21st Century Astronomy 6E
6th Edition
ISBN: 9780393690675
Author: Laura Kay, Stacy Palen, George Blumenthal
Publisher: W. W. Norton
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Chapter 13, Problem 38QP
(a)
To determine
The mass of the companion
(b)
To determine
The semi-major axis of the orbit
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In 2004 astronomers reported the discovery of a large Jupiter-sized planet orbiting very
close to the star HD 179949 (hence the term "hot Jupiter"). The orbit was just 1/9 the
distance of Mercury from our Sun, and it takes the planet only 3.09 days to make one orbit
(assumed to be circular).
a. What is the mass of the star? Express your answer in kilograms and as a multiple of
our Sun's mass.
b. How fast (in km/s) is this planet moving?
A. Use the definition of the center of mass to determine the maximum “wobble” velocity of a star of mass M caused by a planet of mass m orbiting at a distance r from the star with a period T. B. Thanks to Kepler, we know that the mass, period, and distance of an orbiting object are actually related. Use Newton’s version of Kepler’s Third Law to determine the maximum “wobble” velocity in terms of M, m, and r.
2. In a binary star system, the two stars orbit the center of mass of the system. If both stars have the same mass M, the
center of mass will be midway between the stars. Assume the orbital radius is R. Answer in terms of G, M and R.
a. Determine the orbital speed of the stars. (Your answer should
not be the same as the answer for 1b.)
Note that the Earth and moon also actually both orbit around the center of mass of the Earth-moon system.
However, the Earth is so much more massive than the moon that the center of mass is very close to the center of the
Earth and as an approximation we say the moon orbits around the Earth.
b. A space probe is exactly halfway between the two stars. Determine the minimum speed the space probe must
be moving so that it can escape from the binary star-that is the space probe can just reach an infinite distance
away from the stars. This speed is called the escape speed.
C.
*
We want to just reach infinity. This means we will have zero speed at infinity.…
Chapter 13 Solutions
21st Century Astronomy 6E
Ch. 13.1 - Prob. 13.1CYUCh. 13.2 - Prob. 13.2CYUCh. 13.3 - Prob. 13.3CYUCh. 13.4 - Prob. 13.4CYUCh. 13 - Prob. 1QPCh. 13 - Prob. 2QPCh. 13 - Prob. 3QPCh. 13 - Prob. 4QPCh. 13 - Prob. 5QPCh. 13 - Prob. 6QP
Ch. 13 - Prob. 7QPCh. 13 - Prob. 8QPCh. 13 - Prob. 9QPCh. 13 - Prob. 10QPCh. 13 - Prob. 11QPCh. 13 - Prob. 12QPCh. 13 - Prob. 13QPCh. 13 - Prob. 14QPCh. 13 - Prob. 15QPCh. 13 - Prob. 16QPCh. 13 - Prob. 17QPCh. 13 - Prob. 18QPCh. 13 - Prob. 19QPCh. 13 - Prob. 20QPCh. 13 - Prob. 21QPCh. 13 - Prob. 22QPCh. 13 - Prob. 23QPCh. 13 - Prob. 24QPCh. 13 - Prob. 25QPCh. 13 - Prob. 26QPCh. 13 - Prob. 27QPCh. 13 - Prob. 28QPCh. 13 - Prob. 29QPCh. 13 - Prob. 30QPCh. 13 - Prob. 31QPCh. 13 - Prob. 32QPCh. 13 - Prob. 33QPCh. 13 - Prob. 34QPCh. 13 - Prob. 35QPCh. 13 - Prob. 36QPCh. 13 - Prob. 37QPCh. 13 - Prob. 38QPCh. 13 - Prob. 39QPCh. 13 - Prob. 40QPCh. 13 - Prob. 41QPCh. 13 - Prob. 42QPCh. 13 - Prob. 43QPCh. 13 - Prob. 44QPCh. 13 - Prob. 45QP
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