final_exam_review

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====Live clarifications from the June 10 exam==== * **Problem 2C**: "in the opposite direction" means "opposite to the motion of the ship" * **Problem 2D**: "how long" means "find the duration (time) of" * **Problem 3**: in **parts D, E, F** use "mass of the electron", not the mass of <sup>90</sup>Y found in **part B** ======Final exam review====== The following questions and problems are courtesy of Justin Dunlap {{:final_review_with_work_shown.pdf|}} =====Practice Questions, June 5 Lecture===== //MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. // ====Review question 1==== A monochromatic beam of light of wavelength 600 nm is incident normally on a diffraction grating with a slit spacing of $1.70\times 10^{-4}\,$cm. What is the angle for the first order maximum? * [....] A) 10.5$^\circ$ * [....] B) 12.1$^\circ$ * [ <color green>X</color> ] C) 20.7$^\circ$ * [....] D) 16.4$^\circ$ * [....] E) 18.2$^\circ$ Steps: - <color green>For the 1<sup>st</sup> maximum, each slit gains one extra wavelength of path compared to the adjacent slit</color> - <color green>$\lambda=600\,{\text{nm}}$ $=0.6\,\mu{\text m}$</color> - <color green>$\sin\theta=\frac{1\,\cdot\,\lambda}{D} =\frac{0.6\,\mu{\text m}}{1.7\,\mu{\text m}}$ $=0.353$</color> - <color green>$\theta=\arcsin 0.353$ $=20.7^\circ$</color> ---- ====Review question 2==== A proton and an electron are both accelerated to the same final speed. If $\lambda_p$ is the de Broglie wavelength of the proton and $\lambda_e$ is the de Broglie wavelength of the electron, then * [....] A) $\;\lambda_p > \lambda_e$ * [....] B) $\;\lambda_p = \lambda_e$ * [ <color green>X</color> ] C) $\;\lambda_p < \lambda_e$ * [....] D) Not enough data to answer this question * [....] E) None of the above <color green>de Broglie wavelength equation: (λ)=h/p. If p=m*v then the electron's de Broglie wavelength is as follows: (λ)=h/(me*v) while the proton's de Broglie wavelength is (λ)=h/(mp*v). Since the proton and the electron share the same final speed (v) then the the electron will have a larger wavelength than the proton due to its smaller mass in the denominator.</color> ---- ====Review question 3==== When a beam of light, which is traveling in glass, strikes an air boundary, there is * [....] A) a 60$^\circ$ phase change in the reflected beam. * [....] B) a 45$^\circ$ phase change in the reflected beam. * [....] C) a 180$^\circ$ phase change in the reflected beam. * [ <color green>X</color> ] D) no phase change in the reflected beam. * [....] E) a 90$^\circ$ phase change in the reflected beam. <color green>If n1>n2 then there is no phase change; however, if n1<n2 then the phase changes by 1/2 lambda. In this case the beam of light is traveling in glass which has a greater index of refraction than air does. </color> ---- ====Review question 4==== A certain astronomical telescope has a diameter of 5.60 m. What is the minimum angle of resolution for this telescope at a wavelength of 620 nm? * [....] A) $\;3.11 \times 10^{-7}\,$rad * [....] B) $\;2.70 \times 10^{-7}\,$rad * [ <color green>X</color> ] C) $\;1.35 \times 10^{-7}\,$rad * [....] D) $\;1.11 \times 10^{-7}\,$rad * [....] E) $\;4.05 \times 10^{-7}\,$rad <color green>Use Rayleigh criterion and small-angle approximation ($\sin\theta_\text{min}=$ $\tan\theta_\text{min}$ $=\theta_\text{min}$, in radians):</color> * <color green>$\theta_\text{min}=$ $\frac{1.22\lambda}{D}=$ $\frac{1.22\,\cdot\,620\times 10^{-9}\,{\text m}}{5.6\,{\text m}}=$ $1.35\times 10^{-7}\,$rad</color> ---- ====Review question 5==== Two spaceships approach Earth from the same direction. One has a speed of 0.21c and the other a speed of 0.34c, both relative to Earth. What is the speed of one spaceship relative to the other? * [....] A) $\;0.16\,c$ * [....] B) $\;0.13\,c$ * [....] C) $\;0.18\,c$ * [....] D) $\;0.15\,c$ * [ <color green>X</color> ] E) $\;0.14\,c$ * <color green>$v_{13}=0.21\,c$</color> * <color green>$v_{23}=0.34\,c$</color> * <color green>$v_{21}=?$</color> * <color green>$v_{21}=\frac{v_{23} - v_{13}}{1-\frac{v_{23}\,\cdot\,v_{13}}{c^2}}$ $=\frac{0.34\,c -0.21\,c}{1-\frac{(0.34\,c)(0.21\,c)}{c^2}}$ $=\frac{0.13\,c}{0.9286} =0.14\,c$</color> ---- ====Review question 6==== Light of wavelength 687 nm is incident on a single slit 0.75 mm wide. At what distance from the slit should a screen be placed if the second dark fringe in the diffraction pattern is to be 1.7 mm from the center of the screen? * [....] A) 1.9 m * [ <color green>X</color> ] B) 0.93 m * [....] C) 1.1 m * [....] D) 0.39 m * [....] E) 1.5 m - <color green>Use the single-slit diffraction equation for dark fringes: </color> * <color green>$W\sin(\theta)=m\lambda$ to find the angle at which the light bends. </color> * <color green>Second dark fringe means $m=2$. </color> * <color green>Therefore $\theta=\arcsin\big(\!\frac{2\lambda}{W}\!\big)$. </color> - <color green>Then plug in $\theta$ into the linear distance equation </color> * <color green>$y=L\tan\theta$) to solve for the screen to slit distance, //L//. </color> - <color green>The final formula is </color> * <color green>$L=\frac{y}{\tan\left[\arcsin\big(\!\frac{2\lambda}{W}\!\big)\right]}$</color> ---- ====Review question 7==== You are moving at a speed $\frac{2}{3}c$ relative to Randy, and Randy shines a light toward you. At what speed do you see the light passing you by? * [ <color green>X</color> ] A) $\;\;\;\;c$ <color white>$\frac{.}{\big|}$</color> * [....] B) $\;\frac{1}{3}c$ <color white>$\frac{.}{\big|}$</color> * [....] C) $\;\frac{4}{3}c$ <color white>$\frac{.}{\big|}$</color> * [....] D) $\;\frac{2}{3}c$ <color white>$\frac{.}{\big|}$</color> * [....] E) It depends on whether you are moving towards or away from Randy. <color blue>The speed of light in a vacuum $\left(c=3.00\times 10^8\frac{\text m}{\text s}\right)$ is the same in all inertial frames of reference.</color> ---- ====Review question 8==== The first atomic explosion released approximately $1.0\times 10^{14}\,$J of energy. How much matter had to be changed into energy to produce this yield? * [....] A) 23 g * [ <color green>X</color> ] B) 1.1 g * [....] C) 13 g * [....] D) 0.45 kg * [....] E) 1.1 kg <color green>Steps: </color> * <color green>$E=mc^2$ </color> * <color green>$m=\frac{E}{c^2}$ </color> * <color green>$m=\frac{10^{14}\,{\text J}}{(3\times 10^8\,\frac{\text m}{\text s})^2}$ $\approx 1.1\,$g</color> ---- ====Review question 9==== The frequency of a light beam is doubled. Which one of the following is correct for the momentum of the photons in that beam of light? * [....] A) It stays the same. * [....] B) It is quadrupled. * [....] C) It is halved. * [....] D) It is reduced by one-fourth. * [ <color green>X</color> ] E) It is doubled. <color blue>Establish the relationship between the frequency and the momentum of a photon:</color> * <color blue>$\lambda = \frac{c}{f}$ </color> * <color blue>$p = \frac{h}{\lambda}$ $= \frac{hf}{c}$ $=\left(\frac{h}{c}\right)\cdot f$</color> * <color blue>the ratio in brackets is the Planck constant divided by the speed of light constant, it is also a constant</color> ---- ====Review question 10==== The wavelength of light in a thin film is 360 nm and the wavelength of the same light in vacuum is 469 nm. What is the index of refraction for the film? * [ <color green>X</color> ] A) 1.30 * [....] B) 1.10 * [....] C) 1.70 * [....] D) 1.50 * [....] E) 1.90 <color green>Use the fact that the wavelength of light in a material is inversely proportional to its refraction index</color> * <color green>$\frac{\lambda_\text{vac}}{\lambda_\text{film}} =$ $\frac{n_{\,\text{film}}}{n_\text{vac}}$</color> * <color green>$n_\text{vac}=1$</color> * <color green>$n_{\,\text{film}} = \frac{\lambda_\text{vac}}{\lambda_\text{film}}$ $= \frac{469\,{\text{nm}}}{360\,{\text{nm}}} = 1.3$</color> ---- ====Review question 11==== What mass would have to be condensed to a radius of $1.0\times 10^{-15}\,$m (the order of magnitude of the radius of an atomic nucleus) in order for it to become a [[wp>black hole]]? The [[wp>Gravitational constant]] is $G=$ $6.67384\times 10^{-11}\frac{ {\text N}\cdot{\text m}^2}{ {\text{kg}^2}}$ * [ <color green>X</color> ] A) $\;6.7\times 10^{11}\,$kg * [....] B) $\;8.3\times 10^{15}\,$kg * [....] C) $\;1.1\times 10^{-6}\,$kg * [....] D) $\;2.2\times 10^3\,$kg * [....] E) $\;4.1\times 10^{-8}\,$kg <color green>Use the Schwarzschild radius formula:</color> * <color green>$R = \frac{2GM}{c^2}$</color> * <color green>$M = \frac{c^2}{2G}R$ $= \frac{(3 \times 10^8\,\frac{\text m}{\text s})^2}{2\,\cdot\,6.7 \times 10^{-11}\,\frac{ {\text m}^3}{ {\text{kg}}\cdot{\text s}^2}}\cdot 10^{-15}\,{\text m}$ $= 6.7 \times 10^{-11}\,$kg</color> ---- ====Review question 12==== In a two-slit experiment, the slit separation is $3.00\times 10^{-5}\,$m. The interference pattern is created on a screen that is 2.00 m away from the slits. If the 7<sup>th</sup> bright fringe on the screen is a linear distance of 10.0 cm away from the central fringe, what is the wavelength of the light? * [....] A) 224 nm * [....] B) 204 nm * [....] C) 234 nm * [ <color green>X</color> ] D) 214 nm * [....] E) 100 nm <color green></color> ---- ====Review question 13==== The surface temperature of the star is 6000 K. What is the wavelength associated with the light emitted by this star? * [....] A) 907 nm * [....] B) 492 nm * [ <color green>X</color> ] C) 850 nm * [....] D) 502 nm * [....] E) 311 nm <color brown>Find the peak frequency: f=(5.88*10^8)(6000)=3.53*10^14 HZ solve for the wavelength using the calculated frequency and the speed of light constant, c. λ= (3*10^8)/(3.53*10^14)=8.5*10^-7m=850nm. </color> ---- ====Review question 14==== An electron is moving with the speed of $1780\frac{\text m}{\text s}$. What is its de Broglie wavelength? * [....] A) 502 nm * [....] B) 302 nm * [....] C) 205 nm * [....] D) 420 nm * [ <color green>X</color> ] E) 409 nm <color green></color> ---- ====Review question 15==== If the distance between the slits in Young's two-slit experiment is decreased, which one of the following statements is true of the interference pattern? * [....] A) The distance between the maxima decreases. * [....] B) The distance between the maxima stays the same. * [ <color green>X</color> ] C) The distance between the minima increases. * [....] D) The distance between the minima stays the same. * [....] E) Impossible to tell without knowing the wavelength of light in use. <color green></color> ---- =====Practice Problems, June 5 Lecture===== //Solve three of the four problems (cross out the one you do not want graded). Show all of your work to receive full credit, most importantly show all the formulas you used to find the final answers. No credit will be awarded if an answer is given without work shown.// ---- ====Review problem 1==== You observe a sheet of a soap film ($n\,=\,1.33$) illuminated by a 500-nm-wavelength light on the same side of the film as you. * a) If the beam of light is incident normal to the film, what is the minimum thickness of the film that will make it look bright? (4 pts) * <color green> 94 nm</color> * <color blue> $\lambda_\text{film} =$ $\frac{\lambda_\text{vac}}{n}$ $=\frac{500\,{\text{nm}}}{1.33} \approx$ $\frac{3}{4}\!\cdot\!500\,{\text{nm}}$ $= 376\,{\text{nm}}$ </color> * <color blue>To get a bright spot, the round-trip within the film should give a half-wavelength delay, since the dissimilar reflection is already introducing a half-a-period delay:</color> * <color blue> $2t = \frac{\lambda_\text{film}}{2}\;$, where //t// is the thickness of the film </color> * <color blue>$t = \frac{\lambda_\text{film}}{4}$ $= \frac{376\,{\text{nm}}}{4} = 94\,$nm</color> * b) At this thickness, what is one other wavelength of light <color darkorange>(in vacuum)</color> that would work to get a bright appearance? (4 pts) * <color green> 166 nm</color> * <color blue> The next bright fringe happens when the round-trip delay in the film is $\frac{3}{2}$ wavelengths (in addition to the the half-cycle delay due to dissimilar reflections)</color> * <color blue> $2t = \frac{3}{2}\lambda_\text{film}$</color> * <color blue> $\lambda_\text{film}=$ $\frac{4t}{3}$</color> * <color blue> $\lambda_\text{vac}=$ $n\lambda_\text{film}=$ $\frac{4nt}{3}=$ $\frac{4\,\cdot\,1.33\,\cdot\,94\,{\text{nm}}}{3}=$ $166.7\,$nm</color> * c) What is the next thickness that would make it look bright (using the original wavelength of light)? (4 pts) * <color green> 282 nm</color> * <color blue>Assuming $\lambda_\text{vac}=500\,$nm, the next bright fringe will again require $\frac{3}{2}$ wavelengths worth of round-trip delay in the film, to add up to two full wavelengths after adding the dissimilar reflection effect</color> * <color blue> $2t = \frac{3}{2}\lambda_\text{film}$</color> * <color blue> $t = \frac{3}{4}\lambda_\text{film}$</color> * <color blue> $t = \frac{3}{4}\!\cdot\!376\,{\text{nm}}=$ $282\,{\text{nm}}$</color> * <color blue>Alternatively, use the thin-film formula with $m=1$</color> * <color blue>$\frac{2nt}{\;\;\lambda_\text{vac}} - \frac{1}{2} = m$</color> * <color blue>$\frac{2nt}{\;\;\lambda_\text{vac}} = m + \frac{1}{2}$</color> * <color blue>$t = \frac{\lambda_\text{vac}}{2n}\left(m+\frac{1}{2}\right)$ $=\frac{\lambda_\text{vac}}{2n}\left(\frac{3}{2}\right)=$ $\frac{500\,{\text{nm}}\,\cdot\,3}{2\,\cdot\,1.33\,\cdot\,2}=$ $282\,{\text{nm}}$</color> * d) What would you observe at the top of the film where the soap is very thin? Why? (3 pts) * <color blue> As $t\rightarrow 0$, the delay due to extra round-trip path becomes small (close to zero).</color> * <color blue>The extra $\pi$ phase delay due to dissimilar reflection will lead to destructive interference, i.e. a dark spot</color> * <color green>We can use both the dark and the bright fringe equations to see what happens as the thickness //t// becomes very small.</color> * <color green>The equation that allows for an integer //m// will make sense in the case when //t// is small.</color> * <color green>Bright fringe: $\,\frac{2nt}{\lambda_\text{vac}}-\frac{1}{2} =m$</color> * <color green>As //t// becomes smaller (//t// -> 0), we have $\;m\rightarrow -\frac{1}{2}$. </color> * <color green>This is a non-integer number, so it is impossible for bright fringes to occur: the top of the film cannot appear bright.</color> * <color green>Dark fringe: $\,\frac{2nt}{\lambda_\text{vac}} =m$</color> * <color green>As //t// becomes smaller (//t// -> 0), we have $\;m\rightarrow 0$</color> * <color green> This gives an integer //m// for dark fringes so the top of the film can appear dark. Therefore, if the film is very thin, it will look dark.</color> ---- ====Review problem 2==== The [[wp>Large Hadron Collider]] (LHC) at [[wp>CERN]] in Europe can accelerate protons (rest mass, $m_p = 1.673\times 10^{-27}\,$kg, $c = 2.998\times 10^8\frac{\text m}{\text s}$) to a speed of $v = 0.999999991\,c$. * a) What is the rest energy (in MeV) of a proton? (1 eV $= 1.602\times 10^{-19}\,$J) (5pts) * <color blue>Use the Einstein formula and unit conversion:</color> * <color green>$E_0=m_pc^2$ $=1.673\times 10^{-27}\,{\text{kg}}\cdot(2.998\times 10^8\frac{\text m}{\text s})^2=$ $1.504\times 10^{-10}\,{\text J}$</color> * <color green>$\frac{1.504\times 10^{-10}\,{\text J}}{1.602\times 10^{-19}\,\frac{\text J}{\text{eV}}}=$ $9.39\times 10^8\,{\text{eV}}$ $=938.6\,$MeV</color> * b) Find the magnitude of the relativistic momentum of the protons in the LHC. Comparing it with the rest nonrelativistic value, is it bigger or smaller? By what factor? (5 pts) * <color blue>$\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{1}{\sqrt{1-0.999999991^2}}\approx$ $\frac{1}{\sqrt{2\,\cdot\,9\times 10^{-9}}}$ $=7454$</color> * <color blue>$p_\text{rel}=\frac{m_pv}{\sqrt{1-0.999999991^2}}$ $=7454\cdot 1.673\times 10^{-27}\,{\text{kg}}\cdot 0.999999991\,c$ $=3.74\times 10^{-15}\frac{\text{kg}\cdot{\text m}}{\text s}$</color> * <color blue>$p_\text{non-rel}=m_pv$ $=1.673\times 10^{-27}\,{\text{kg}}\cdot 0.999999991\,c$ $=5.01\times 10^{-19}\frac{\text{kg}\cdot{\text m}}{\text s}$ </color> * <color blue>The relativistic value is larger by a factor of 7454</color> * c) Compute a protons total energy in the LHC (in GeV). How much more energy is this compared to the rest energy? (5 pts) * <color blue>$E=\frac{m_pc^2}{\sqrt{1-\frac{v^2}{c^2}}}$ $=7454\cdot 1.673\times 10^{-27}\,{\text{kg}}\cdot c^2$ $=1.121\times 10^{-6}\,{\text J}$ $=6994\,$GeV</color> * <color green>$\frac{E}{E_0}=$ $\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=$ $\frac{6994\,{\text{GeV}}}{0.9386\,{\text{GeV}}}$ $=7454$</color> ---- ====Review problem 3==== Match the definitions and descriptions with the best term or phrase given below (1.5 pts each): | Coherent light | Special Relativity | Quantum mechanics | | Incoherent light | General Relativity | Blackbody radiation | | Bright fringe | Principle of Equivalence | Wien's displacement law | | Dark fringe | Proper length | Planck constant | | Monochromatic light | Proper time | Quantized energy | | Superposition | Time dilation | Photoelectric effect | | Interference | Length contraction | Photon | | Young's Two-Slit Experiment | Relativistic momentum | Ultraviolet catastrophe | | Phase change due to reflection | Speed of light | Work function | | Air wedge | Relativistic energy | Cutoff frequency | | Newton's rings | Rest mass | Rest mass of a photon | | Thin film effects | Rest energy | Photon scattering | | Diffraction | Relativistic kinetic energy | Compton effect | | Single slit diffraction | Antimatter | de Broglie wavelength | | Resolution | Positron | Wave-particle duality | | Rayleigh's Criterion | Electron | Heisenberg uncertainty principle | | Diffraction grating | Proton | Quantum tunneling | | Reflection grating | Antiproton | Schwarzschild radius | | Black hole | | | - When a positron and this encounter each other, pure energy is given off in the form of gamma rays. * <color green>Electron</color> - In other words, all physical experiments conducted in a uniform gravitational field and in an accelerated frame of reference give identical results. * <color green>Principle of Equivalence</color> - Evidence suggesting matter has this quantity (S.I. unit: meter) is a principle of __quantum mechanics__. * <color red><del>Length Contraction?</del></color> -> <color green>de Broglie wavelength</color> - A constant that defines the smallest values, such as energy, a system in the universe can have. * <color green>Planck constant</color> - Light striking a metal ultimately resulting in a measurable current. * <color green>Photoelectric Effect</color> - A problem with classical physics suggesting that a heated object would release large amounts of high frequency light. * <color green>Ultraviolet Catastrophe</color> - Light is bounced off this surface made of many grooves, resulting in interference. * <color green>Reflection Grating</color> - Closer than this and the escape velocity would exceed that of light. * <color green>Schwarzschild Radius</color> - The label given to the observation that if the energy of an event is known, how well you can measure time for that event is affected. * <color green>Heisenberg uncertainty principle </color> - A rainbow observed from a puddle in a parking lot is most likely due to this. * <color green>Thin Film Effects</color> ---- ====Review problem 4==== Kepler-62 (K-62) is a star 1,200.0 light years from Earth as measured by Earth astronomers. It appears there are at least two planets approximately the mass of the Earth in the habitable zone of K-62. You are tired of this planet and design a spaceship to get you to K-62. - What is the distance to K-62 in km as measured by Earth observers? (1 year $= 3.154\times 10^7\,$s) (3 pts) * <color green>$L_0=c\Delta t_\text{light}=$ $c\cdot (1200 \cdot 3.154\times 10^7\,{\text s})$ $=1.135\times 10^{19}\,$m </color> * <color green>divide by 1000 to get km: $L_0=1.135\times 10^{16}\,$km</color> - Is the distance calculated in the previous part a proper length? Justify your answer. (3 pts) * <color green>Yes, because the distance is measured by an observer at rest</color> - How many years would pass according to you if you traveled to K-62 at a speed of $v = 0.99990\,c$? (3 pts) * <color green>We have two unknowns in the time dilation equation: $\Delta t$ and $\Delta t_0$</color> * <color green>We can find $\Delta t$ using prior information and can plug that into the time dilation equation to solve for $\Delta t_0$. </color> * <color green>First find $\Delta t$ $= \frac{L_0}{v}$ $=\frac{1200\,{\text{ly}}}{0.99990\,c} =1200.12\,$years</color> * <color green>Re-arrange the time dilation equation to solve for $\Delta t_0$.</color> * <color green>$\Delta t$ $=\frac{\Delta t_0}{\sqrt{1-\frac{v^2}{c^2}}}$</color> * <color green>$\Delta t_0$ $=\Delta t\sqrt{1-\frac{v^2}{c^2}}$</color> * <color green>$\Delta t_0$ $=1200.12\,{\text y}\cdot\sqrt{1-0.99990^2}$ $=1200.12\,{\text y} \cdot 0.01412= 16.97\,$years</color> - According to you, what is the distance you traveled on your trip? (3 pts) * <color green>Technically, in your own frame, you traveled zero km, because you stayed on the ship.</color> * <color green>I guess, what the problem is asking is this: knowing your speed, what would you compute as your traveled distance?</color> * <color green>Must find the contracted length, we know the proper length from part 1 (we re-assured ourselves of this in part 2)</color> * <color green>$L_\text{contracted}=$ $L_0\sqrt{1-\frac{v^2}{c^2}}$</color> * <color green>$L_\text{contracted}=$ $1.135\times 10^{19}\,{\text m}\times 0.01412=$ $1.6026x10^{17}\,{\text m}=$ $1.6026x10^{14}\,{\text{km}}$ $=16.97\,{\text{ly}}$</color> * <color green>this is ~ two orders of magnitude less than the distance on earth</color> * <color red>Alternatively,</color> <color green>you can just multiply your speed by the proper time of travel:</color> * <color green>$L_\text{trav}=v\Delta t_0=$ $0.99990\,c\cdot 16.97\,{\text{y}}$ $\approx 16.97\,$light years (ly).</color> * <color green>This is exactly the same as $L_\text{contracted}$ found above:</color> * <color green>$L_\text{trav}$ $=v\Delta t_0=v\frac{L_0}{v}\sqrt{1-\frac{v^2}{c^2}}$ $=L_0\sqrt{1-\frac{v^2}{c^2}}$ $=L_\text{contracted}$</color> * <color green>Note, that the magnitude of the speed of spaceship relative to Earth is the same as the speed of Earth relative to ship</color> - Say you can only get your spaceship up to 0.99900 c, how many more years will you age on your trip compared to the trip in part 3? (3 pts) * <color green>Using the re-arranged time dilation equation from part 3</color> * <color green>$\Delta t_0$ $=\Delta t\sqrt{1-\frac{v^2}{c^2}}$</color> * <color green>$\Delta t_0$ $=1200.12\,{\text y}\cdot\sqrt{1-0.99990^2}$ $=1200.12\,{\text y} \cdot 0.04471= 53.65\,$years</color> * <color green>Take the difference (to answer "how many more years..."): </color> * <color green>$53.65-16.97 = 36.67\,$years more!!! </color> ^ Parameter ^ Kepler-62//b// ^ Kepler-62//c// ^ Kepler-62//d// ^ Kepler-62//e// ^ Kepler-62//f// ^ | $T_0$ (BJD-2454900) | 103.9189 $\pm$ 0.0009 | 67.651 $\pm$ 0.008 | 113.8117 $\pm$ 0.0008 | 83.404 $\pm$ 0.003 | 522.710 $\pm$ 0.006 | | $P$ (days) | 5.714932 $\pm$ 0.000009 | 12.4417 $\pm$ 0.0001 | 18.16406 $\pm$ 0.00002 | 122.3874 $\pm$ 0.0008 | 267.291 $\pm$ 0.005 | | Transit duration (hours) | 2.31 $\pm$ 0.09 | 3.02 $\pm$ 0.09 | 2.97 $\pm$ 0.09 | 6.92 $\pm$ 0.16 | 7.46 $\pm$ 0.20 | | Depth (%) | 0.043 $\pm$ 0.001 | 0.007 $\pm$ 0.001 | 0.092 $\pm$ 0.002 | 0.070 $\pm$ 0.003 | 0.042 $\pm$ 0.004 | | $R_p/R^*$ | 0.0188 $\pm$ 0.0003 | 0.0077 $\pm$ 0.0004 | 0.0278 $\pm$ 0.0006 | 0.0232 $\pm$ 0.0003 | 0.0203 $\pm$ 0.0008 | | $a/R^*$ | 18.7 $\pm$ 0.5 | 31.4 $\pm$ 0.8 | 40.4 $\pm$ 1.0 | 144 $\pm$ 4 | 243 $\pm$ 6 | | $b$ | 0.25 $\pm$ 0.13 | 0.16 $\pm$ 0.09 | 0.22 $\pm$ 0.13 | 0.06 $\pm$ 0.05 | 0.41 $\pm$ 0.14 | | $i$ | 89.2 $\pm$ 0.4 | 89.7 $\pm$ 0.2 | 89.7 $\pm$ 0.3 | 89.98 $\pm$ 0.02 | 89.90 $\pm$ 0.03 | | $e\cos\omega$ | 0.01 $\pm$ 0.17 | $-$0.05 $\pm$ 0.14 | $-$0.03 $\pm$ 0.24 | 0.05 $\pm$ 0.17 | $-$0.05 $\pm$ 0.14 | | $e\sin\omega$ | $-$0.07 $\pm$ 0.06 | $-$0.18 $\pm$ 0.11 | 0.09 $\pm$ 0.09 | $-$0.12 $\pm$ 0.02 | $-$0.08 $\pm$ 0.10 | | $a$ (AU) | 0.0553 $\pm$ 0.0005 | 0.0929 $\pm$ 0.0009 | 0.120 $\pm$ 0.001 | 0.427 $\pm$ 0.004 | 0.718 $\pm$ 0.007 | | $R_p\;(R_\oplus)$ | 1.31 $\pm$ 0.04 | 0.54 $\pm$ 0.03 | 1.95 $\pm$ 0.07 | 1.61 $\pm$ 0.05 | 1.41 $\pm$ 0.07 | | $T_\text{eq}$ (K) | 750 $\pm$ 41 | 578 $\pm$ 31 | 510 $\pm$ 28 | 270 $\pm$ 15 | 208 $\pm$ 11 | ^ A table you might want to take with you on the trip, but will not do you any good for this test. ^^^^^^ | //Adapted from//: \\ W.J. Borucki //et al//., "Kepler-62: A Five-Planet System with Planets of 1.4 and 1.6 Earth Radii in the Habitable Zone", \\ Science **340** (6132), 587-590 (3 May 2013)\\ [[http://www.sciencemag.org/content/340/6132/587|DOI:10.1126/science.1234702]] ||||||

final_exam_review.1402428465.txt.gz · Last modified: 2014/06/10 19:27 by wikimanager

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