Правка раздела «Solution»

Ismail правка от 27 февр. 2025 г., 09:43

правка #13300 предыдущая #13299 ← раньше позже →

en/2.6.53.md +11 −11

		@@ -1,57 +1,57 @@
			### Statement

			$2.6.53.$ The famous physicist F. Dyson suggested that it would be possible to fully utilize the energy of stars if space civilizations could surround stars with spherical shells. Find the stress in the material of a stationary homogeneous shell that would surround the Sun according to this assumption, with its radius equal to the radius of the Earth’s orbit. The density of the shell material $ρ =4·10^3 kg/m^3$ .

			### Solution
		−
		+	The force on a small area of a sphere can be defined as the mechanical stress multiplied by the area of this area:
			$$
			σdS=dF
			$$
		−
		+	From stereometry we know that the area of a sphere is determined by the product of the solid angle and the square of the radius of this sphere:
			$$
			S=R²Ω
			$$
		−
		+	The thickness of the sphere is unknown to us, although it is not required. Assuming the solid angle of the sphere to be constant, we take the derivative of the area:
			$$
			dS=2RdRΩ
			$$

			$$
			dR=\frac{dS}{2RΩ}
			$$
		−
		+	Force on a small fraction of the sphere's mass:
			$$
			dF=g*dm
			$$
		−
		+	Applying Gauss's theorem, we obtain the flow of the gravitational field through the area of the sphere:
			$$
			\oint gdS=-4πGM
			$$

			$$
			-4πR²g=-4πGM
			$$
		−
		+	Let's find the gravitational field strength:
			$$
			g=\frac{GM}{R²}
			$$
		−
		+	Mass of a small part of the sphere:
			$$
			dm=ρdV
			$$
		−
		+	The derivative of volume, which can be written through the derivative of area:
			$$
			dV=4πR²dR=\frac{RdS}{2}
			$$
		−
		+	Let's substitute the force into the very first equation we wrote down:
			$$
			σdS=\frac{GM}{R²}ρ\frac{RdS}{2}
			$$
		−
		+	From here we get the working formula for mechanical stress:
			$$
			σ=\frac{GMρ}{2R}
			$$
		−
		+	Substituting the numbers, we get:
			$$
			\boxed{σ=1.8*10¹²Pa}
			$$

			#### Answer
			$$
			σ=1.8*10¹²Pa
			$$
		ещё строк без изменений 5

	@@ -1,57 +1,57 @@
	### Statement		### Statement

	$2.6.53.$ The famous physicist F. Dyson suggested that it would be possible to fully utilize the energy of stars if space civilizations could surround stars with spherical shells. Find the stress in the material of a stationary homogeneous shell that would surround the Sun according to this assumption, with its radius equal to the radius of the Earth’s orbit. The density of the shell material $ρ =4·10^3 kg/m^3$ .		$2.6.53.$ The famous physicist F. Dyson suggested that it would be possible to fully utilize the energy of stars if space civilizations could surround stars with spherical shells. Find the stress in the material of a stationary homogeneous shell that would surround the Sun according to this assumption, with its radius equal to the radius of the Earth’s orbit. The density of the shell material $ρ =4·10^3 kg/m^3$ .

	### Solution		### Solution
			The force on a small area of a sphere can be defined as the mechanical stress multiplied by the area of this area:
	$$		$$
	σdS=dF		σdS=dF
	$$		$$
			From stereometry we know that the area of a sphere is determined by the product of the solid angle and the square of the radius of this sphere:
	$$		$$
	S=R²Ω		S=R²Ω
	$$		$$
			The thickness of the sphere is unknown to us, although it is not required. Assuming the solid angle of the sphere to be constant, we take the derivative of the area:
	$$		$$
	dS=2RdRΩ		dS=2RdRΩ
	$$		$$

	$$		$$
	dR=\frac{dS}{2RΩ}		dR=\frac{dS}{2RΩ}
	$$		$$
			Force on a small fraction of the sphere's mass:
	$$		$$
	dF=g*dm		dF=g*dm
	$$		$$
			Applying Gauss's theorem, we obtain the flow of the gravitational field through the area of the sphere:
	$$		$$
	\oint gdS=-4πGM		\oint gdS=-4πGM
	$$		$$

	$$		$$
	-4πR²g=-4πGM		-4πR²g=-4πGM
	$$		$$
			Let's find the gravitational field strength:
	$$		$$
	g=\frac{GM}{R²}		g=\frac{GM}{R²}
	$$		$$
			Mass of a small part of the sphere:
	$$		$$
	dm=ρdV		dm=ρdV
	$$		$$
			The derivative of volume, which can be written through the derivative of area:
	$$		$$
	dV=4πR²dR=\frac{RdS}{2}		dV=4πR²dR=\frac{RdS}{2}
	$$		$$
			Let's substitute the force into the very first equation we wrote down:
	$$		$$
	σdS=\frac{GM}{R²}ρ\frac{RdS}{2}		σdS=\frac{GM}{R²}ρ\frac{RdS}{2}
	$$		$$
			From here we get the working formula for mechanical stress:
	$$		$$
	σ=\frac{GMρ}{2R}		σ=\frac{GMρ}{2R}
	$$		$$
			Substituting the numbers, we get:
	$$		$$
	\boxed{σ=1.8*10¹²Pa}		\boxed{σ=1.8*10¹²Pa}
	$$		$$

	#### Answer		#### Answer
	$$		$$
	σ=1.8*10¹²Pa		σ=1.8*10¹²Pa
	$$		$$
	ещё строк без изменений 5