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### 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 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 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:
−$$
−\oint gdS=-4πGM
−$$
+The force on an infinitesimal area element of the shell is given by:
+\[
+\sigma \, dS = dF
+\]
+where \( \sigma \) is the mechanical stress, and \( dS \) is the differential surface area.
−$$
−-4πR²g=-4πGM
−$$
−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:
−$$
−\boxed{σ=1.8*10¹²Pa}
−$$
+From geometry, the area of a spherical cap under a constant solid angle \( \Omega \) is:
+\[
+S = R^2 \Omega
+\]
+Taking the differential:
+\[
+dS = 2R \, dR \, \Omega \quad \Rightarrow \quad dR = \frac{dS}{2R \Omega}
+\]
+The gravitational force on a differential mass element is:
+\[
+dF = g \, dm
+\]
+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:
+\[
+dm = \rho \, dV
+\]
+
+Assuming the shell is thin, the differential volume can be written in terms of \( dS \):
+\[
+dV = 4\pi R^2 \, dR = \frac{R \, dS}{2} \quad \Rightarrow \quad dm = \rho \cdot \frac{R \, dS}{2}
+\]
+
+Substitute into the expression for \( dF \):
+\[
+dF = \frac{GM}{R^2} \cdot \rho \cdot \frac{R \, dS}{2}
+\]
+
+Now substitute into the original stress equation:
+\[
+\sigma \, dS = \frac{GM}{R^2} \cdot \rho \cdot \frac{R \, dS}{2}
+\quad \Rightarrow \quad
+\sigma = \frac{GM \rho}{2R}
+\]
+
+Substituting known values:
+
+
+\[
+\sigma = \frac{(6.674 \times 10^{-11})(1.989 \times 10^{30})(4 \times 10^3)}{2 \times (1.496 \times 10^{11})}
+\approx \boxed{1.8 \times 10^{12} \, \text{Pa}}
+\]
+
#### Answer
+
$$
−σ=1.8*10¹²Pa
+\sigma = 1.8 \times 10^{12} \, \text{Pa}
$$