UNIVERSE (LOOSELEAF):STARS+GALAXIES
6th Edition
ISBN: 9781319115043
Author: Freedman
Publisher: MAC HIGHER
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Chapter 4, Problem 30Q
To determine
The reason why does Venus have its largest angular diameter when it is new and smallest angular diameter when it is full.
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The average Earth-Moon distance is 3.84 X 10^5 km, while the Earth-Sun is 1.496 X 10^8 km. Since the radius of the Moon is
1.74 X 10^3 km and that of the Sun is 6.96 X 10^5 km.
a) Calculate the angular radius of the Moon and the Sun, qmax, according to the following figure.
D
Bax
R
b) Calculate the solid angle of the Moon and the Sun as seen from Earth.
(c) Interpret its results; Would this be enough to explain the occurrence of total solar eclipses?
6
Chapter 4 Solutions
UNIVERSE (LOOSELEAF):STARS+GALAXIES
Ch. 4 - Prob. 1QCh. 4 - Prob. 2QCh. 4 - Prob. 3QCh. 4 - Prob. 4QCh. 4 - Prob. 5QCh. 4 - Prob. 6QCh. 4 - Prob. 7QCh. 4 - Prob. 8QCh. 4 - Prob. 9QCh. 4 - Prob. 10Q
Ch. 4 - Prob. 11QCh. 4 - Prob. 12QCh. 4 - Prob. 13QCh. 4 - Prob. 14QCh. 4 - Prob. 15QCh. 4 - Prob. 16QCh. 4 - Prob. 17QCh. 4 - Prob. 18QCh. 4 - Prob. 19QCh. 4 - Prob. 20QCh. 4 - Prob. 21QCh. 4 - Prob. 22QCh. 4 - Prob. 23QCh. 4 - Prob. 24QCh. 4 - Prob. 25QCh. 4 - Prob. 26QCh. 4 - Prob. 27QCh. 4 - Prob. 28QCh. 4 - Prob. 29QCh. 4 - Prob. 30QCh. 4 - Prob. 31QCh. 4 - Prob. 32QCh. 4 - Prob. 33QCh. 4 - Prob. 34QCh. 4 - Prob. 35QCh. 4 - Prob. 36QCh. 4 - Prob. 37QCh. 4 - Prob. 38QCh. 4 - Prob. 39QCh. 4 - Prob. 40QCh. 4 - Prob. 41QCh. 4 - Prob. 42QCh. 4 - Prob. 43QCh. 4 - Prob. 44QCh. 4 - Prob. 45QCh. 4 - Prob. 46QCh. 4 - Prob. 47QCh. 4 - Prob. 48QCh. 4 - Prob. 49QCh. 4 - Prob. 50QCh. 4 - Prob. 51QCh. 4 - Prob. 52QCh. 4 - Prob. 53QCh. 4 - Prob. 54QCh. 4 - Prob. 55QCh. 4 - Prob. 56QCh. 4 - Prob. 57QCh. 4 - Prob. 58Q
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- Why does the moon rotate aroud the Sun, and is it important?arrow_forwardPlz give brief explanation within 15 minutes will definitely upvotearrow_forwardDuring a retrograde loop of Mars, would you expect Mars to be brighter than usual in the sky, about average in brightness, or fainter than usual in the sky? Explain.arrow_forward
- Repeat Problem 2 for Pluto. In other words, ignoring the motion of Earth, how far across the sky would Pluto move in 1 Earth day? Assume Pluto is in a circular orbit around the Sun at its average distance of 39.3 AU.arrow_forwardPhobos, one of the moons of Mars, is about 25 km in diameter and orbits about 6000 km above the surface of the planet. What is the angular diameter of Phobos as seen from Mars? (Hint: See Reasoning with Numbers 3-1.)arrow_forwardAt perigee, the Moon is closer than average by 21,100 km. At apogee, the Moon is further than average by 21,100 km. Is the angular diameter more or less at perigee than apogee? What is the angular diameter of the Moon at perigee? At apogee? By how much greater a percentage is the angular diameter larger or smaller at perigee than at the average distance? At apogee? (Hint: The Moons average distance from Earth is given in this chapter.)arrow_forward
- Give four ways to demonstrate that Earth is spherical.arrow_forwardWhat do you see in this image that indicates this planet formed far from the Sun?arrow_forwardVenus’s average distance from the Sun is 0.72 AU and Saturn’s is 9.54 AU. Calculate the circular orbital velocity of Venus and Saturn around the Sun. (Notes: The mass of the Sun is 1.991030 kg. An AU is 1.501011 m.)arrow_forward
- 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 smallarrow_forwardIf 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 sensearrow_forwardThe Great Red Spot is a massive, oval, ruddy spot that is prominent in photographs of Jupiter. How may this main characteristic be applied to predict the planet's rotation period?arrow_forward
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