Question

moon is seen to be of (1/2)⁰diameter from the earth. what must be the relative size compared to the earth ?

Answer 1

Using relation,
$$\begin{gathered} \mathrm{\theta }=\frac{\mathrm{l}}{\mathrm{r}} \\ \mathrm{here}, \\ \mathrm{l}=2{\mathrm{R}}_{\mathrm{m}} \\ \mathrm{and}\mathrm{r}=60{\mathrm{R}}_{\mathrm{e}} \\ \mathrm{So}, \\ \mathrm{\theta }=\frac{2{\mathrm{R}}_{\mathrm{m}}}{60{\mathrm{R}}_{\mathrm{e}}} \\ \mathrm{also}\mathrm{given}, \\ \mathrm{\theta }=(\frac{1}{2}{)}^0 \\ \mathrm{So},\frac{2{\mathrm{R}}_{\mathrm{m}}}{60{\mathrm{R}}_{\mathrm{e}}} \\ =(\frac{1}{2}{)}^0=\frac{1}{2}\times \frac{\mathrm{\pi }}{180}\mathrm{rad} \\ \mathrm{or}\frac{{\mathrm{R}}_{\mathrm{m}}}{{\mathrm{R}}_{\mathrm{e}}}=\frac{1}{2}\times \frac{\mathrm{\pi }}{180}\times \frac{60}{2}=\frac{\mathrm{\pi }}{12}=\frac{11}{42}=1/4\left(\mathrm{approx}\right) \\ \mathrm{So}, \\ \mathrm{Radius}\mathrm{of}\mathrm{moon}\mathrm{is}\mathrm{approximately}1/4\mathrm{th}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{earth}. \end{gathered}$$