Wiemy, że:

$r_{zs}\left[m\right]$

$S\left[\frac{W}{m^2}\right]$

$\Delta t\left[s\right]$

$c\left[\frac{m}{s}\right]$

Teraz zamieniamy waty ja jednostki podstawowe:

$\left[W\right]=\left[\frac{J}{s}\right]=\left[\frac{kg\cdot m^2}{s^3}\right]$

$a)$

$\Delta M_s=\frac{\left[m^2\right]\cdot \left[\frac{W}{m^2}\right]\cdot \left[s\right]}{{\left(\left[\frac{m}{s}\right]\right)}^2}=\frac{\left[{\sout{m}}^2\right]\cdot \left[\frac{W}{{\sout{m}}^2}\right]\cdot \left[s\right]}{{\left(\left[\frac{m}{s}\right]\right)}^2}=\frac{\left[W\right]\cdot \left[s\right]}{\left[\frac{m^2}{s^2}\right]}=\frac{\left[\frac{kg\cdot m^2}{s^3}\right]\cdot \left[s\right]}{\left[\frac{m^2}{s^2}\right]}=\frac{\left[\frac{kg\cdot m^2}{s^{\sout{3}}}\right]\cdot \left[\sout{s}\right]}{\left[\frac{m^2}{s^2}\right]}=\frac{\left[kg\cdot \frac{m^2}{s^2}\right]}{\left[\frac{m^2}{s^2}\right]}=\frac{\begin{array}{c}kg\cdot \left(\frac{{\sout{m}}^2}{{\sout{s}}^2}\right)\end{array}}{\begin{array}{c}\frac{{\sout{m}}^2}{{\sout{s}}^2}\end{array}}=\left[kg\right]$

$b)$

Korzystamy ze wzoru podanego w zadaniu:

$\Delta M_s=\frac{4\pi r_{zs}^2\cdot S\cdot \Delta t}{c^2}$

Wartości stałej słonecznej S, prędkości światła c i odległości Słońca od Ziemi r_zs wyszukujemy w tablicach fizycznych:

$S=1360,8\frac{W}{m^2}$

$r_{zs}=149,6\cdot {10}^{6}km=149,6\cdot {10}^{9}m$

$c=3\cdot {10}^8\frac{m}{s}$

Dane podstawiamy do wzoru:

$\Delta M_s=\frac{4\cdot 3,14\cdot {\left(149,6\cdot {10}^{9}m\right)}^2\cdot 1360,8\frac{W}{m^2}\cdot 1s}{{\left(3\cdot {10}^8\frac{m}{s}\right)}^2}=\frac{17158,22\frac{W\cdot s}{m^2}\cdot 2,24\cdot {10}^{22}{m}^2}{9\cdot {10}^{16}\frac{m^2}{s^2}}$

Upraszczamy jednostki:

$\Delta M_s=\frac{1,7158\cdot {10}^4\frac{W\cdot s}{{\left(\sout{m}\right)}^2}\cdot 2,24\cdot {10}^{4+18}{\left(\sout{m}\right)}^2}{9\cdot {10}^{4+12}\frac{m^2}{s^2}}=\frac{1,7158\cdot {10}^4\cdot 2,24\cdot {10}^4\cdot {10}^{18}W\cdot s}{9\cdot {10}^4\cdot {10}^{12}\frac{m^2}{s^2}}=\frac{1,7158\cdot {\sout{10}}^4\cdot 2,24\cdot {10}^4\cdot {10}^{18}W\cdot s}{9\cdot {\sout{10}}^4\cdot {10}^{12}\frac{m^2}{s^2}}=0,427\cdot \frac{{10}^4\cdot {10}^{18}}{{10}^{12}}\frac{\frac{kg\cdot m^2}{s^3}\cdot s}{\frac{m^2}{s^2}}=$

$=0,427\cdot {10}^{4+18-12}\frac{\frac{kg\cdot {\sout{m}}^2}{{\sout{s}}^3}\cdot \sout{s}}{\frac{{\sout{m}}^2}{{\sout{s}}^2}}=0,427\cdot {10}^{10}kg=4,27\cdot {10}^{9}kg$

$\Delta M_s=4,3\cdot {10}^{9}kg$