Universe
11th Edition
ISBN: 9781319039448
Author: Robert Geller, Roger Freedman, William J. Kaufmann
Publisher: W. H. Freeman
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Chapter 3, Problem 1CLC
To determine
The estimated size of the Sun if Aristarchus had estimated the Sun to be
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d. Diameter of the Sun
I h
A
d.
Standing on the beach at sunset, you extend the tip of your finger at your full arm length from
your face, covering the Sun. Upon moving your finger around, you find that only about half of its
width is needed to completely cover the Sun's diameter. You measure your finger width to be 0.5
inches. You know your arm length to be 28.0 inches. You have be told that the Sun is
approximately 93 million miles away. Use this information to determine the approximate
diameter of the Sun, filling in the table below with the proper quantities measure in meters.
1 = ½ finger
g = eye level
height
d = object
h = Diameter
f= object
height from
level to top of eye level
X = Arm
A =
angle
width
length
distance
from eye
of the Sun
object
The Sun occupies an angle of approximately 0.5 degrees to us on Earth, at a distance of 150 million kilometers from the Sun. Given this information, calculate the Sun's radius.
Suppose, hypothetically, that the Earth orbited the Sun at half its current distance. (That is, at 1/2 AU instead of 1 AU). What would be the length of the year? What else would be different?
Chapter 3 Solutions
Universe
Ch. 3 - Prob. 1CCCh. 3 - Prob. 2CCCh. 3 - Prob. 3CCCh. 3 - Prob. 4CCCh. 3 - Prob. 5CCCh. 3 - Prob. 6CCCh. 3 - Prob. 7CCCh. 3 - Prob. 8CCCh. 3 - Prob. 9CCCh. 3 - Prob. 10CC
Ch. 3 - Prob. 11CCCh. 3 - Prob. 12CCCh. 3 - Prob. 13CCCh. 3 - Prob. 14CCCh. 3 - Prob. 1CLCCh. 3 - Prob. 1QCh. 3 - Prob. 2QCh. 3 - Prob. 3QCh. 3 - Prob. 4QCh. 3 - Prob. 5QCh. 3 - Prob. 6QCh. 3 - Prob. 7QCh. 3 - Prob. 8QCh. 3 - Prob. 9QCh. 3 - Prob. 10QCh. 3 - Prob. 11QCh. 3 - Prob. 12QCh. 3 - Prob. 13QCh. 3 - Prob. 14QCh. 3 - Prob. 15QCh. 3 - Prob. 16QCh. 3 - Prob. 17QCh. 3 - Prob. 18QCh. 3 - Prob. 19QCh. 3 - Prob. 20QCh. 3 - Prob. 21QCh. 3 - Prob. 22QCh. 3 - Prob. 23QCh. 3 - Prob. 24QCh. 3 - Prob. 25QCh. 3 - Prob. 26QCh. 3 - Prob. 27QCh. 3 - Prob. 28QCh. 3 - Prob. 29QCh. 3 - Prob. 30QCh. 3 - Prob. 31QCh. 3 - Prob. 32QCh. 3 - Prob. 33QCh. 3 - Prob. 34QCh. 3 - Prob. 35QCh. 3 - Prob. 36QCh. 3 - Prob. 37QCh. 3 - Prob. 38QCh. 3 - Prob. 39QCh. 3 - Prob. 40QCh. 3 - Prob. 41QCh. 3 - Prob. 42QCh. 3 - Prob. 43QCh. 3 - Prob. 44QCh. 3 - Prob. 46QCh. 3 - Prob. 47QCh. 3 - Prob. 48QCh. 3 - Prob. 49QCh. 3 - Prob. 50QCh. 3 - Prob. 53Q
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- A planet's speed in orbit is given by V = (30 km/s)[(2/r)-(1/a)]0.5 where V is the planet's velocity, r is the distance in AU's from the Sun at that instant, and a is the semimajor axis of its orbit. Calculate the Earth's velocity in its orbit (assume it is circular): What is the velocity of Mars at a distance of 1.41 AU from the Sun? What is the spacecraft's velocity when it is 1 AU from the Sun (after launch from the Earth)? What additional velocity does the launch burn have to give to the spacecraft? (i.e. What is the difference between the Earth's velocity and the velocity the spacecraft needs to have?) How fast will the spacecraft be traveling when it reaches Mars? Does the spacecraft need to gain or lose velocity to go into the same orbit as Mars?arrow_forwardThe angle between two lines drawn from a point on Earth to two opposite sides of the Moon make an angle of 0.5 degrees. If you do the same thing for the two opposite ends of Andromeda (as shown above), you find an angle of 5 degrees. Let's assume Andromeda and the Moon are equally far away from our location on Earth (of course that's wrong, but how are we supposed to know?) - then how much larger would the diameter of Andromeda be (as indicated by the arrows at the top), compared to the diameter of the Moon? Pick the answer that's closest to what you get under this hypothetical assumption: A. Equal Diameter B. Twice C. Five times D. Ten timesarrow_forwardFrom a distance of 300 km above the surface of the Moon, what is the angular diameter in arc seconds of an astronaut in a space suit who has a linear diameter of 0.80 m as seen from above?arrow_forward
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