MIDTERM EXAM - Version A ASTRONOMY 2 - J. Brodie Spring, 2002 Section I - Multiple Choice [2 pts each] 1. d 2. d 3. d 4. b 5. b 6. c 7. b Section II - True or False [1 pts each] 1. F 2. F 3. F 4. T 5. F 6. T 7. T 8. F 9. F 10. T Section III - Short Answer [3 pts each] 1. Describe when it is appropriate to use Kepler?s third law in its simple form, p2 = d3, and when, on the other hand, it is necessary to use the complete form of this equation, P2 = 4 ?2 a3 / [G (M1 + M2)], explaining which units to use. 1- The simple equation for orbits is good when we are in the solar system and one of the masses involved is the sun and the other is one of the planets or comets except for Jupiter. The units are AU for distance and years for time. The second equation is more general and is used when we have satellites going around planets and the two masses have to be taken into consideration. The units are meter for distance and second for time, kilos for mass or any other set of units as long as they match the ones in the constant G. 2. Space Ghost is lounging at his friend Brak?s house when he intercepts a message that intergalactic thugs are boarding a spaceship and plan to arrive and beat up Brak. Space Ghost?s watch and thug spaceships?s clock both say 3:00. The thugs fly in their rocket at nearly the speed of light toward Brack?s house, and arrive when their clock says 4:00. If Space Ghost takes a nap until his watch says 4:00, does he wake up before, at the same time as, or after the Brack attack ? Explain why. 2- Space ghost wakes up before because his watch in the planet ( not moving frame of reference) is moving faster than the on in the space ship. Since the space ship is moving at relativistic speeds close to the speed of light then the time in the space ship will go by slower and when the watch in the ship hits 4:00 then on the planet space ghost will be awake because the time there went by faster. 3. If the Earth had three times as large a radius, but its mass were the same, would the force exerted by the Earth?s gravity on you change ? If yes, what would the new acceleration of gravity on you be ? Be clear on your explanations, using equations where necessary. 3- If the radius goes up by three and mass remains the same then the force will change (decrease) by 9 times and therefore the new acceleration( gravity = g ) will decrease by 9 also ( g/9). If mass remains the same the force is related by the inverse square of the radius. F= G m1 m2 / r2 and F = m a 4. Draw a picture with the correct relative positions of Sun, Earth, and Moon during a solar eclipse. 4- Positions as follow : sun-------moon------earth Section III - Solve these two problems and show all your work. [17 pts each] 1) The following questions refer to the diagram below. The levels represent energy levels in a hydrogen atom. Each level is labelled with its energy (above the ground state of Level 1) in units of electron/volts (eV). The labelled transitions represent an electron moving between energy levels. Answer to the following questions and explain your answers. a) Which transition represents an electron that absorbs a photon with 10.2 eV of energy ? ans: Transition C b) Which transition represents the emission of a photon with the longest wavelength ? ans: Transition E c) Which transition, as shown, is not possible ? ans: Transition D d) If an electron goes from the energy level 3 to the energy level 2, does it emit or absorb a photon ? ans: Emits a photon What is the energy of this photon ? And, finally, what is the corresponding frequency of this photon ? ans: Energy = 12.1eV - 10.2eV = 1.9eV Frequency = E/h = 1.9eV/(4.131 x 10e-15 eV*s) = 4.60 x 10e14 Hz 2) Twin sisters, Gwen and Jackie, were both 20 years old in the year 2000. Jackie took off on a round trip to Vega, 25 light-years away. She travelled at an average speed very close to the speed of light, v = 0.9999c. Answer the following 3 questions explaining your answers and showing all your calculations. a) According to Gwen back on the Earth, about how long does it take Jackie to reach Vega? Show why. ans: About 25 years b) Which of the following best describes the situation according to Jackie ? ans: She stays still, while Earth rushes away from her at 0.9999c and Vega rushes toward her at 0.9999c. She sees the distance from Earth to Vega shortened considerably from 25 light-years, and therefore it takes far less that 25 years for Vega to reach her. c) Which of the following correctly describes the situation when Jackie returns to Earth ? (In this problem ages refer to biological ages = time passed since the two sisters were born.) None of the answers given were correct. ans: t_moving = t_rest * sqrt(1-(v/c)^2) 50 yrs * sqrt(1-(0.9999c/c)^2) = 0.7 yrs. Age = 20yrs + 0.7 yrs = 20.7 yrs. Section IV - Choose any TWO (and only two) of the following 4 problems. Show all your work on the back of this page. (15 pts each) 1) The planet Kripton is more than 200 times as massive as the Earth (Mk = 200 ME ), and a body on its surface would weight 5 times more as it would on the Earth?s surface, (Wk = 5 WE ). What is the radius of the planet Kripton as a function of the Earth?s radius ? (Hint: let the gravitational forces on the two planets guide your thinking). W_k = 5*W_e, where W_k is the weight on Krypton and W_e is the weight on Earth Weight is equivalent to gravitational force: W_k = G*M_k*M_body/R_k^2 , where G is the gravitational constant, M_k is the mass of Kripton, M_body is the mass of the body and R_k is the radius of Krypton W_e = G*M_e*M_body/R_e^2 , where G is the gravitational constant, M_e is the mass of Earth, M_body is the mass of the body and R_e is the radius of Earth Rearranging the equations gives: W_k*R_k^2 = G*M_k*M_body W_e*R_e^2 = G*M_e*M_body taking the ratio of the two equations gives: W_e*R_e^2 = G*M_e*M_body --------- ------------ W_k*R_k^2 = G*M_k*M_body As you can see, the G and M_body terms cancel leaving: W_e*R_e^2 M_e --------- = --- W_k*R_k^2 M_k Rearranging gives: (R_e/R_k)^2 = (M_e/M_k)*(W_k/W_e) From the given information we know W_k = 5*W_e and M_k = 200*M_e. Substituting this information into the equation gives: (R_e/R_k)^2 = (M_e/200*M_e)*(5*W_e/W_e) = 1/40 R_e/R_k = sqrt(1/40) = 0.025 Solving for R_k: R_k = 6.32*R_e 2) Consider a binary star system in which a star A has an apparent magnitude of mA = 9 and star B has an apparent magnitude of mB = 5. Which star appears fainter and how many times fainter ? If Both stars have the same radius R, how many times hotter (larger temperature) is one than the other ? (Hint: find their relative luminosities, LA / LB). Star A appears fainter (larger apparent magnitude corresponds to less brightness) mA- mB = -2.5 log(Ba/Bb) 9-5 = -2.5log(Ba/Bb), where Bb is the brightness of Star B and Ba is the brightness of Star A. 4 = -2.5log(Ba/Bb) -1.6 = log(Ba/Bb) Raising both sides to the power of 10 gives: 10^-1.6 = Ba/Bb 0.025 = Ba/Bb Ba = 0.025Bb = 1/40 * Bb Therefore Star A has 1/40th the brightness of Star B In general, luminosity, L = B*4pi*D^2 In this problem, since the stars make up a binary star system, their distances, D are the same. Take the ratio of the luminosity of star B to the luminosity of star A. (Note that in this ratio the 4pi and D^2 terms cancel in the numerator and denominator.) L(b)/L(a) = Bb/Ba, where L(a) and L(b) are the Luminosities of Stars A & B respectively. From the calculation of brightness, Bb/Ba = 40. Therefore L(b)/L(a) = 40 as well. The other equation for luminosity is L = 4pi*R^2*sigma*T^4 In this question the radii of the two stars are the same. Again taking the ratio of the luminosity of star B to that of star A gives: (Note that the radii, sigma and the 4pi terms cancel out in the numerator and denominator.) So, L(b)/L(a) = T(b)^4/T(a)^4, where T(b) and T(a) are the temperatures of stars B & A respectively. From above, we already know that L(b)/L(a) = 40. Substituting this into the above eqn. gives 40 = T(b)^4/T(a)^4 Taking the quarter root of both sides gives: 2.51 = T(b)/T(a) or T(b) = 2.51T(a) 3) As measured in the laboratory, the prominent F_alpha line of atom Fantomat has a wavelength of lambda_lab= 895.362 nm. But in the spectral line of Fantomat spectrum of the star Splendor, this line has a wavelength lambda_ob= 894.998 nm. What is Splendor's radial velocity (in km/s and miles/s) ? Is Splendor receding or approaching us ? the radial velocity (v_r) divided by the speed of light (c) is: v_r/c = (lambda_ob - lambda_lab)/lambda_lab = (894.998 nm - 895.362 nm)/895.362 nm = -4*10e-4 (note: units of nm cancel out) Solving for the radial velocity gives: v_r = -4*10e-4 * c = -4*10e-4 * 3*10e5 km/s = -121.96 km/s Since the velocity is negative (i.e. lambda_ob < lambda_lab), then Splendor must be approaching. Converting to miles per sec: -121.96 km/s * (1 mile/1.6 km) = -75 miles/s 4) The planet Saturn is approximately ten times as far away from the Sun as the Earth's distance from the Sun, (that is aSaturn = 10 aEarth , where a is the average distance from the Sun). Estimate the time in Earth years for Saturn to take to make one revolution (that is a complete orbit) around the Sun. The period of Saturn orbit is obtained using Kepler's third law. a(saturn) = 10*a(earth) P^2 = k*a^3 (k=1 since Mass of the Sun >> Mass Saturn and P is in units of years and a is in units of AU). Noting that a(earth) = 1 AU and substituting into the above eq'n: P^2 = (10a(earth))^3 = (10AU^3) = 1000 AU^3 P = sqrt(1000 AU^3) = 31.62 yrs. For those of you who chose the more complex version of Kepler's third law: a(saturn) must be converted to units of meters to match the units of G. P^2 = 4*pi^2*a^3/(G*(Msun + Msaturn)), where G = 6.67 * 10e-11 m^3/(kg*s^2), a = 1.427 * 10e12 m (from textbook), Msun = 1.99 * 10e30 kg and Msaturn = 5.69 * 10e26 kg Carefully substituting gives P = 29.46 yrs