Difference between revisions of "Quantities large and small"

1 pc = $3.18 \cdot 10^{16} $m.

$\alpha _\Gamma = arctg\left( {\frac{{30\;\mbox{\it kpc}}}{{700\;\mbox{\it kpc}}}} \right) = 2^ \circ 28' = 5\alpha _ \odot $

Average human lifespan is approximately 70 years. It is easy to demonstrate that during this period a star's shift is too small to be visible by the naked eye. Assuming that the average tangential velocity of a star is $100$ km/s and distance is 100 ly (distance to nearest star, $\alpha$ Centauri, is 4 ly), the star would shift by $\alpha = 48''$ in 70 years. Let's recall for comparison that Moon's angular size is $30'$.

$2.28 \cdot 10^6$ years.

Assuming that the distance to the galaxy is $R_e$ at the moment of emission, the light will reach us in $t=R_e/c$. During this period of time the galaxy will cover the distance $\Delta R = Vt = VR_e /c$. The distance to the galaxy at the moment of observation is $R_e (1 + V/c) = R$ and, thus, $R_e = {R}/(1 + V/c)$.

See figure:

As is well known, $N_{} = N_0 e^{ - \lambda t}$, where $\tau = 1/\lambda$ is half--lifetime. Thus, the number $\left[ {U^{235} } \right]$ of $U^{235}$ nuclei and the number $\left[ {U^{238} } \right]$ of $U^{238}$ nuclei are $$ \left[ {U^{235} } \right] = \left[ {U^{235} } \right]_0 e^{ - T/\tau (U^{235} )}, $$ $$ \left[ {U^{238} } \right] = \left[ {U^{238} } \right]_0 e^{ - T/\tau (U^{238} )}, $$ where $T$ is the age of the Universe. Then, $$ {{\left[ {U^{235} } \right]} \over {\left[ {U^{238} } \right]}} = {{\left[ {U^{235} } \right]_0 } \over {\left[ {U^{238} } \right]_0 }}e^{ - T\left( {{1 \over {\tau (U^{235} )}} - {1 \over {\tau (U^{238} )}}} \right)}; $$ $$ \ln {{\left[ {U^{235} } \right]} \over {\left[ {U^{238} } \right]}} = \ln {{\left[ {U^{235} } \right]_0 } \over {\left[ {U^{238} } \right]_0 }} - T\left( {{1 \over {\tau (U^{235} )}} - {1 \over {\tau (U^{238} )}}} \right); $$ $$\displaystyle T = {{\ln {{\left[ {U^{235} } \right]_0 } \over {\left[ {U^{238} } \right]_0 }} - \ln {{\left[ {U^{235} } \right]} \over {\left[ {U^{238} } \right]}}} \over {\left( {{1 \over {\tau (U^{235} )}} - {1 \over {\tau (U^{238} )}}} \right)}}, $$ $$ T = {{\ln \left( {{{\left[ {U^{235} } \right]_0 } \over {\left[ {U^{238} } \right]_0 }}{{\left[ {U^{238} } \right]} \over {\left[ {U^{235} } \right]}}} \right)} \over {\ln 2\left( {{1 \over {t_{1/2} (U^{235} )}} - {1 \over {t_{1/2} (U^{238} )}}} \right)}}, $$ and, finally, $$ T = 9.6 \cdot 10^9 \mbox{\it years}. $$

Given the period of Sun's revolution around the Galaxy and distance to its center, the mass of matter inside its orbit could be calculated as $$M_G \approx {{4\pi ^2 R_G^3 } \over {GT_ \odot ^2 }} \approx 2\cdot10^{44}\mbox{\it ã} \approx 10^{11} M_ \odot.$$

$\rho = {{NM_ \odot } \over {L^3 }} = {{10^{11} \times 2 \cdot 10^{33} } \over {\left( {3 \cdot 10^{24} } \right)^3 }} = 0.7 \cdot 10^{ - 29}\mbox{\it g/cm}^3.$

$\displaystyle a_0 = {{\hbar ^2 } \over {m_e e^2 }} = {{\lambda _c } \over \alpha }$

The mean free path is $l \sim 1/n\sigma $, where $\sigma$ is the cross--section of hydrogen atom: $\sigma \sim 10^{ - 15} \mbox{\it cm}^2$. Let's assume that average density of matter in interstellar space is equal to critical density: $\rho _{cr} \sim 10^{ - 29} \mbox{\it g/cm}^3$, then $n = {{\rho _{cr} } / {m_p }}$, and, thus, $l \sim 10^{18} \mbox{\it cm}\sim 1 \mbox{\it pc}$.

$\Delta l \approx 1.5 \times 10^3 \mbox{\it m},~\Delta t \approx 5 \times 10^{-6} c$.

Consider the page with dimensions $20 \times 15 \mbox{\it cm}$ and mass $m=2\mbox{\it g}$. To turn it, one needs to spend the energy $$E_p = mgh = 10^{ - 3} \mbox{\it J},$$ where $h$ is the distance to the center of mass of the page.
Let's estimate the upper limit of energy collected by optical telescopes in the XX century. It is known that a star with zero apparent magnitude gives on average $\sim 10^6 $ photons (estimate this) \underline{(?in unit time?)}. Let's assume also that we observe the star with magnitude 6 on a 5-meter telescope every night for 8 hours during the entire XX century. The energy collected by the telescope is $$E \approx 10^{ - 4} \mbox{\it J}.$$ Thus, even this overestimated value is by order of magnitude smaller than the energy needed to turn a page of this book.

Quantum gravity effects are negligible when the Compton wavelength $\lambda_c = h/mc$ of a particle is much larger than its Schwarzschild radius $r_s=2mG/c^2$. The role of quantum gravity is determined by the parameter $$ {\lambda_c\over r_s}\sim {m_{Pl}^2\over m^2}. $$ For electron, for example, this parameter is $$ {\lambda_c\over r_s}\sim {m_{Pl}^2\over m^2}\sim 10^{45}, $$ so quantum gravity effects are completely negligible. The same is true for all other Standard Model particles.