The solar constant for the Earth is \(1368~\mathrm{W\,m^{-2}}\). Consider Jupiter with mass \(=320\) times Earth and distance from the Sun \(=5.2\) times Earth’s. The solar constant for Jupiter is ______________________ \(\mathrm{W\,m^{-2}}\). (Round off to the nearest integer)
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For any planet: \(S\propto r^{-2}\). A quick mental check: doubling distance \(\Rightarrow\) flux quartered; \(5\times\) farther \(\Rightarrow\) \(\sim1/25\).
Step 1 (Physics). Solar constant \(S\) at distance \(r\) from the Sun obeys the inverse-square law: \[ S(r)=\frac{L_\odot}{4\pi r^2}. \] Planet mass does not affect the received flux at TOA.
Step 2 (Scale from Earth to Jupiter). If \(S_E=1368~\mathrm{W\,m^{-2}}\) at \(r_E\), then at \(r_J=5.2\,r_E\): \[ S_J=S_E\left(\frac{r_E}{r_J}\right)^2 =1368\left(\frac{1}{5.2}\right)^2 =1368\times\frac{1}{27.04}. \]
Step 3 (Compute and round). \(S_J\approx 1368/27.04\approx 50.6~\mathrm{W\,m^{-2}}\Rightarrow \boxed{51~\mathrm{W\,m^{-2}}}.\)
Step 4 (Sense-check). Jupiter is \(\sim 5\times\) farther, so flux should be \(\sim 1/25\) of Earth’s: \(1368/25\approx 54.7\ \mathrm{W\,m^{-2}}\) — close to the precise value. Final Answer: \[ \boxed{51~\mathrm{W\,m^{-2}}} \]
