(a) How many astronomical units (AU) make 1 parsec?
(b) Consider a sunlike star at a distance of 2 parsecs. When it is seen through a telescope with 100 magnification, what should be the angular size of the star? Sun appears to be (1/2)° from the earth. Due to atmospheric fluctuations, eye cannot resolve objects smaller than 1 arc minute.
(c) Mars has approximately half of the earth's diameter. When it is closest to the earth it is at about 1/2 AU from the earth. Calculate what size it will appear when seen through the same telescope.
(a) By definition,
1 parsec $=$ Distance at which 1 AU long arc subtends an angle of 1 s .
$\therefore \quad 1 \text { parsec }=\left(\frac{1 \mathrm{AU}}{1 \operatorname{arc~sec}}\right)$
$1 \mathrm{deg}=3600 \text { arc sec }$
$\therefore \quad 1 \text { parsec }=\frac{\pi}{3600 \times 180} \mathrm{rad}$
$\therefore \quad 1$ parsec $=\frac{3600 \times 180}{\pi} \mathrm{AU}=206265 \mathrm{AU} \approx 2 \times 10^5 \mathrm{AU}$
(b) Sun's diameter is $\left(\frac{1}{2}\right)^{\circ}$ at 1 AU .
Therefore, at 1 parsec, star is $\frac{1 / 2}{2 \times 10^5}$ degree in diameter $=15 \times 10^{-5} \mathrm{arc} \,\mathrm{min}$.
With 100 magnification, it should look $15 \times 10^{-3} \mathrm{arc\,min}$. However, due to atmospheric fluctuations, it will still look of about 1 arc min. It cannot be magnified using telescope.
(c) Given that $$ \frac{D_{\text {mars }}}{D_{\text {earth }}}=\frac{1}{2} $$ ...(i)
where $D$ represents diameter.
From answer 25(e)
we know that,
$$ \begin{aligned} & \frac{D_{\text {earth }}}{D_{\text {sun }}}=\frac{1}{100} \\\\ & \frac{D_{\text {mars }}}{D_{\text {sun }}}=\frac{1}{2} \times \frac{1}{100}\,\,\,\, \text{[from Eq. (i)]}\end{aligned} $$
$\begin{aligned} & \text { At } 1 \mathrm{AU} \text { sun's diameter }=\left(\frac{1}{2}\right)^{\circ} \\\\ & \therefore \quad \text { mar's diameter }=\frac{1}{2} \times \frac{1}{200}=\frac{1}{400} \end{aligned}$
$\text { At } \frac{1}{2} \mathrm{AU}, \text { mar's diameter }=\frac{1}{400} \times 2^{\circ}=\left(\frac{1}{200}\right)^{\circ}$
With 100 magnification, Mar's diameter $=\frac{1}{200} \times 100^{\circ}=\left(\frac{1}{2}\right)^{\circ}=30^{\prime}$
This is larger than resolution limit due to atmospheric fluctuations. Hence, it looks magnified.
Einstein’s mass-energy relation emerging out of his famous theory of relativity relates mass ($m$) to energy ($E$) as $ E = mc^2 $, where $c$ is the speed of light in vacuum. At the nuclear level, the magnitudes of energy are very small. The energy at the nuclear level is usually measured in MeV, where $1$ MeV $= 1.6 \times 10^{-13}$ J; the masses are measured in unified atomic mass unit ($u$) where, $1u = 1.67 \times 10^{-27}$ kg.
(a) Show that the energy equivalent of $1u$ is $931.5$ MeV.
(b) A student writes the relation as $1u = 931.5$ MeV. The teacher points out that the relation is dimensionally incorrect. Write the correct relation.
In this problem, we have to apply Einstein’s mass-energy relation. $E = mc^2$, to calculate the energy equivalent of the given mass.
(a) We know that
$$ 1 \text{ amu} = 1u = 1.67 \times 10^{-27} \text{kg} $$
Applying $ E = mc^2 $
Energy $E$ = $( 1.67 \times 10^{-27})( 3 \times 10^8)^2$ J
= $ 1.67 \times 9 \times 10^{-11} $ J
$$ E = \dfrac{ 1.67 \times 9 \times 10^{-11}}{ 1.6 \times 10^{-13}} \text{MeV} \approx 939.4 \text{MeV} \approx 931.5 \text{MeV} $$
(b) The dimensionally correct relation is
$$ 1 \text{ amu} \times c^2 = 1u \times c^2 = 931.5 \text{MeV} $$