorem
The **Pi Theorem** is a key concept in dimensional analysis. It helps us systematically create dimensionless groups. This theorem states that if you have a problem involving different physical variables, you can reduce these variables to a set of dimensionless groups using fewer parameters.

For example, if a problem depends on the number of variables, including the dimensions of these variables, the Pi Theorem guides us to selecting repeating variables that cover every dimension involved. Then, using these as a base, we can form the dimensionless groups. These are called Pi groups (like \(\Pi_1\)and \(\Pi_2\)in this exercise). With the Pi Theorem, a complex problem can be broken into more manageable dimensionless components, greatly simplifying analysis and comparison.
Physical Dimensions
**Physical dimensions** are core elements to understanding how quantities relate to each other physically. Every quantity has dimensions that can be expressed in terms of certain basic units - mass (\(M\)), length (\(L\)), and time (\(T\)). For example, force is measured in terms of these three as \([F] = MLT^{-2}\).

Understanding the dimensions behind quantities helps us ensure that equations accurately represent natural laws. It is fundamental to the process of dimensional analysis. By carefully examining the physical dimensions of each quantity, we can confirm that similar dimensions on either side of an equation agree correctly. In the exercise under discussion, analyzing the dimensions of variables like - mean path length, - density, - speedhelps form meaningful groups by confirming constructs such as Pi groups indeed make sense dimensionally.
Force Dynamics
When looking into **force dynamics**, we are exploring how different physical variables interact to result in a force. In the context of an upper atmosphere satellite, this involves many contributing factors like the density of the atmosphere or speed of molecules.

Here, the problem examines how the force upon a satellite can be related to various factors without directly invoking specific measurements: - mean path length, - density, - diameter, and - molecular speed. The dynamics of force are tied to these elements through dimensionless analysis, showing their interplay without pinpointing numerical values.
Such an approach helps focus on understanding the forces' behavior and interactions straightforwardly, allowing for easy experimentation and verification.

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Most popular questions from this chapter

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