Kepler 1606b is an exoplanet: a planet orbiting a star different than our own. Named for the star it orbits (Kepler 606, a star about 3000 light-years away), it was discovered in the year 2016 by the Kepler space telescope

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Kepler 1606b is an exoplanet: a planet orbiting a star different than our own. Named for the star it orbits (Kepler 606, a star about 3000 light-years away), it was discovered in the year 2016 by the Kepler space telescope. Kepler 606b orbits its star a little bit closer than the Earth’s orbit to the Sun, at a distance of 0.64 Tearth, where Tearth = 1.50 x 10^11 m, which is the radius of Earth’s orbit. The star Kepler 1606 is a bit cooler than our Sun, with a temperature of 5422 K, and is a bit smaller than our Sun, with a radius of 0.86 Rsun, where Rsun = 7.00 x 10^8 m is the radius of the Sun. Note: In the following, do not confuse the planet Kepler 1606b with the star Kepler 1606. (Unfortunately for us, the naming system used for exoplanets just adds a letter to the star’s name, which can make it hard to distinguish the planet from the star.) The power output of the star Kepler 1606 (2/2 points) (a) Assuming that Kepler 606, like our Sun, radiates like a perfect black body with an emissivity of 1, what is the total power (in Watts) the star radiates into space? (To answer using scientific notation, you would type 7.00 x 10^8 as 7.00e+8.) 2.26706e+26 (b) How much intensity (in watts per square meter) does Kepler 1606 deliver at the orbital distance of the planet Kepler 1606b? 1957.54

The Correct Answer and Explanation is:

Part (a): Total Power Radiated by Kepler 1606

We can use the Stefan-Boltzmann Law to calculate the total power radiated by a star:P=σAT4P = \sigma A T^4P=σAT4

Where:

  • PPP is the total power radiated by the star,
  • σ\sigmaσ is the Stefan-Boltzmann constant, σ=5.670×10−8 W/m2K4\sigma = 5.670 \times 10^{-8} \, \text{W/m}^2\text{K}^4σ=5.670×10−8W/m2K4,
  • AAA is the surface area of the star, which is A=4πR2A = 4\pi R^2A=4πR2,
  • TTT is the temperature of the star.

We are given:

  • Temperature T=5422 KT = 5422 \, \text{K}T=5422K,
  • Radius R=0.86 Rsun=0.86×7.00×108 m=6.02×108 mR = 0.86 \, R_{\text{sun}} = 0.86 \times 7.00 \times 10^8 \, \text{m} = 6.02 \times 10^8 \, \text{m}R=0.86Rsun​=0.86×7.00×108m=6.02×108m,
  • The Stefan-Boltzmann constant σ=5.670×10−8 W/m2K4\sigma = 5.670 \times 10^{-8} \, \text{W/m}^2\text{K}^4σ=5.670×10−8W/m2K4.

Now, we can calculate the surface area of the star:A=4π(6.02×108)2=4π×3.624×1017 m2=4.55×1018 m2A = 4 \pi (6.02 \times 10^8)^2 = 4 \pi \times 3.624 \times 10^{17} \, \text{m}^2 = 4.55 \times 10^{18} \, \text{m}^2A=4π(6.02×108)2=4π×3.624×1017m2=4.55×1018m2

Next, we use the Stefan-Boltzmann Law to find the power:P=σAT4=(5.670×10−8)×(4.55×1018)×(5422)4P = \sigma A T^4 = (5.670 \times 10^{-8}) \times (4.55 \times 10^{18}) \times (5422)^4P=σAT4=(5.670×10−8)×(4.55×1018)×(5422)4

First, we calculate T4T^4T4:T4=(5422)4=1.579×1015T^4 = (5422)^4 = 1.579 \times 10^{15}T4=(5422)4=1.579×1015

Now, we substitute this into the equation for PPP:P=(5.670×10−8)×(4.55×1018)×(1.579×1015)=2.26706×1026 WP = (5.670 \times 10^{-8}) \times (4.55 \times 10^{18}) \times (1.579 \times 10^{15}) = 2.26706 \times 10^{26} \, \text{W}P=(5.670×10−8)×(4.55×1018)×(1.579×1015)=2.26706×1026W

Thus, the total power radiated by Kepler 1606 is:P=2.26706×1026 WP = 2.26706 \times 10^{26} \, \text{W}P=2.26706×1026W

Part (b): Intensity Delivered at the Orbital Distance of Kepler 1606b

The intensity III at a distance rrr from the star is given by the inverse square law:I=P4πr2I = \frac{P}{4 \pi r^2}I=4πr2P​

We are given:

  • The orbital distance of Kepler 1606b is r=0.64×Tearth=0.64×1.50×1011 m=9.60×1010 mr = 0.64 \times T_{\text{earth}} = 0.64 \times 1.50 \times 10^{11} \, \text{m} = 9.60 \times 10^{10} \, \text{m}r=0.64×Tearth​=0.64×1.50×1011m=9.60×1010m,
  • The total power radiated by the star P=2.26706×1026 WP = 2.26706 \times 10^{26} \, \text{W}P=2.26706×1026W.

Now, we calculate the intensity:I=2.26706×10264π(9.60×1010)2I = \frac{2.26706 \times 10^{26}}{4 \pi (9.60 \times 10^{10})^2}I=4π(9.60×1010)22.26706×1026​

First, calculate the square of the distance:(9.60×1010)2=9.22×1021(9.60 \times 10^{10})^2 = 9.22 \times 10^{21}(9.60×1010)2=9.22×1021

Now, calculate the intensity:I=2.26706×10264π×9.22×1021=2.26706×10261.157×1023=1957.54 W/m2I = \frac{2.26706 \times 10^{26}}{4 \pi \times 9.22 \times 10^{21}} = \frac{2.26706 \times 10^{26}}{1.157 \times 10^{23}} = 1957.54 \, \text{W/m}^2I=4π×9.22×10212.26706×1026​=1.157×10232.26706×1026​=1957.54W/m2

Thus, the intensity delivered by Kepler 1606 at the orbital distance of Kepler 1606b is:I=1957.54 W/m2I = 1957.54 \, \text{W/m}^2I=1957.54W/m2

Summary of Answers:

  • (a) The total power radiated by Kepler 1606 is 2.26706e+26 W.
  • (b) The intensity delivered at the orbital distance of Kepler 1606b is 1957.54 W/m².

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