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Aspherical-satelite-orbiting-Earth-is-lighted-on-one-side-by-the-sun-with-intensity-340W-m-2-If-the-radius-of-the-satelite-is-1m-what-power-is-incident-upon-it-




Question Number 132195 by aurpeyz last updated on 12/Feb/21
Aspherical satelite orbiting Earth is  lighted on one side by the sun with intensity  340W/m^2 . If the radius of the   satelite is 1m. what power is   incident upon it?
$${Aspherical}\:{satelite}\:{orbiting}\:{Earth}\:{is} \\ $$$${lighted}\:{on}\:{one}\:{side}\:{by}\:{the}\:{sun}\:{with}\:{intensity} \\ $$$$\mathrm{340}{W}/{m}^{\mathrm{2}} .\:{If}\:{the}\:{radius}\:{of}\:{the}\: \\ $$$${satelite}\:{is}\:\mathrm{1}{m}.\:{what}\:{power}\:{is}\: \\ $$$${incident}\:{upon}\:{it}? \\ $$
Answered by ajfour last updated on 12/Feb/21
P_(incident) =I_(incident) (A_(projected) )  P_(in) =(((340W)/m^2 ))[π(1m)^2 ]  P_(in) =340π W ≈ 1068.1415W
$${P}_{{incident}} ={I}_{{incident}} \left({A}_{{projected}} \right) \\ $$$${P}_{{in}} =\left(\frac{\mathrm{340}{W}}{{m}^{\mathrm{2}} }\right)\left[\pi\left(\mathrm{1}{m}\right)^{\mathrm{2}} \right] \\ $$$${P}_{{in}} =\mathrm{340}\pi\:{W}\:\approx\:\mathrm{1068}.\mathrm{1415}{W} \\ $$
Commented by aurpeyz last updated on 12/Feb/21
Area of sphere is supposed to be  4πr^2  ?
$${Area}\:{of}\:{sphere}\:{is}\:{supposed}\:{to}\:{be} \\ $$$$\mathrm{4}\pi{r}^{\mathrm{2}} \:? \\ $$
Commented by ajfour last updated on 12/Feb/21
but projected area of the  sphere is that of a disc of  the same radius, and power  received by sphere is equal to  the power received by the disc.
$${but}\:{projected}\:{area}\:{of}\:{the} \\ $$$${sphere}\:{is}\:{that}\:{of}\:{a}\:{disc}\:{of} \\ $$$${the}\:{same}\:{radius},\:{and}\:{power} \\ $$$${received}\:{by}\:{sphere}\:{is}\:{equal}\:{to} \\ $$$${the}\:{power}\:{received}\:{by}\:{the}\:{disc}. \\ $$
Commented by aurpeyz last updated on 12/Feb/21
thank you Sir. youre correct
$${thank}\:{you}\:{Sir}.\:{youre}\:{correct} \\ $$

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