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Download PHYS 1311: In Class Problems Chapter 5 Solutions Feb. 23, 2016
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PHYS 1311: In Class Problems Chapter 5 Solutions Feb. 23, 2016 Problem 1. Given only the distance between the Earth and Moon (REM = 3.84 × 108 m), determine the mass of the Earth and the mass of the Sun. How can we measure REM ? Solution In my notes, I derived the orbital period T for the circular orbit of an object of mass m around a much more massive object M with the two separated by a distance R. This was found to be: 2πR3/2 T = √ . (1) GM So, for the case of the Moon orbiting the Earth, M = ME and T = TM ∼ 30 days. The only piece of information missing is the Earth-Moon distance, REM . One way to determine this is to bounce a radar or laser pulse off of the moon and measure the time it takes for the light pulse to travel to the Moon and back, ∆t. Since light travels at c, REM = c∆t/2. It turns out REM = 3.84 × 108 m, so ∆t = 2.56 s. Converting 30 days to seconds, gives TM = 2.592 × 106 s. Solving Eq. (1) for the Earth’s mass gives ME = 3 4π 2 REM 4π 2 (3.84 × 108 )3 = 5.0 × 1024 kg. = 2 GTM (6.67 × 10−11 )(2.592 × 106 )2 (2) Repeating the same process for the Sun, the period is one year, or TE = 3.1536 × 107 s. The Sun-Earth distance is defined as an astronomical unit (AU) and 1 AU = 1.496×1011 m. This can be used to find the time for a photon to travel from the Sun to the Earth, which is about 500 s or 8.3 mins. So, from the motion of the Earth around the Sun, we can estimate the mass of the Sun MS = 3 4π 2 (1.496 × 1011 )3 4π 2 RSE = = 2.0 × 1030 kg. GTE2 (6.67 × 10−11 )(3.1536 × 107 )2 (3) Problem 2. Given the answers in problem 1, what is the location of the center-of-mass of the Sun-Earth system? Is this the center-of-mass of the Solar System? If not, what is a better way to estimate the Solar System center-of-mass? (Compare to Sun’s radius, R = 0.696 × 109 m). How can this information be used to detect possible planets around other stars? Solution The equation for the center of mass of two objects is rcm = m1 r1 + m2 r2 . m1 + m2 (4) So, taking the origin at the center of the Sun and using the Earth as the other mass, we find 1 0 + (5.0 × 1024 )(1.496 × 1011 ) = 0.38 × 106 m. (5) 2.0 × 1030 + 5.0 × 1024 This is inside the Sun, but the Earth is not the most massive planet, Jupiter is. So, repeating but using the mass of Jupiter (1.9 × 1027 kg) and the distance from the Sun to Jupiter (7.78 × 1011 m) gives SJ Rcm = 0.74 × 109 m. (6) SE Rcm = This location is outside the Sun. While using all of the planets would give an even better estimate, the true center of mass of the Solar System is close to this result. What does this mean? The planets do NOT orbit about the center of the Sun. They orbit about the center of mass of the Solar System. Likewise, the Sun orbits about the Solar System center of mass, but with a period nearly the same as the orbital period of Jupiter, 11.78 years. An observer in another star system could likely not detect any of our 8 planets due to the Sun’s overpowering luminosity, but if they observed long enough (at least 6 years), they would see the Sun “wobble” and therefore deduce a massive object must be orbiting it. If the observers happen to view the Solar System edge-on, the Sun would display simple harmonic motion. This same logic is applied to the observations of other stars to try to detect if they are “wobbling.” Actually, we cannot physically observe the stars moving about their system’s center of mass. What is observed is a frequency shift, called the Doppler effect, of the radiation emitted by the star. A topic we will discuss in Chapter 8. More than 5400 exoplanets have been detected, most with this method. 2