question_answer

How fast (in\[{{m}^{2}}/s\]) is area swept out by (i) the radius from sun to earth (ii) the radius from earth to moon? Sun to earth distance\[=1.496\times {{10}^{11}}m,\]earth to moon distance\[=\text{3}.\text{845}\times \text{1}{{0}^{\text{8}}}\] m and period of revolution of moon\[=27\frac{1}{3}\]days (Hint: Area swept\[\frac{dA}{dt}=\frac{\pi {{R}^{2}}}{T}\])

Answer:

(i) \[R=1.496\times {{10}^{11}}m;\] \[T=365\,\,days\,\,=365\times 24\times 60\times 60s\] Area swept out by radius from sun to earth, \[\frac{dA}{dt}=\frac{\pi {{R}^{2}}}{T}\] \[=\frac{\pi \times {{(1.496\times {{10}^{11}})}^{2}}}{365\times 24\times 60\times 60}=2.23\times {{10}^{15}}{{m}^{2}}{{s}^{-1}}\] (ii) \[R=3.845\times {{10}^{8}}m;\] \[T=27\frac{1}{3}days\,=\frac{82}{3}\times 24\times 60\times 60s\] Area swept out from earth to moon, \[\frac{dA}{dt}=\frac{\pi {{R}^{2}}}{T}\] \[=\frac{\pi {{(3.845\times {{10}^{8}})}^{2}}}{\frac{82}{3}\times 24\times 60\times 60}=1.97\times {{10}^{11}}{{m}^{2}}{{s}^{-1}}\]