[OTO] Simplified calculations of last task

2 files changed, 26 insertions, 38 deletions

diff --git a/Year 3/General theory of relativity/Project.pdf b/Year 3/General theory of relativity/Project.pdf
index 0c2dfc2..bd8a456 100644
--- a/Year 3/General theory of relativity/Project.pdf
+++ b/Year 3/General theory of relativity/Project.pdf
diff --git a/Year 3/General theory of relativity/Project.tex b/Year 3/General theory of relativity/Project.tex
index 6250fca..195b2ae 100644
--- a/Year 3/General theory of relativity/Project.tex
+++ b/Year 3/General theory of relativity/Project.tex
@@ -264,20 +264,20 @@ $0\text{MI}080002\mathbf{\underline{8}}$, следователно ще разг
 
 Ще използваме, че:
 \begin{itemize}
-	\item гравитационния радиус на слънцето е $\alpha_\text{Слънце} = 2.953$ km
+	\item гравитационния радиус на слънцето е $\alpha_\text{Слънце} \approx 2.9$ km
 	\item средния радиус на Нептун е $r_\text{средно,Нептун} = 4.5 \cdot 10^9$ km
-	\item ексцентритетът на Нептун е $e = 0.008678$
-	\item Нептун прави една обиколка (около Слънцето) на всеки $164.8$ години
+	\item ексцентритетът на Нептун е $e \approx 0.008$
+	\item Нептун прави една обиколка (около Слънцето) на всеки $164$ години
 \end{itemize}
 
 По формулата на Айнщайн:
 
 \begin{align*}
-	\Delta \varphi & \approx \frac{3 \pi \alpha_\text{Слънце}}{r_\text{средно,Нептун} (1 - e^2)} \approx \frac{3 \pi \cdot 2.953}{4.5 \cdot 10^9 \cdot (1 - 0.008678^2)} \approx \pi \frac{8859 \cdot 10^{-3}}{4.5 \cdot 10^9 \cdot (1 - 0.000075307684)} \\
-	& \approx \pi \frac{8859}{4.5 \cdot 10^{12} \cdot 0.999924692316} \approx 6.1852145315485203527881859... \cdot 10^{-9} \; \text{ радиани на обиколка} \\
+	\Delta \varphi & \approx \frac{3 \pi \alpha_\text{Слънце}}{r_\text{средно,Нептун} (1 - e^2)} \approx \frac{3 \pi \cdot 2.9}{4.5 \cdot 10^9 \cdot (1 - 0.008^2)} = \pi \frac{87}{45 \cdot 10^9 \cdot (1 - 0.000064)} \\
+	& \approx \pi \frac{1.9333}{10^{9} \cdot 0.999936} \approx \pi \cdot 1.9334 \cdot 10^{-9} \; \text{ радиани на обиколка} \\
 	& \text{Превръщайки  към дъгови секунди на 100 години}: \\
-	& \approx 0.0000000061852145315485203527881859... \cdot \frac{180}{\pi} 3600 \cdot \frac{100}{164.8} \\
-	& \approx 0.00077414567775884732... \, ''
+	& = \cancel{\pi} \cdot 1.9334 \cdot 10^{-9} \cdot \frac{180}{\cancel{\pi}} 3600 \cdot \frac{100}{164} = 1.9334 \cdot 10^{-9} \cdot 18 \cdot 36 \cdot 10^3 \cdot \frac{10^2}{164} \\
+	& = 1.9334 \cdot 10^{-4} \cdot 8 \cdot 9 \cdot 9 \cdot \frac{1}{164} = 1.9334 \cdot 10^{-4} \cdot \frac{81}{20.5} \approx 0.00076 \, ''
 \end{align*}
 
 По точната формула:
@@ -287,51 +287,39 @@ $0\text{MI}080002\mathbf{\underline{8}}$, следователно ще разг
 \end{align*}
 
 \begin{align*}
-	& r_\text{min} = (1 - e) r_\text{средно,Нептун} = (1 - 0.008678) \cdot 4.5 \cdot 10^9 = 0.991322 \cdot 4.5 \cdot 10^9 = 4.460949 \cdot 10^9 \\
-	& r_\text{max} = (1 + e) r_\text{средно,Нептун} = (1 + 0.008678) \cdot 4.5 \cdot 10^9 = 1.008678 \cdot 4.5 \cdot 10^9 = 4.539051 \cdot 10^9 \\
-	& \frac{\alpha_\text{Слънце}}{r_\text{min}} = \frac{2.953}{4.460949 \cdot 10^9} = \frac{2953}{4460949 \cdot 10^3}
-	\qquad \frac{\alpha_\text{Слънце}}{r_\text{max}} = \frac{2.953}{4.539051 \cdot 10^9} = \frac{2953}{4539051 \cdot 10^3} \\
+	& r_\text{min} = (1 - e) r_\text{средно,Нептун} = (1 - 0.008) \cdot 4.5 \cdot 10^9 = 0.992 \cdot 4.5 \cdot 10^9 \approx 4.4 \cdot 10^9 \\
+	& r_\text{max} = (1 + e) r_\text{средно,Нептун} = (1 + 0.008) \cdot 4.5 \cdot 10^9 = 1.008 \cdot 4.5 \cdot 10^9 \approx 4.5 \cdot 10^9 \\
+	& \frac{\alpha_\text{Слънце}}{r_\text{min}} = \frac{2.9}{4.4 \cdot 10^9} = \frac{29}{44 \cdot 10^9}
+	\qquad \frac{\alpha_\text{Слънце}}{r_\text{max}} = \frac{2.9}{4.5 \cdot 10^9} = \frac{29}{45 \cdot 10^9} \\
 \end{align*}
 
 \begin{align*}
-	\sqrt{1 - \frac{\alpha_\text{Слънце}}{r_\text{min}} - \frac{2 \alpha_\text{Слънце}}{r_\text{max}}} & = \sqrt{1 - \frac{2953}{4460949 \cdot 10^3} - 2 \frac{2953}{4539051 \cdot 10^3}} \\
-	& = \sqrt{\frac{4460949 \cdot 10^3 \cdot 4539051 \cdot 10^3 - 2953 \cdot 4539051 - 2 \cdot 2953 \cdot 4460949}{4460949 \cdot 10^3 \cdot 4539051 \cdot 10^3}} \\
-	& = \sqrt{\frac{1}{4460949 \cdot 10^3 \cdot 4539051 \cdot 10^3}} \sqrt{20248474979648817603}
+	\sqrt{1 - \frac{\alpha_\text{Слънце}}{r_\text{min}} - \frac{2 \alpha_\text{Слънце}}{r_\text{max}}} & = \sqrt{1 - \frac{29}{44 \cdot 10^9} - 2 \frac{29}{45 \cdot 10^9}} \\
+	& = \sqrt{\frac{44 \cdot \cancel{10^9} \cdot 45 \cdot 10^9 - 29 \cdot 45 \cdot \cancel{10^9} - 2 \cdot 29 \cdot 44 \cdot \cancel{10^9}}{44 \cdot \cancel{10^9} \cdot 45 \cdot 10^9}} \\
+	& = \sqrt{\frac{1980 \cdot 10^9 - 1305 - 2 \cdot 1276}{1980 \cdot 10^9}} \\
+	& = \sqrt{\frac{1980 \cdot 10^{9} - 3857}{1980 \cdot 10^{9}}} \approx 0.99999999902
 \end{align*}
 
 \begin{align*}
-	\sqrt{1 - \frac{2 \alpha_\text{Слънце}}{r_\text{min}} - \frac{\alpha_\text{Слънце}}{r_\text{max}}} & = \sqrt{1 - 2 \frac{2953}{4460949 \cdot 10^3} - \frac{2953}{4539051 \cdot 10^3}} \\
-	& = \sqrt{\frac{4460949 \cdot 10^3 \cdot 4539051 \cdot 10^3 - 2 \cdot 2953 \cdot 4539051 - 2953 \cdot 4460949}{4460949 \cdot 10^3 \cdot 4539051 \cdot 10^3}} \\
-	& = \sqrt{\frac{1}{4460949 \cdot 10^3 \cdot 4539051 \cdot 10^3}} \sqrt{20248474979418182397}
+	\sqrt{1 - \frac{2 \alpha_\text{Слънце}}{r_\text{min}} - \frac{\alpha_\text{Слънце}}{r_\text{max}}} & = \sqrt{1 - 2 \frac{29}{44 \cdot 10^9} - \frac{29}{45 \cdot 10^9}} \\
+	& = \sqrt{\frac{44 \cdot \cancel{10^9} \cdot 45 \cdot 10^9 - 2 \cdot 29 \cdot 45 \cdot \cancel{10^9} - 29 \cdot 44 \cdot \cancel{10^9}}{44 \cdot \cancel{10^9} \cdot 45 \cdot 10^9}} \\
+	& = \sqrt{\frac{1980 \cdot 10^9 - 2 \cdot 1305 - 1276}{1980 \cdot 10^9}} \\
+	& = \sqrt{\frac{1980 \cdot 10^{9} - 3886}{1980 \cdot 10^{9}}} \approx 0.99999999901
 \end{align*}
 
 \begin{align*}
 	& M \left( \sqrt{1 - \frac{\alpha_\text{Слънце}}{r_\text{min}} - \frac{2 \alpha_\text{Слънце}}{r_\text{max}}}, \sqrt{1 - \frac{2 \alpha_\text{Слънце}}{r_\text{min}} - \frac{\alpha_\text{Слънце}}{r_\text{max}}} \right) \\
-	& = M \left( \sqrt{\frac{1}{4460949 \cdot 10^3 \cdot 4539051 \cdot 10^3}} \sqrt{20248474979648817603} \; , \right. \\
-	& \qquad\quad \left. \sqrt{\frac{1}{4460949 \cdot 10^3 \cdot 4539051 \cdot 10^3}} \sqrt{20248474979418182397} \right) \\
-	& = \sqrt{\frac{1}{4460949 \cdot 4539051 \cdot 10^{6}}} M \left( \sqrt{20248474979648817603}, \sqrt{20248474979418182397} \right) \\
-	& = 2.2223059021530026666489... \cdot 10^{-10} \cdot    
-	    4.4998305500911364873496... \cdot 10^9 \\
-	& = 0.999999999015592532735669488682770254067475544... = M_\text{стойност}
+	& \approx M (0.99999999902, 0.99999999901) \\
+	& = M \left(\sqrt{0.99999999902 \cdot 0.99999999901}, \frac{0.99999999902 + 0.99999999901}{2} \right) \\
+	& \approx M(0.999999999014, 0.999999999015) \\
+	& \approx 0.9999999990145
 \end{align*}
 
 \begin{align*}
-	\frac{2 \pi}{M_\text{стойност}} = 6.28318531336480101760716936984332803180490350...
-\end{align*}
-
-\begin{align*}
-	\Delta \varphi & = 6.28318531336480101760716936984332803180490350... - 2 \pi \\
-	& = 6.1852145406818826032843222... \cdot 10^{-9} \; \text{ радиани на обиколка} \\
+	\Delta \varphi & = \frac{2 \pi}{0.9999999990145} - 2 \pi = 2 \pi \left( \frac{1}{0.9999999990145} - 1 \right) \approx 2 \pi \cdot 0.00000000098 \; \text{ радиани на обиколка} \\
 	& \text{Превръщайки  към дъгови секунди на 100 години}: \\
-	& = 0.0000000061852145406818826032843222... \cdot \frac{180}{\pi} 3600 \cdot \frac{100}{164.8} \\
-	& = 0.00077414567890198518... \, ''
-\end{align*}
-
-Приближението на Айнщайн е достатъчно точно за 11 цифри след запетаята (различаващите цифри са удебелени и подчертани):
-
-\begin{align*}
-	\text{По Айнщайн: } & \Delta \varphi \approx 0.00077414567\mathbf{\underline{775884732}}... \, '' \\
-	\text{По точната: } & \Delta \varphi =       0.00077414567\mathbf{\underline{890198518}}... \, ''
+	& = 2 \cancel{\pi} \cdot 0.00000000098 \cdot \frac{180}{\cancel{\pi}} 3600 \cdot \frac{100}{164} = 0.00000000196 \cdot 10^{5} \cdot \frac{81}{20.5} \\
+	& \approx 0.00077 \, ''
 \end{align*}
 
 \end{document}
\ No newline at end of file