Edits to “Statement”, “Solution”, “Answer”

astrosander edited Mar 31, 2025, 3:23 PM

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revision #13966 parent #13965 ← older

en/2.6.53.md +10 −10

		@@ -1,12 +1,12 @@
			### Statement

		−	$2.6.53.$ The famous physicist Freeman Dyson suggested that an advanced space civilization could fully harness the energy of a star by surrounding it with a spherical shell. Find the mechanical stress in a stationary, homogeneous shell that surrounds the Sun in this manner, assuming its radius equals the Earth's orbital radius. The density of the shell material is $ \rho = 4 \times 10^3 \, \text{kg/m}^3 $.
		+	$2.6.53.$ The famous physicist Freeman Dyson suggested that an advanced space civilization could fully harness the energy of a star by surrounding it with a spherical shell. Find the mechanical stress in a stationary, homogeneous shell that surrounds the Sun in this manner, assuming its radius equals the Earth's orbital radius. The density of the shell material is $ \rho = 4 \times 10^3 \\, \text{kg/m}^3 $.

			### Solution

			The force on an infinitesimal area element of the shell is given by:
			\[
		−	\sigma \, dS = dF
		+	\sigma \\, dS = dF
			\]
			where $ \sigma $ is the mechanical stress, and $ dS $ is the differential surface area.

			From geometry, the area of a spherical cap under a constant solid angle $ \Omega $ is:
			\[
			S = R^2 \Omega
		@@ -16,36 +16,36 @@Solution
			\]
			Taking the differential:
			\[
		−	dS = 2R \, dR \, \Omega \quad \Rightarrow \quad dR = \frac{dS}{2R \Omega}
		+	dS = 2R \\, dR \\, \Omega \quad \Rightarrow \quad dR = \frac{dS}{2R \Omega}
			\]

			The gravitational force on a differential mass element is:
			\[
		−	dF = g \, dm
		+	dF = g \\, dm
			\]
			By Gauss’s Law for gravity:
			\[
			\oint \vec{g} \cdot d\vec{S} = -4\pi G M \quad \Rightarrow \quad g = \frac{GM}{R^2}
			\]

			The mass of the differential shell element is:
			\[
		−	dm = \rho \, dV
		+	dm = \rho \\, dV
			\]

			Assuming the shell is thin, the differential volume can be written in terms of $ dS $:
			\[
		−	dV = 4\pi R^2 \, dR = \frac{R \, dS}{2} \quad \Rightarrow \quad dm = \rho \cdot \frac{R \, dS}{2}
		+	dV = 4\pi R^2 \\, dR = \frac{R \\, dS}{2} \quad \Rightarrow \quad dm = \rho \cdot \frac{R \\, dS}{2}
			\]

			Substitute into the expression for $ dF $:
			\[
		−	dF = \frac{GM}{R^2} \cdot \rho \cdot \frac{R \, dS}{2}
		+	dF = \frac{GM}{R^2} \cdot \rho \cdot \frac{R \\, dS}{2}
			\]

			Now substitute into the original stress equation:
			\[
		−	\sigma \, dS = \frac{GM}{R^2} \cdot \rho \cdot \frac{R \, dS}{2}
		+	\sigma \\, dS = \frac{GM}{R^2} \cdot \rho \cdot \frac{R \\, dS}{2}
			\quad \Rightarrow \quad
			\sigma = \frac{GM \rho}{2R}
			\]

			Substituting known values:

		@@ -55,11 +55,11 @@Solution

			\[
			\sigma = \frac{(6.674 \times 10^{-11})(1.989 \times 10^{30})(4 \times 10^3)}{2 \times (1.496 \times 10^{11})}
		−	\approx \boxed{1.8 \times 10^{12} \, \text{Pa}}
		+	\approx \boxed{1.8 \times 10^{12} \\, \text{Pa}}
			\]

			#### Answer

			$$
		−	\sigma = 1.8 \times 10^{12} \, \text{Pa}
		+	\sigma = 1.8 \times 10^{12} \\, \text{Pa}
			$$

	@@ -1,12 +1,12 @@
	### Statement		### Statement

	$2.6.53.$ The famous physicist Freeman Dyson suggested that an advanced space civilization could fully harness the energy of a star by surrounding it with a spherical shell. Find the mechanical stress in a stationary, homogeneous shell that surrounds the Sun in this manner, assuming its radius equals the Earth's orbital radius. The density of the shell material is $ \rho = 4 \times 10^3 \, \text{kg/m}^3 $.		$2.6.53.$ The famous physicist Freeman Dyson suggested that an advanced space civilization could fully harness the energy of a star by surrounding it with a spherical shell. Find the mechanical stress in a stationary, homogeneous shell that surrounds the Sun in this manner, assuming its radius equals the Earth's orbital radius. The density of the shell material is $ \rho = 4 \times 10^3 \\, \text{kg/m}^3 $.

	### Solution		### Solution

	The force on an infinitesimal area element of the shell is given by:		The force on an infinitesimal area element of the shell is given by:
	\[		\[
	\sigma \, dS = dF		\sigma \\, dS = dF
	\]		\]
	where $ \sigma $ is the mechanical stress, and $ dS $ is the differential surface area.		where $ \sigma $ is the mechanical stress, and $ dS $ is the differential surface area.

	From geometry, the area of a spherical cap under a constant solid angle $ \Omega $ is:		From geometry, the area of a spherical cap under a constant solid angle $ \Omega $ is:
	\[		\[
	S = R^2 \Omega		S = R^2 \Omega
	@@ -16,36 +16,36 @@Solution
	\]		\]
	Taking the differential:		Taking the differential:
	\[		\[
	dS = 2R \, dR \, \Omega \quad \Rightarrow \quad dR = \frac{dS}{2R \Omega}		dS = 2R \\, dR \\, \Omega \quad \Rightarrow \quad dR = \frac{dS}{2R \Omega}
	\]		\]

	The gravitational force on a differential mass element is:		The gravitational force on a differential mass element is:
	\[		\[
	dF = g \, dm		dF = g \\, dm
	\]		\]
	By Gauss’s Law for gravity:		By Gauss’s Law for gravity:
	\[		\[
	\oint \vec{g} \cdot d\vec{S} = -4\pi G M \quad \Rightarrow \quad g = \frac{GM}{R^2}		\oint \vec{g} \cdot d\vec{S} = -4\pi G M \quad \Rightarrow \quad g = \frac{GM}{R^2}
	\]		\]

	The mass of the differential shell element is:		The mass of the differential shell element is:
	\[		\[
	dm = \rho \, dV		dm = \rho \\, dV
	\]		\]

	Assuming the shell is thin, the differential volume can be written in terms of $ dS $:		Assuming the shell is thin, the differential volume can be written in terms of $ dS $:
	\[		\[
	dV = 4\pi R^2 \, dR = \frac{R \, dS}{2} \quad \Rightarrow \quad dm = \rho \cdot \frac{R \, dS}{2}		dV = 4\pi R^2 \\, dR = \frac{R \\, dS}{2} \quad \Rightarrow \quad dm = \rho \cdot \frac{R \\, dS}{2}
	\]		\]

	Substitute into the expression for $ dF $:		Substitute into the expression for $ dF $:
	\[		\[
	dF = \frac{GM}{R^2} \cdot \rho \cdot \frac{R \, dS}{2}		dF = \frac{GM}{R^2} \cdot \rho \cdot \frac{R \\, dS}{2}
	\]		\]

	Now substitute into the original stress equation:		Now substitute into the original stress equation:
	\[		\[
	\sigma \, dS = \frac{GM}{R^2} \cdot \rho \cdot \frac{R \, dS}{2}		\sigma \\, dS = \frac{GM}{R^2} \cdot \rho \cdot \frac{R \\, dS}{2}
	\quad \Rightarrow \quad		\quad \Rightarrow \quad
	\sigma = \frac{GM \rho}{2R}		\sigma = \frac{GM \rho}{2R}
	\]		\]

	Substituting known values:		Substituting known values:

	@@ -55,11 +55,11 @@Solution

	\[		\[
	\sigma = \frac{(6.674 \times 10^{-11})(1.989 \times 10^{30})(4 \times 10^3)}{2 \times (1.496 \times 10^{11})}		\sigma = \frac{(6.674 \times 10^{-11})(1.989 \times 10^{30})(4 \times 10^3)}{2 \times (1.496 \times 10^{11})}
	\approx \boxed{1.8 \times 10^{12} \, \text{Pa}}		\approx \boxed{1.8 \times 10^{12} \\, \text{Pa}}
	\]		\]

	#### Answer		#### Answer

	$$		$$
	\sigma = 1.8 \times 10^{12} \, \text{Pa}		\sigma = 1.8 \times 10^{12} \\, \text{Pa}
	$$		$$