The Cosmic Perspective Fundamentals (2nd Edition)
The Cosmic Perspective Fundamentals (2nd Edition)
2nd Edition
ISBN: 9780133889567
Author: Jeffrey O. Bennett, Megan O. Donahue, Nicholas Schneider, Mark Voit
Publisher: PEARSON
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Chapter 6, Problem 8QQ

Choose the best answer to each of the following. Explain your reasoning.

The asteroid belt lies between the orbits of (a) Earth and Mars. (b) Mars and Jupiter. (c) Jupiter and Saturn.

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Procedure  Table 11.1 presents Djup and Pday for the major Jovian satellites. First use these data and the equation above to calculate Jupiter’s mass in kilograms (kg). Enter your results in the table for each satellite. Next calculate the average Jupiter mass (Mjup, av) and enter the result in the table. Finally, calculate the percent difference (PD) using Mjup, av and the standard value for Jupiter’s mass (1.9 X 1027 kg). In the calculation of PD you can ignore 1027 because it will appear in both numerator and denominator. ________________________________________________________ Table 11.1 Calculated values for Jupiter’s Mass Satellite               Djup       Pday         Mjup Io                       2.95       1.77 Europa              4.69        3.55 Ganymede        7.50        7.15 Callisto           13.15       16.7 __________________________________________________________ Average Jupiter Mass =                              Percent Difference =
We need to create a scale model of the solar system (by shrinking the sun down to the size of a basketball or ~30cm). First, we will need to scale down actual solar system dimensions (planet diameters and average orbital radiuses) by converting our units. There are two blank spaces in the table below. We will effectively fill in the missing data in the next set of questions. Use the example below to help you. Example: What is the scaled diameter of Mercury if the Sun is scaled to the size of a basketball (30 cm)? The actual diameter of Mercury is 4879 km The Sun's diameter is 1392000 km If the Sun is to be reduced to the size of a basketball, then the conversion we need for this equation will be: 30cm1392000km Here is how we run the conversion:      4879km×30cm1392000km=0.105cm    or    0.11cm if we were to round our answer. This means that if the sun in our model is the size of a basketball, Mercury is the size of a grain of sand. We can also see by looking at the table, that we would…
Suppose there were a planet in our Solar System orbiting at a distance of 0.5 AU from theSun, and having ten times the mass and four times the radius of Earth. For reference, theEarth has a mass of 5.97 × 1024 kg and a radius of 6,378 km a) Calculate the density of this hypothetical planet.b) Based on your answer from part a), what do you think this planet would be made of?Explain your reasoning.c) Do this planet’s properties agree with the condensation theory for the formation of ourSolar System? Why or why not?
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