+26387 Nullstellen von x+3+ln(x+3)
siehe: https://de.wikipedia.org/wiki/Fixpunktiteration
Wir setzen z= x+3
Die Gleichung muss zuerst in eine Fixpunktgleichung, also in eine Gleichung der Form
$$z= \varphi(z)$$
umgeformt werden.
$$\\ z + \ln(z) = 0 \\
z= -\ln(z) \quad | \quad e^{()} \\
e^z = e^{ -\ln(z)} \\\\
e^z = \dfrac{1} { e^{ \ln(z)} }\\\\
e^z = \dfrac{1} { z } \\\\
z = \dfrac{1} { e^z } \\$$
und erhalten die Gleichung:
$$\boxed{z = \dfrac{1} { e^z } \small{\text{ mit } x = z-3}} \\
\small{\text{
Die Iterationsformel lautet nun:
$
z_{i+1} = \dfrac{1} { e^{z_i} }
$
}}$$
Wir starten die Iteration mit
$$\small{
$
z_0 = 1 \quad ( x = z_0-3 = 1-3 = -2)\\\\
z_1 = \dfrac{1}{e^1} = 0.36787944117 \\
z_2 = \dfrac{1}{e^{0.36787944117 } } =0.69220062756\\
z_3 = \dfrac{1}{e^{0.69220062756} } =0.50047350056\\
z_4 = \dfrac{1}{e^{0.50047350056} } =0.60624353509\\
z_5 = \dfrac{1}{e^{0.60624353509} } =0.54539578598\\
z_6 = \dfrac{1}{e^{0.54539578598} } =0.57961233550\\
z_7 = \dfrac{1}{e^{0.57961233550} } =0.56011546136\\
z_8 = \dfrac{1}{e^{0.56011546136} } =0.57114311508\\
z_9 = \dfrac{1}{e^{0.57114311508} } =0.56487934739\\
$
}}$$
$$\\ \small{
$
z_{10} = \dfrac{1}{e^{0.56487934739} } =0.56842872503\\
z_{11} = \dfrac{1}{e^{0.56842872503} } =0.56641473315\\
z_{12} = \dfrac{1}{e^{0.56641473315} } =0.56755663733\\
z_{13} = \dfrac{1}{e^{0.56755663733} } =0.56690891192\\
z_{14} = \dfrac{1}{e^{0.56690891192} } =0.56727623218\\
z_{15} = \dfrac{1}{e^{0.56727623218} } =0.56706789839\\
z_{16} = \dfrac{1}{e^{0.56706789839} } =0.56718605010\\
z_{17} = \dfrac{1}{e^{0.56718605010} } =0.56711904006\\
z_{18} = \dfrac{1}{e^{0.56711904006} } =0.56715704400\\
z_{19} = \dfrac{1}{e^{0.56715704400} } =0.56713549021\\
z_{20} = \dfrac{1}{e^{0.56713549021} } =0.56714771426\\
\dots
$
}}$$
$$\\z = 0.56714329041 \\
x = z - 3= 0.56714329041 - 3\\
\textcolor[rgb]{1,0,0}{x = -2.43285670959}$$
+26387 Gliese 581e is an exoplanet with a mass of 1.9 Earths that orbits a red dwarf star at a distance of 5 x1010 m (0.33 au). If its orbital period is 124 days, find the mass of the star in kg. Divide your answer by the suns mass to see how much more or less massive the star is than our sun.
$$1 \, au = 149\ 597\ 870\ 700 \; m$$
We have:
$$\small{\text{$T = 124\ days =124 * 24 * 60 * 60\ s = 1.07136*10^7\ s $}}$$ and
$$\small{\text{$a = 0.33 au \approx 5 * 10^{10}\ m$}}$$ and
$$\small{\text{
$
m_{Exoplanet}= 1.9* m_{earth} \qquad m_{earth} = 5.97219 * 10^{24}\ kg
$
}}$$ see: https://en.wikipedia.org/wiki/Earth_mass
$$\small{\text{
$
m_{Exoplanet}= 1.13471610000* 10^{25}\ kg
$
}}$$
see: https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion
$$\boxed{
\dfrac{T^2}{a^3} = \dfrac{ 4 \pi^2}{G(M+m)}
} \qquad
\boxed { M=\dfrac{ 4 \pi^2a^3}{G T^2 } - m
}$$
where
T is the orbital period of the orbiting body,
M is the mass of the star,
G is the universal gravitational constant and
a is the radius, i.e. the semi-major axis of the ellipse.
m is the mass of the orbiting body.
The gravitational constant is : see: https://en.wikipedia.org/wiki/Gravitational_constant
$$\small{\text{
$
G = 6.67384 \cdot 10^{-11} \dfrac{m^3}{kg \cdot s^2}
$
}}$$
$$\small{\text{
$
M=\dfrac{ 4 \pi^2 (5 * 10^{10}\ m)^3}{
6.67384 \cdot 10^{-11} \dfrac{m^3}{kg \cdot s^2}
* (1.07136*10^7\ s) ^2
} - 1.13471610000* 10^{25}\ kg
$
}}\\\\\\
\small{\text{
$
M=\dfrac{ 4 \pi^2 *5^3 * 10^{30}}{
6.67384 \cdot 10^{-11}* 1.07136^2*10^{14
}}\ kg
- 1.13471610000* 10^{25}\ kg
$
}}\\\\\\
\small{\text{
$
M=\dfrac{ 4 \pi^2 *5^3 * 10^{30}\cdot 10^{11}\cdot 10^{-14
} }{
6.67384 * 1.07136^2}\ kg
- 1.13471610000* 10^{25}\ kg
$
}}\\\\\\
\small{\text{
$
M=\dfrac{ 4 \pi^2 *5^3 * 10^{27
} }{
6.67384 * 1.07136^2}\ kg
- 1.13471610000* 10^{25}\ kg
$
}}\\\\
\small{\text{
$
M=6.44203535336 \cdot 10^{29}\ kg
- 1.13471610000* 10^{25}\ kg
$
}}\\
\small{\text{
$
M=10^{25} (6.44203535336 \cdot 10^{4} - 1.13471610000 )\ kg
$
}}\\
\small{\text{
$
M=64419.2188175 \cdot 10^{25} \ kg
$
}}\\
\small{\text{
$
M_{star}=6.44192188175 \cdot 10^{29} \ kg
$
}}$$
Solar mass see: https://en.wikipedia.org/wiki/Solar_mass
$$\small{\text{
$
M_\odot= 1.98855 \cdot 10^{30}\ kg
$
}}\\\\
\small{\text{
$
\dfrac{ M_{star} }{ M_\odot }
= \dfrac{6.44192188175 \cdot 10^{29} \ kg }
{1.98855 \cdot 10^{30}\ kg }
$
}}\\\\
\small{\text{
$
= \dfrac{6.44192188175 \cdot 10^{-1}}
{1.98855}$
}}\\\\
\small{\text{
$
= 3.23950711913 \cdot 10^{-1}
$
}}\\
\small{\text{
$
= 0.323950711913 $
}}\\
\small{\text{
$
M_{star}= 0.324\ * M_\odot $
}}$$
