{"id":265653,"date":"2025-07-22T08:09:31","date_gmt":"2025-07-22T08:09:31","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=265653"},"modified":"2025-07-22T08:09:33","modified_gmt":"2025-07-22T08:09:33","slug":"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","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/22\/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\/","title":{"rendered":"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"},"content":{"rendered":"\n<p><br>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&#8217;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&#8217;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&#8217;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<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Part (a): Total Power Radiated by Kepler 1606<\/h3>\n\n\n\n<p>We can use the <strong>Stefan-Boltzmann Law<\/strong> to calculate the total power radiated by a star:P=\u03c3AT4P = \\sigma A T^4P=\u03c3AT4<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>PPP is the total power radiated by the star,<\/li>\n\n\n\n<li>\u03c3\\sigma\u03c3 is the Stefan-Boltzmann constant, \u03c3=5.670\u00d710\u22128\u2009W\/m2K4\\sigma = 5.670 \\times 10^{-8} \\, \\text{W\/m}^2\\text{K}^4\u03c3=5.670\u00d710\u22128W\/m2K4,<\/li>\n\n\n\n<li>AAA is the surface area of the star, which is A=4\u03c0R2A = 4\\pi R^2A=4\u03c0R2,<\/li>\n\n\n\n<li>TTT is the temperature of the star.<\/li>\n<\/ul>\n\n\n\n<p>We are given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Temperature T=5422\u2009KT = 5422 \\, \\text{K}T=5422K,<\/li>\n\n\n\n<li>Radius R=0.86\u2009Rsun=0.86\u00d77.00\u00d7108\u2009m=6.02\u00d7108\u2009mR = 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\u200b=0.86\u00d77.00\u00d7108m=6.02\u00d7108m,<\/li>\n\n\n\n<li>The Stefan-Boltzmann constant \u03c3=5.670\u00d710\u22128\u2009W\/m2K4\\sigma = 5.670 \\times 10^{-8} \\, \\text{W\/m}^2\\text{K}^4\u03c3=5.670\u00d710\u22128W\/m2K4.<\/li>\n<\/ul>\n\n\n\n<p>Now, we can calculate the surface area of the star:A=4\u03c0(6.02\u00d7108)2=4\u03c0\u00d73.624\u00d71017\u2009m2=4.55\u00d71018\u2009m2A = 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=4\u03c0(6.02\u00d7108)2=4\u03c0\u00d73.624\u00d71017m2=4.55\u00d71018m2<\/p>\n\n\n\n<p>Next, we use the Stefan-Boltzmann Law to find the power:P=\u03c3AT4=(5.670\u00d710\u22128)\u00d7(4.55\u00d71018)\u00d7(5422)4P = \\sigma A T^4 = (5.670 \\times 10^{-8}) \\times (4.55 \\times 10^{18}) \\times (5422)^4P=\u03c3AT4=(5.670\u00d710\u22128)\u00d7(4.55\u00d71018)\u00d7(5422)4<\/p>\n\n\n\n<p>First, we calculate T4T^4T4:T4=(5422)4=1.579\u00d71015T^4 = (5422)^4 = 1.579 \\times 10^{15}T4=(5422)4=1.579\u00d71015<\/p>\n\n\n\n<p>Now, we substitute this into the equation for PPP:P=(5.670\u00d710\u22128)\u00d7(4.55\u00d71018)\u00d7(1.579\u00d71015)=2.26706\u00d71026\u2009WP = (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\u00d710\u22128)\u00d7(4.55\u00d71018)\u00d7(1.579\u00d71015)=2.26706\u00d71026W<\/p>\n\n\n\n<p>Thus, the total power radiated by Kepler 1606 is:P=2.26706\u00d71026\u2009WP = 2.26706 \\times 10^{26} \\, \\text{W}P=2.26706\u00d71026W<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Part (b): Intensity Delivered at the Orbital Distance of Kepler 1606b<\/h3>\n\n\n\n<p>The intensity III at a distance rrr from the star is given by the inverse square law:I=P4\u03c0r2I = \\frac{P}{4 \\pi r^2}I=4\u03c0r2P\u200b<\/p>\n\n\n\n<p>We are given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The orbital distance of Kepler 1606b is r=0.64\u00d7Tearth=0.64\u00d71.50\u00d71011\u2009m=9.60\u00d71010\u2009mr = 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\u00d7Tearth\u200b=0.64\u00d71.50\u00d71011m=9.60\u00d71010m,<\/li>\n\n\n\n<li>The total power radiated by the star P=2.26706\u00d71026\u2009WP = 2.26706 \\times 10^{26} \\, \\text{W}P=2.26706\u00d71026W.<\/li>\n<\/ul>\n\n\n\n<p>Now, we calculate the intensity:I=2.26706\u00d710264\u03c0(9.60\u00d71010)2I = \\frac{2.26706 \\times 10^{26}}{4 \\pi (9.60 \\times 10^{10})^2}I=4\u03c0(9.60\u00d71010)22.26706\u00d71026\u200b<\/p>\n\n\n\n<p>First, calculate the square of the distance:(9.60\u00d71010)2=9.22\u00d71021(9.60 \\times 10^{10})^2 = 9.22 \\times 10^{21}(9.60\u00d71010)2=9.22\u00d71021<\/p>\n\n\n\n<p>Now, calculate the intensity:I=2.26706\u00d710264\u03c0\u00d79.22\u00d71021=2.26706\u00d710261.157\u00d71023=1957.54\u2009W\/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=4\u03c0\u00d79.22\u00d710212.26706\u00d71026\u200b=1.157\u00d710232.26706\u00d71026\u200b=1957.54W\/m2<\/p>\n\n\n\n<p>Thus, the intensity delivered by Kepler 1606 at the orbital distance of Kepler 1606b is:I=1957.54\u2009W\/m2I = 1957.54 \\, \\text{W\/m}^2I=1957.54W\/m2<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Summary of Answers:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>(a) The total power radiated by Kepler 1606 is <strong>2.26706e+26 W<\/strong>.<\/li>\n\n\n\n<li>(b) The intensity delivered at the orbital distance of Kepler 1606b is <strong>1957.54 W\/m\u00b2<\/strong>.<\/li>\n<\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner6-1495.jpeg\" alt=\"\" class=\"wp-image-265660\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>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&#8217;s orbit to the [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-265653","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/265653","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=265653"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/265653\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=265653"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=265653"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=265653"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}