Homework 2 Solutions


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1 Homework 2 Solutions Astronomy 7 Spring 2009 Problems: 1. How fast must the moon be traveling to star in a circular orbit? F gravity = F centripetal (1) GM Earth M Moon = M Moonv 2 (2) d 2 Moon d Moon GM v = (3) d Moon M Earth = kg, d Moon = m (4) v = m kg s 24 kg 2 (5) m = 1.0 km/s = km/hr (6) 2. Your book (Box 4 4) shows an analysis of the orbit of Jupiter s moon, Io. Look up information about Jupiter s other large moons: Europa, Ganymede, and Callisto. Demonstrate that these data are in agreement with Newton s form of Kepler s 3rd Law. Newton s form of Kepler s 3rd Law is given by: M 1 + M 2 = 4π2 r 3 From our textbook, we have the following information: Europa: r = m P = days Ganymede: r = m P = days Callisto: r = m P = days GP 2 (7) 11 kg s2 r 3 = (8) m 3 P 2 In Box 44, M 1 + M 2 = kg, which was equal to Jupiter s mass within the precision of the value found. If the data we have for Europa, Ganymede, and Callisto are 1
2 correct, then they should have the same value for M 1 + M 2 : Europa: M 1 + M 2 11 kg s2 ( m) 3 = m 3 ( s) 2 (9) = kg = kg (10) Ganymede: M 1 + M 2 11 kg s2 ( m) 3 = m 3 ( s) 2 (11) = kg = kg (12) Callisto: M 1 + M 2 11 kg s2 ( m) 3 = m 3 ( s) 2 (13) = kg = kg (14) 3. How far would you have to go from Earth to be completely beyond the pull of its gravity? Explain. It is not possible to travel so far away from Earth that you are beyond the pull of its gravity. Since F gravity = GMm r 2, (15) gravity only goes to zero infinitely far from the Earth. As an aside, gravity travels at the speed of light. So if the sun disappeared right now, we wouldn t know about it in any way for about 8.5 minutes. Since the Earth formed about 4.6 billion years ago, an alien in a galaxy 5 or 10 billion light years away wouldn t feel the gravity of the Earth, because its effects haven t gotten there yet! 4. What is the evidence that there is a supermassive black hole at the center of our Galaxy? How have scientists managed to determine the mass of this black hole? The strongest piece of evidence for a black hole at the center of the Galaxy comes from observing orbits of stars. Some of these stars are moving extremely rapidly (around 2% the speed of light), and even turn around and change directions suddenly. They appear to orbit empty space. By measuring the size of their orbits and their orbital period, scientists use Newton s version of Kepler s Third Law to calculate the mass of the central object. The answer is M. 5. Show that the form of Newton s modification to Kepler s 3rd law is equivalent to M = rv 2 /G. (Assume that the mass of the Sun is much less than the mass of the Galaxy inside the Sun s orbit.) 2
3 Newton s version of Kepler s 3rd law is: (M 1 + M 2 )P 2 = 4π2 r 3 In this case, we can assume that M 1 (the mass of the Sun) is much less than M 2 (the Galaxy s mass inside the Sun s orbit), so M 1 + M 2 M 2 = M. We can then rearrange the above formula to: G (16) M = r ( ) 2 2πr (17) G P v = 2πr (18) P M = rv2 G 6. The supermassive black hole at the center of our Milky Way is M. (a) The star designated SO2, with a measured semimajor axis of 1023 AU has the shortest orbit. What is the timescale of this orbit? We use Newton s form of Kepler s 3rd Law, where one mass is much larger than the other (as in problem 5): (19) P 2 = 4π2 r 3 GM An easy way to do this is to divide by the same equation for the solar system: (20) SO2 therefore has a period of 17 years. (1 yr) 2 = 4π2 (1 AU) 3 GM (21) Pyr 2 = r3 AU (22) M M P yr = r3 AU (23) M M = (24) = 17.0 (25) (b) Star SO1 has an orbital period of 63 years, what is the semimajor axis of the orbit of this star? 3
4 For this we use Newton s form of Kepler s 3rd Law in the following form: S01 has a semimajor of about 2450 AU. r AU = 3 MM P 2 yr (26) = (27) = (28) 7. You are give a traffic ticket for going through a red light (wavelength 700 nm). You tell the police officer that because you were approaching the light, the Doppler effect caused a blueshift that made the light appear green (wavelength 500 nm). How fast would you have had to be going for this to be true? Would the speeding ticket be justified? Explain. The Doppler shift formula gives you the speed v given a shift in wavelength λ: λ λ = v c Note that the absolute value signs are there so that you get a speed rather than a velocity. When λ is negative, v is also negative, which means the source and observer are getting closer. Since we just want to know how fast we are moving, we use the absolute value of the difference. Here λ = 500 nm 700 nm = 200 nm. The actual wavelength of the light is 700 nm so λ = 700 nm. 200 nm v = 700 nm 3.0 m 108 s = m km = s hr That s 191 million miles per hour! Given that you are traveling at nearly 30% the speed of light, you deserve a speeding ticket. If you invented your vehicle, you probably also deserve a prize of some kind! 8. Imagine a planet like the Earth orbiting a star 4 the Sun s mass. If the semimajor axis of the planet s orbit is 1 AU, what would the planet s sidereal period be? (Hint: use Newton s form of Kepler s 3rd law and compare with the case of the Earth orbiting the Sun.) Like in problem 6, we have Newton s form of Kepler s 3rd law and we can use the case in the solar system to simplify it: (29) (30) (31) r 3 = GMP 2 4π 2 (32) (1 AU) 3 = GM (1 yr) 2 4π 2 (33) rau 3 = M M Pyr 2 (34) 4 (35)
5 Since r = 1 AU and M = 4 M here: The planet would have a sidereal period of 0.5 year. 1 = 4 P 2 yr (36) P = 1/2 yr (37) 9. A satellite is said to be a geosynchronous orbit if it appears to always remain over the exact same spot on the rotating Earth. (a) What is the period of this orbit? The satellite needs to have the same orbital period as the Earth s rotation period in an inertial frame. That means we need the Earth s rotation period relative to distant stars and not relative to the Sun in relation to which we are moving. That is called the sidereal period. For the Earth, that is 23 hrs 56 min s. (b) At what distance from the center of the Earth must such a satellite be place into orbit? Using Newton s form of Kepler s 3rd law that we ve already seen many times in this problem set, we find: r 3 = GMP 2 (38) 4π m r = 3 kg s 24 kg ( s) 2 2 (39) 4π 2 = km (40) (c) Explain why the orbit must be in the place of the Earth s equator. If we want the satellite to be above the same spot on the Earth, then it must have a circular orbit so that its speed is constant (and not variable as it is in an eccentric orbit). Then there are three types of orbit: the special orbit that is parallel and above the equator, an orbit parallel to the equator but at a different latitude, and an orbit at an angle to the equator. We will explain why the last two of these will not yield the desired result of the satellite remaining over the exact same spot from the point of view of someone on Earth. In the case of an orbit parallel to the equator but at a different latitude Consider the figure below. In the case of the orbits passing through the equator, at all times, half the planet is above the orbit and half is below. This means the gravitational force from the parts of the planet above and below the orbit are balanced and there is not net force on the satellite in the direction of the planet s rotation axis. In the case of the orbit not passing through the equator, more of the planet is below the orbit so there is a net force pulling the satellite towards the equator. This means that the orbit is not stable, so the satellite will move towards the equator with time. 5
6 Figure 1: The red line is an orbit of the equator. The blue line is an orbit passing through the equator but inclined to it. The green line is an orbit parallel but above the equator. The straight lines are all of the same length. In the case of the inclined orbit, the orbit is stable because half of the mass of the planet is always both below and above the orbit. However, from the point of view of an observer on the Earth, the satellite will appear to move north and south over the course of one sidereal day. To see this, consider what you see when you are at the equator. If you start the day at the point where the satellite is above the equator, then it will be directly overhead. As time passes, you rotate towards the right on the figure below. You stay at the equator, but the satellite moves towards the north pole, so from your point of view it also moves northward in the sky. Half a day later, the satellite moves towards and then away from the southern pole. Therefore, in order for the satellite to remain in exactly the same spot from the point of view of an observer on Earth, it must both pass through the equator and not be inclined to it. It must therefore orbit above the equator. 6
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