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, \[\frac{dA}{dt}=\frac{\pi {{R}^{2}}}{T}\] \[=\frac{\pi {{(3.845\times {{10}^{8}})}^{2}}}{\frac{82}{3}\times 24\times 60\times 60}=1.97\times {{10}^{11}}{{m}^{2}}{{s}^{-1}}\]
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