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Mobile App Chapter 9: Problem 3
Find the gravitational potential energy of a sphere with a \(1 / r^{2}\) density distribution. Take the total mass of the sphere to be \(M,\) and let the density \(\rho(r)=\rho_{0} / r^{2}\) out to a radius, \(R\).
Short Answer
Expert verified
-\frac{GM^2}{2R}
Step by step solution
01
Define the Gravitational Potential Energy
Gravitational potential energy (GPE) for a continuous mass distribution is given by: \[ U = -\frac{G}{2} \int \frac{\rho(r) \rho(r') dV dV'}{| \textbf{r} - \textbf{r}' |} \], where \(G\) is the gravitational constant and \( \rho(r) \) is the density distribution.
02
Express density distribution
Given the density distribution \( \rho(r) = \frac{\rho_0}{r^2} \), we first need to determine the constant \( \rho_0 \) such that the total mass \( M \) is accurate.
03
Total Mass Calculation
Calculate the total mass \( M \) of the sphere. Total mass is \( \int_0^R \rho(r) dV = M \). Therefore, \( M = \int_0^R \frac{\rho_0}{r^2} \cdot 4\pi r^2 dr \). Solving this integral: \[ M = \rho_0 4\pi \int_0^R dr = 4\pi \rho_0 R \Rightarrow \rho_0 = \frac{M}{4\pi R} \].
04
Insert \( \rho_0 \) into Density Function
Now insert \( \rho_0 = \frac{M}{4\pi R} \) back into the density function: \[ \rho(r) = \frac{M}{4\pi R r^2} \].
05
Calculate Gravitational Potential Energy
Substitute the density back into the potential energy integral: \[ U = -\frac{G}{2} \int_0^R \int_0^R \frac{\left( \frac{M}{4\pi R r^2} \right) \left( \frac{M}{4\pi R r'^2} \right) 4\pi r^2 dr 4\pi r'^2 dr'}{| \textbf{r} - \textbf{r}' |} \].
06
Solve for Gravitational Potential Energy
Simplify and solve the potential energy integral after placing the values: \[ U = -\frac{GM^2}{2} \int_0^R \left( \frac{4\pi^2 r^2 dr}{4\pi R r^2} \right) = -\frac{GM^2}{2R} \int_0^R dr = -\frac{GM^2}{2R} (R) = -\frac{GM^2}{2R} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
gravitational potential energy
Gravitational potential energy (GPE) is a measure of the energy stored due to the gravitational attraction between objects.
For a system with continuous mass distribution, calculating the GPE involves integrating over all mass elements.
The integral form of GPE for a continuous mass distribution is:
\[ U = -\frac{G}{2} \int \frac{\rho(r) \rho(r') dV dV'}{| \textbf{r} - \textbf{r}' |} \], where:
In simple terms, it sums up the gravitational interactions within the system.
For a system with continuous mass distribution, calculating the GPE involves integrating over all mass elements.
The integral form of GPE for a continuous mass distribution is:
\[ U = -\frac{G}{2} \int \frac{\rho(r) \rho(r') dV dV'}{| \textbf{r} - \textbf{r}' |} \], where:
- \(G\) is the gravitational constant.
- \(\rho(r)\) and \(\rho(r')\) are local mass densities.
- \(dV\) and \(dV'\) are volume elements.
- \( | \textbf{r} - \textbf{r}' |\) is the distance between the volume elements.
In simple terms, it sums up the gravitational interactions within the system.
density distribution
Density distribution describes how mass is spread out in a system. Here, we have a specific density distribution given by: \[ \rho(r) = \frac{\rho_0}{r^2} \]
This means density decreases with the square of the distance from the center.
This distribution helps in understanding how mass varies across different regions of the sphere.
This means density decreases with the square of the distance from the center.
- \(\rho_0\) is a constant to be determined.
- \(r\) is the distance from the center.
This distribution helps in understanding how mass varies across different regions of the sphere.
integral calculation
Integral calculations in this context help find quantities like total mass and potential energy.
We start by determining the total mass \(M\):
\[ M = \int_0^R \rho(r) dV = \int_0^R \frac{\rho_0}{r^2} \cdot 4 \pi r^2 dr = \rho_0 4\pi \int_0^R dr \Rightarrow \rho_0 = \frac{M}{4\pi R} \]
Solving this helps us understand how much mass is contained within a given radius \(R\).
We start by determining the total mass \(M\):
\[ M = \int_0^R \rho(r) dV = \int_0^R \frac{\rho_0}{r^2} \cdot 4 \pi r^2 dr = \rho_0 4\pi \int_0^R dr \Rightarrow \rho_0 = \frac{M}{4\pi R} \]
- \( dV = 4\pi r^2 dr\) covers the spherical volume element.
Solving this helps us understand how much mass is contained within a given radius \(R\).
sphere mass distribution
Mass distribution within a sphere using density \( \rho(r) = \frac{ \rho_0 } { r^2 } \) affects gravitational potential energy:
\[ \rho(r) = \frac{ M } { 4\pi R r^2 } \]
This formulation simplifies our integral for gravitational potential energy.
\[ \rho(r) = \frac{ M } { 4\pi R r^2 } \]
- Substituting \( \rho_0 = \frac{ M } { 4 \pi R } \) into the density function gives the required expression.
This formulation simplifies our integral for gravitational potential energy.
continuous mass distribution
In continuous mass distribution, the mass isn't just concentrated at points but is spread out continuously.
This is depicted by the density function \( \rho(r) = \frac{ \rho_0 } { r^2 } \).
To find the gravitational potential energy, we replace \( \rho(r) \) in our integral:
\[ U = -\frac{ G M^2 } { 2 R } \]
Here, the integral incorporates the continuous mass distribution to calculate the GPE of the entire system.
This is depicted by the density function \( \rho(r) = \frac{ \rho_0 } { r^2 } \).
To find the gravitational potential energy, we replace \( \rho(r) \) in our integral:
\[ U = -\frac{ G M^2 } { 2 R } \]
Here, the integral incorporates the continuous mass distribution to calculate the GPE of the entire system.
