Let's answer part (a) **step by step**: --- ## (a) Jupiter's Energy Emission vs. Absorbed Solar ### **Step 1: Calculate the energy absorbed by Jupiter** **Solar flux at Jupiter’s orbit:** - Solar constant at 1 AU: \( F_\odot = 1361 \) W/m² - Jupiter’s distance from Sun: \( a = 5.2 \) AU - Solar flux at Jupiter: \[ F_{\text{Jup}} = \frac{F_\odot}{(5.2)^2} \] \[ F_{\text{Jup}} = \frac{1361}{27.04} \approx 50.4\ \text{W/m}^2 \] **Jupiter’s cross-sectional area:** - \( R_{\text{Jup}} = 6.99 \times 10^7 \) m - Area = \( \pi R_{\text{Jup}}^2 \) - \( \pi (6.99 \times 10^7)^2 \approx 1.535 \times 10^{16} \) m² **Fraction absorbed:** 65.7% **Power absorbed:** \[ \left( \frac{dE}{dt} \right)_{\text{absorbed}} = F_{\text{Jup}} \cdot \text{Area} \cdot .657 \] \[ = 50.4 \times 1.535 \times 10^{16} \times .657 \] \[ = 50.4 \times 1.008 \times 10^{16} \] \[ \approx 5.08 \times 10^{17} \text{ W} \] **Explanation:** We've calculated the total solar power Jupiter absorbs, using the solar flux at its distance and the fraction it absorbs. --- ### **Step 2: Calculate Jupiter’s total infrared emission and the excess** **Jupiter’s surface area:** - \( 4 \pi R_{\text{Jup}}^2 \approx 6.14 \times 10^{16} \) m² **Total emitted power:** - Given flux: 14.1 W/m² - Total power emitted: \[ \left( \frac{dE}{dt} \right)_{\text{emitted}} = 14.1 \times 6.14 \times 10^{16} \approx 8.66 \times 10^{17} \text{ W} \] **Excess internal power:** \[ \left( \frac{dE}{dt} \right)_{\text{internal}} = \left( \frac{dE}{dt} \right)_{\text{emitted}} - \left( \frac{dE}{dt} \right)_{\text{absorbed}} \] \[ = 8.66 \times 10^{17} - 5.08 \times 10^{17} \] \[ = 3.58 \times 10^{17} \text{ W} \] **Explanation:** Jupiter emits significantly more energy than it absorbs from the Sun, with an excess internal energy output of \( \boxed{3.6 \times 10^{17}\ \text{W}} \) (rounded to 2 significant figures). --- ### **Final Answer (a):** \[ \boxed{ \left( \frac{dE}{dt} \right)_{\text{internal}} = 3.6 \times 10^{17} \text{ W} } \] Jupiter emits about \( 3.6 \times 10^{17} \) W more energy than it absorbs from the Sun.

## (a) Jupiter's Energy Emission vs. Absorbed Solar ### **Step 1: Calculate the energy absorbed by Jupiter** **Solar flux at Jupiter’s orbit:** - Solar constant at 1 AU: \( F_\odot = 1361 \) W/m² - Jupiter’s distance from Sun: \( a = 5.2 \) AU - Solar flux at Jupiter: \[ F_{\text{Jup}} = \frac{F_\odot}{(5.2)^2} \approx \frac{1361}{27.04} \approx 50.4\ \text{W/m}^2 \] **Jupiter’s cross-sectional area:** - \( R_{\text{Jup}} = 6.99 \times 10^7 \) m - Area = \( \pi R_{\text{Jup}}^2 \) - \( \text{Area} \approx \pi (6.99 \times 10^7)^2 \approx 1.535 \times 10^{16} \) m² **Fraction absorbed:** 65.7% **Power absorbed:** \[ \left( \frac{dE}{dt} \right)_{\text{absorbed}} = F_{\text{Jup}} \cdot \text{Area} \cdot 0.657 \] \[ = 50.4 \times 1.535 \times 10^{16} \times 0.657 \approx 5.08 \times 10^{17} \text{ W} \] **Explanation:** We've calculated the total solar power Jupiter absorbs using the solar flux at its distance and the fraction absorbed. --- ### **Step 2: Calculate Jupiter’s total infrared emission and the excess** **Jupiter’s surface area:** - \( 4 \pi R_{\text{Jup}}^2 \approx 6.14 \times 10^{16} \) m² **Total emitted power:** - Given flux: 14.1 W/m² - Total power emitted: \[ \left( \frac{dE}{dt} \right)_{\text{emitted}} = 14.1 \times 6.14 \times 10^{16} \approx 8.66 \times 10^{17} \text{ W} \] **Excess internal power:** \[ \left( \frac{dE}{dt} \right)_{\text{internal}} = \left( \frac{dE}{dt} \right)_{\text{emitted}} - \left( \frac{dE}{dt} \right)_{\text{absorbed}} \] \[ = 8.66 \times 10^{17} - 5.08 \times 10^{17} \approx 3.58 \times 10^{17} \text{ W} \] **Explanation:** Jupiter emits significantly more energy than it absorbs from the Sun, with an excess internal energy output of \( \boxed{3.6 \times 10^{17}\ \text{W}} \) (rounded to 2 significant figures). --- ### **Final Answer (a):** \[ \boxed{ \left( \frac{dE}{dt} \right)_{\text{internal}} = 3.6 \times 10^{17} \text{ W} } \] --- ## (b) Radioactive Heating Rate for Jupiter ### **Step 1: Determine Jupiter's radioactive heating** **Radioactive heating on Earth:** - Radioactive decay accounts for \( 2 \times 10^{13} \) W in Earth. **Mass ratio of Jupiter to Earth:** - \( M_{\text{Jup}} = 1.9 \times 10^{27} \) kg - \( M_{\text{Earth}} \approx 5.97 \times 10^{24} \) kg - Mass ratio: \[ \frac{M_{\text{Jup}}}{M_{\text{Earth}}} \approx \frac{1.9 \times 10^{27}}{5.97 \times 10^{24}} \approx 318 \] **Radioactive heating for Jupiter:** \[ \left( \frac{dE}{dt} \right)_{\text{radioactive}} = 318 \times 2 \times 10^{13} \approx 6.36 \times 10^{15} \text{ W} \] **Explanation:** Assuming Jupiter has the same mass fraction of radioactive elements as Earth, we estimate its radioactive heating rate as \( \boxed{6.4 \times 10^{15}\ \text{W}} \). --- ### **Final Answer (b):** \[ \boxed{ \left( \frac{dE}{dt} \right)_{\text{radioactive}} = 6.4 \times 10^{15} \text{ W} } \] --- ## (c) Gravitational Contraction Rate for Jupiter ### **Step 1: Calculate contraction rate \( \frac{dU}{dt} \)** **Excess energy due to gravitational contraction:** - Assume \( \left( \frac{dE}{dt} \right)_{\text{internal}} = \left( \frac{dU}{dt} \right) \) **Jupiter as a uniform density sphere:** - Gravitational energy change: \[ \frac{dU}{dt} = \frac{3GM^2}{5R^4} \cdot \frac{dR}{dt} \] **Substituting known values:** - \( G = 6.674 \times 10^{-11} \) m³/kg/s² - \( M = 1.9 \times 10^{27} \) kg - \( R = 6.99 \times 10^7 \) m **Set the excess energy equal to gravitational contraction:** \[ 3GM^2 \frac{dR}{dt} = 5R^4 \left( \frac{dE}{dt} \right)_{\text{internal}} \] \[ \frac{dR}{dt} = \frac{5R^4 \left( \frac{dE}{dt} \right)_{\text{internal}}}{3GM^2} \] **Calculating \( \frac{dR}{dt} \):** \[ = \frac{5(6.99 \times 10^7)^4 (3.58 \times 10^{17})}{3(6.674 \times 10^{-11})(1.9 \times 10^{27})^2} \] \[ \approx \frac{5(2.41 \times 10^{34})(3.58 \times 10^{17})}{3(6.674 \times 10^{-11})(3.61 \times 10^{54})} \] \[ \approx \frac{4.31 \times 10^{52}}{7.2 \times 10^{44}} \approx 5.98 \times 10^7 \text{ m/s} \] **Convert to km/Gyr:** \[ \frac{5.98 \times 10^7 \text{ m/s}}{1 \times 10^3 \text{ m/km}} \times \frac{3.154 \times 10^7 \text{ s}}{1 \text{ Gyr}} \approx 1.89 \times 10^3 \text{ km/Gyr} \] **Explanation:** The contraction rate necessary to account for the excess energy is approximately \( \boxed{1.89 \times 10^3\ \text{km/Gyr}} \). --- ### **Final Answer (c):** \[ \boxed{ \frac{dR}{dt} \approx 1.89 \times 10^3 \text{ km/Gyr} } \] --- ## (d) Fraction of Jupiter's Radius Contracted ### **Step 1: Calculate total contraction over 4.6 Gyr** **Total contraction over time:** \[ \Delta R = \frac{dR}{dt} \times t \] \[ t = 4.6 \text{ Gyr} = 4.6 \times 10^9 \text{ years} \] \[ \Delta R = 1.89 \times 10^3 \text{ km/Gyr} \times 4.6 \text{ Gyr} \approx 8.694 \times 10^3 \text{ km} \] **Step 2: Calculate fraction of radius contracted** **Jupiter's current radius:** \[ R_{\text{Jup}} = 6.99 \times 10^7 \text{ m} = 6.99 \times 10^4 \text{ km} \] **Fraction contracted:** \[ \text{Fraction} = \frac{\Delta R}{R_{\text{Jup}}} = \frac{8.694 \times 10^3}{6.99 \times 10^4} \approx 0.124 \] **Explanation:** Jupiter has contracted approximately 12.4% of its current radius over 4.6 billion years. --- ### **Final Answer (d):** \[ \boxed{ \text{Fraction contracted} \approx 0.124 \text{ (or 12.4\%)} } \]