Homework5 (1)

GE CRT-SC 101 - Life Beyond Earth: Homework 5 Name: Olga Ivanova Special Instructions for HW#5: As this is the first STEM course at the college level for many of you, please make sure to watch the video I posted with very similar problems to all of these. In that video I show you step-by-step how to solve these problems. I’ve also posted the PDF of the full solutions for that practice video, so you can make sure you understand it. I do NOT expect you to know how to fully solve them on your own at this point, so if you’re stuck it doesn’t mean you’re behind. The goal of this homework is to learn-by-doing , just like scientists do everyday. My video and the grader help session are resources that will help you learn. Make sure you start early and allocate a lot of time for this. HW#5 and HW#6 are the technical foundations for the Signature Project, so it’s crucial to give them your best effort. The problems add up to 100, but you’ll get full credit (i.e., perfect) if you get 70 points. While you could only do a subset of the problems, I strongly recommend you try them all. Not only will you increase your chances of getting perfect, but you’ll be learning key techniques that you’ll be using on your signature project. Instructions: i. Please write neatly . If I can’t read it, it gets zero. The best practice is to rewrite your first draft solutions neatly, but make sure to not leave out any steps. ii. The point values for each question are given in square brackets [N points]. iii. Take scans of your work with a free smartphone app like CamScanner. iPhones can also scan documents in the Notes app. iv. UPLOAD your scanned homework in Canvas by the deadline.

1. The Stefan-Boltzmann Law states that a star’s luminosity, temperature and size are related as: L = 4 πR 2 σT 4 a) If a star has a surface temperature five times higher than that of the Sun and twice the radius, by what factor is the star more luminous than the Sun? [8] Given: 1. Star's surface temperature = 5 * Sun's temperature 2. Star's radius = 2 * Sun's radius L= 3.846×10 26 W R= 695,700 km Stefan-Boltzmann constant is σ= 5.6704×10 −8 W/(m 2 K 4 ) Luminosity for that star= L1= 4 π(2x695,700 km) 2 5.6704×10 −8 W/(m 2 K 4 )x (5x T) 4 Luminosity of the Sun=L 3.846×10 26 W= 4 π( 695,700 km) 2 x 5.6704×10 −8 W/(m 2 K 4 )x T 4 So, to find the factor by which that star is more luminous than the Sun: L1/L= 4 π ( 2 R ) 2 σ ( 5 T ) 4 4 πR 2 σT 4 = 4 π 4 R 2 σ 625 T 4 4 πR 2 σT 4 = 4x 625=2500

c) Which planets would be swallowed up by the Sun when it becomes a red giant? The current radius of the Sun is 695,500 km and the distances to each planet are listed in the Appendix at the end of this problem set. [8] R expanded = 695, 500 km x 346. 41= 240 928 155 km ● Mercury: Average distance ≈ 57.9 million km ● Venus: Average distance ≈ 108.2 million km ● Earth: Average distance ≈ 149.6 million km ● Mars: Average distance ≈ 227.9 million km ● Jupiter: Average distance ≈ 778.5 million km ● Saturn: Average distance ≈ 1.43 billion km ● Uranus: Average distance ≈ 2.88 billion km ● Neptune: Average distance ≈ 4.5 billion km From the distances, we can see that Mercury, Venus, and Earth have orbits that are within the expanded radius of the Sun. Therefore, when the Sun becomes a red giant, Mercury, Venus, Earth and Mars are likely to be swallowed up by the Sun's expanded outer layers.

2. Wien’s Law states that the emission from a Blackbody peaks at a wavelength that becomes shorter with increasing temperature. The relation between this peak wavelength and temperature is: λ max T = 2 . 9 × 10 6 nm K d) If the Sun’s surface temperature is 5777 K, at what wavelength does the Sun’s spectrum peak? What color does this wavelength correspond to? [8] λ max x 5777 K = 2 . 9 × 10 6 nm K λ max = 2 . 9 × ( 10 ) 6 nm K 5777 = 501.39 nm 501.39 nm falls in the green portion of the visible spectrum, the corresponding color is green. Therefore, the peak wavelength of the Sun's spectrum corresponds to green light. e) Stars less massive that the Sun, such as M-type stars, have lower surface temperatures of T ≈ 4000K. If you were on a planet that orbited an M-type star, what color do you think the sky would be? Explain your reasoning. [8] λ max x 4000 K = 2 . 9 × 10 6 nm K λ max= 2.9 x ( 10 ) 6 4000 =725 nm The color of the sky on a planet orbiting an M-type star, which has a surface temperature of

orangish color during the day due to the characteristics of the star's emitted light and the scattering properties of the planet's atmosphere. 2. A planet has recently been detected around the star HD20945. This star is a K-type star and is fainter than the Sun. Its luminosity and mass are L = 0 . 25 L Ⓢ and M = 0 . 85 M Ⓢ , where L Ⓢ and M Ⓢ represent the luminosity and mass of the Sun, respectively. First, let’s determine the boundary of the Habitable Zone around HD20945. The inner and outer edges of the Habitable Zone around a star with luminosity L (in units of L Ⓢ ) are given by the equations: ( Dinner 1 AU ) = 0.95 √ ❑ ( D outer 1 AU ) = 1.4 √ ❑ Here D inner and D outer have units of AU. Recall that 1 AU is the Earth-Sun distance. Using these equations, determine the inner and outer edges of the Habitable Zone around HD20945 in units of AU. [20] ¿ ) = 0.95 √ ❑ = 0.95 √ ❑ =0.95 x √ ❑ = 0.95 x 0.5=0.475 D inner=0.475 AU ( D outer 1 AU ) = 1.4 √ ❑ = 1.4 √ ❑ = 1.4 x 0.5 = 0.7 D outer = 0.7 AU

3. Figure 1 below shows the radial velocity curve of HD20945 made using the Doppler technique. The figure plots time against the observed velocity of the star (measured by the Doppler shift). Using this figure, answer the following questions: a) What is the orbital period of the planet in days? What is the period in Earth years? Recall there are 365 days in an Earth year. [10] Measuring the period in the radial velocity curve from peak to peak, we see the period is 210-70= 140 days P=140 days In years: P= 140 days x 1 year 365 days = 0.3836 years b) Kepler’s Third Law tells us that the orbital period of a planet is determined by how far the planet is from its parent star. A modified version of Kepler’s Third Law can be written as: ( a 1 AU ¿ 3 = M M 1 ( P 1 year ) 2 Here P is the orbital period in years, a is the planet-star distance in AU, and M ٨ is the mass of HD20945 in units of the mass of our Sun. Using the period you determined in part 4a and the given mass of the star, determine the planet-star distance in units of AU. [10] ¿ ) 3 = ¿ =0.5 a=0.5 AU

¿ ) =11.2( V ∗ ¿ 1 m / s ¿ ) ( M ∗ ¿ M ¿¿ P 1 year ) ( 1 AU a ) = 11.2( 1.5 m / s 1 m / s ¿ ( 0.85 M M ) ( 0.3836 ) ¿ ) M planet M Earth = 10.96

Appendix