Universe
11th Edition
ISBN: 9781319039448
Author: Robert Geller, Roger Freedman, William J. Kaufmann
Publisher: W. H. Freeman
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Chapter 8, Problem 12CC
To determine
The reason why radial velocity method fails to determine the presence of an extrasolar planet when its orbit lies perpendicular to the line of sight of Earth and the wobbling star.
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If the semi-major axis, a, is measured in AU and the orbital period, p, is measured in years, then Kepler's 3rd law allows us to calculate the mass of the object they are orbiting using the following equation: M = a3/p2
Furthermore, the mass that is calculated by this equation is given in solar masses (MSun) where, by definition, the Sun's mass is 1 MSun.
Now, suppose I were to tell you that the mass of Jupiter is equal to 4.5e7 MSun.
Does the stated mass of Jupiter make sense?
it is to big or to small or makes sense
If the semi-major axis, a, is measured in AU and the orbital period, p, is measured in years, then Kepler's 3rd law allows us to calculate the mass of the object they are orbiting using the following equation: M = a3/p2
Furthermore, the mass that is calculated by this equation is given in solar masses (MSun) where, by definition, the Sun's mass is 1 MSun.
Now, suppose I were to tell you that the mass of Jupiter is equal to 4.5e7 MSun.
Does the stated mass of Jupiter make sense?
Group of answer choices
- Yes
- No, it's too big.
- No, it's too small
What would be the angular diameter (in arc seconds) of a planet with diameter 8.5 x 105 km and orbital distance from it's star of 175 x 108 km as seen from a planet with. orbital distance from the same star of 70 x 107 km as seen from their closest approach?
Chapter 8 Solutions
Universe
Ch. 8 - Prob. 1CCCh. 8 - Prob. 2CCCh. 8 - Prob. 3CCCh. 8 - Prob. 4CCCh. 8 - Prob. 5CCCh. 8 - Prob. 6CCCh. 8 - Prob. 7CCCh. 8 - Prob. 8CCCh. 8 - Prob. 9CCCh. 8 - Prob. 10CC
Ch. 8 - Prob. 11CCCh. 8 - Prob. 12CCCh. 8 - Prob. 1QCh. 8 - Prob. 2QCh. 8 - Prob. 3QCh. 8 - Prob. 4QCh. 8 - Prob. 5QCh. 8 - Prob. 6QCh. 8 - Prob. 7QCh. 8 - Prob. 8QCh. 8 - Prob. 9QCh. 8 - Prob. 10QCh. 8 - Prob. 11QCh. 8 - Prob. 12QCh. 8 - Prob. 13QCh. 8 - Prob. 14QCh. 8 - Prob. 15QCh. 8 - Prob. 16QCh. 8 - Prob. 17QCh. 8 - Prob. 18QCh. 8 - Prob. 19QCh. 8 - Prob. 20QCh. 8 - Prob. 21QCh. 8 - Prob. 22QCh. 8 - Prob. 23QCh. 8 - Prob. 24QCh. 8 - Prob. 25QCh. 8 - Prob. 26QCh. 8 - Prob. 27QCh. 8 - Prob. 28QCh. 8 - Prob. 29QCh. 8 - Prob. 30QCh. 8 - Prob. 31QCh. 8 - Prob. 32QCh. 8 - Prob. 33QCh. 8 - Prob. 34QCh. 8 - Prob. 35QCh. 8 - Prob. 36QCh. 8 - Prob. 37QCh. 8 - Prob. 38QCh. 8 - Prob. 39QCh. 8 - Prob. 40QCh. 8 - Prob. 41QCh. 8 - Prob. 42QCh. 8 - Prob. 43QCh. 8 - Prob. 44QCh. 8 - Prob. 45QCh. 8 - Prob. 46QCh. 8 - Prob. 47QCh. 8 - Prob. 48QCh. 8 - Prob. 49QCh. 8 - Prob. 50QCh. 8 - Prob. 51QCh. 8 - Prob. 52Q
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- Kepler’s third law says that the orbital period (in years) is proportional to the square root of the cube of the mean distance (in AU) from the Sun (Pa1.5) . For mean distances from 0.1 to 32 AU, calculate and plot a curve showing the expected Keplerian period. For each planet in our solar system, look up the mean distance from the Sun in AU and the orbital period in years and overplot these data on the theoretical Keplerian curve.arrow_forwardSnowT is an exoplanet orbiting star 27 Kood. If the planet’s orbital period is 3.75 days and 27 Kood's constant is 2.9X10^-19 s^2/m^3, what is SnowT orbital radius?arrow_forwardAlthough we say that the Earth orbits the Sun, in reality they are both orbiting the center of mass. It turns out that the equation for the size of the Sun's orbit around the center of mass is to take the full semi-major axis (1 AU) and multiply it by MEarth/(Msun + MEarth), sometimes known as the fulcrum equation. What is the semi-major axis of the Sun's orbit around the center of mass in km?arrow_forward
- Calculate the escape velocity to an orbit of 457 km height from a planet with the radius of 2000 km and the density of 3400 kg⋅m-³. Give your answer in Sl units.arrow_forwardCalculate the time taken by the Earth to fall from the orbit onto the Sun if its instantaneous radial velocity became zero.arrow_forwardA certain binary system consists of two stars that have equal masses and revolve in circular orbits around a fixed point half-way between them. If the orbital velocity of each star is v=186 km/s and the orbital period of each is 11.3 days, calculate the mass M of each star. Give your answer in units of the solar mass, 1.99×1030 kg (e.g. if each planet's mass is 3.98×1030 kg, you would answer "2.00").arrow_forward
- In 2004, astronomers reported the discovery of a large Jupiter-sized planet orbiting very close to the star HD179949. The orbit was just 6.4x106 km (about 9 less than the orbit of Mercury) and the planet takes 3.1 days to make one circular orbit. The mass of the star is Answerx_______1030 kg. (Give the number before the exponent.)arrow_forwardA new mystery planet is detected around our Sun. We measure its position relative to the Sun to be 2 AU at perihelion and 6 AU at aphelion. What is the semimajor axis of this planet's orbit (in AU)? With that information, what is the orbital period of that planet (in years)? If this planet has the same mass as Earth, how does the average force of gravity on the planet by the Sun compare with the average force of gravity on the Earth by the Sun? Please give a numerical ratio of the forces. (Hint: You can take the semimajor axis to represent the average position of the planets) 6:this is all one question with multiples steps. Thank youarrow_forwardExplain the tidal hypothesis.arrow_forward
- The orbital velocity for a sattelite orbiting earth at it's radius is 8 km/s. What would be the orbital velocity around a planet that was 2 times as massive as earth but 1/2 the size?arrow_forwardBarnard’s Star, the second closest star to us, is about 56 trillion (5.61012) km away. Calculate how far it would be using the scale model of the solar system given in Overview of Our Planetary System.arrow_forwardWhat would be the Schwarzschild radius, in light years, if our Milky Way galaxy of 100 billion stars collapsed into a black hole? Compare this to our distance from the center, about 13,000 light years.arrow_forward
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