Universe: Stars And Galaxies
6th Edition
ISBN: 9781319115098
Author: Roger Freedman, Robert Geller, William J. Kaufmann
Publisher: W. H. Freeman
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Chapter 4, Problem 35Q
To determine
The gravitational force that the earth exerts on moon and the moon exerts on earth.
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Directions: Complete the given table by finding the ratio of the planet’s time of the revolution to its radius.
Planet
Average
Radius of
Orbit
Times of
Revolution
R3
T2
T2 /R3
Mercury
5.7869 × 1010
7.605 ×106
Venus
1.081 × 1011
1.941 ×107
Earth
1.496 × 1011
3.156 ×107
What pattern do you observe in the last column of data? Which law of Kepler's does this seem to support?
It is important to have an idea about the distances between and relative sizes of celestial objects in the
solar system. In Part 1 we will pretend to shrink the solar system until its center piece, the Sun, is 67.3 cm
in diameter. This will represent the Sun which is 1,390,000 km in diameter. The scale of our model is thus:
67.3 cm
= 4.84 x 10-5 cm
km
Scale
1, 390, 000 km
To find the size or distance between objects in centimeters for the model, simply multiply the actual size
or distance in kilometers by the scale factor above.
1. Fill in following table:
Quantity
Actual Distance (km) Model Distance (cm)
Diameter of Sun
1,390,000
Diameter of Earth
12,760
Diameter of Moon
3,480
Distance Between Earth and Sun
1.5 x 108
Distance Between Earth and Moon
384,000
Distance to Proxima Centauri
3.97 x 1013
What is the force of gravity between a keen physics student and Earth if the keen physics student is on a
space walk (3.78x10^2) km above Earth's surface?
Mass of keen physics student including space suit = (3.9x10^2) kg
24
ME = 5.98 x 10 kg
TE = 6.38 x 106 m
(Note: the red writing below just means to input your answer in proper scientific notation, with 2 significant
digits)
Note: Your answer is assumed to be reduced to the highest power possible.
Chapter 4 Solutions
Universe: Stars And 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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- The table below presents the semi-major axis (a) and Actual orbital period for all of the major planets in the solar system. Cube for each planet the semi-major axis in Astronomical Units. Then take the square root of this number to get the Calculated orbital period of each planet. Fill in the final row of data for each planet. Table of Data for Kepler’s Third Law: Table of Data for Kepler’s Third Law: Planet aau = Semi-Major Axis (AU) Actual Planet Calculated Planet Period (Yr) Period (Yr) __________ ______________________ ___________ ________________ Mercury 0.39 0.24 Venus 0.72 0.62 Earth 1.00 1.00 Mars 1.52 1.88 Jupiter…arrow_forwardAs an aspiring science fiction author, you are writing about a space-faring race and their home planet, Krypton, which has one moon. This moon takes 1,702,948 seconds to complete an orbit around Krypton. If the distance from the center of the moon to the surface of Krypton is 462.5 x 106 m and the planet has a radius of 37.2 x 106 m, calculate the moon's centripetal acceleration. Your Answer: Answerarrow_forwardWhat is the force of gravity between a keen physics student and Earth if the keen physics student is on a space walk (3.63x10^2) km above Earth's surface? Mass of keen physics student including space suit = (3.9x10^2) kg 24 ME = 5.98 x 10 kg rE = 6.38 x 106 m %3D (Note: the red writing below just means to input your answer in proper scientific notation, with 2 significant digits) Note: Your answer is assumed to be reduced to the highest power possible.arrow_forward
- The moons Prometheus and Pandora orbit Saturn at 139,350 and 141,700 kilometers, respectively. a. Using Newton's version of Kepler's third law, find the orbital periods of the two moons. b. Find the percent difference in their.distances and in their orbital periods. c. Consider the two in a race around Saturn: In one Prometheus orbit, how far behind is Pandora (in units of time)? In how many Prometheus orbits will Pandora have fallen behind by one of its own orbital periods? Convert this number of periods back into units of time. This is how often the satellites pass by each other.arrow_forwardSpeaking of Mercury, approximately how long is one year on the planet closest to the Sun? The sizes of the objects in our model of the solar system are not to scale; however, the relative orbital periods around the Sun are. So you can answer this question by counting the revolutions of Mercury during one Earth year. a) Approximately 90 Earth days b) Approximately 370 Earth days c) Approximately 120 Earth days d) Approximately 50 Earth daysarrow_forwardA planet orbits the Sun every 539.4 years. What is its distance (semi-major axis of the orbit) from the Sun? (Give your answer in SI units and include the unit.)arrow_forward
- Suppose you were given a 3 in diameter ball to represent the Earth and a 1 in diameter ball to represent the Moon. (The actual ratio of Earth diameter to Moon diameter is 3.7 to 1.) The actual average Earth–Moon distance is about 384,000 kilometers, and Earth’s diameter is about 12,800 kilometers. How many “Earth diameters” is the distance from Earth to the Moon? Based on your answer to Question 2, what is the correct scaled distance of the Moon, using the 3-inch ball as Earth? The Sun’s actual diameter is about 1,400,000 kilometers. How many “Earth diameters” is this? Given your 3-inch Earth, how large (i.e what diameter) of a ball would you need to represent the Sun? Give your answer in feet. The average Earth–Sun distance is about 149,600,000 km. To represent this distance to scale, how far away would you have to place your 3-inch Earth from your Sun? Give your answer in feet. Could we use this scale to visualize the solar system instead of just the Earth and Moon? Why or Why…arrow_forwardThe fictional asteroid Lilliput orbits around the Sun one time every 3.605 Earth years. What is the average distance for Lilliput away from the Sun in astronomical units (AU)? One astronomical unit is equal to the average distance the Earth is away from the Sun, 1.496x1011m. The mass of the Sun is 1.989x1030kg.arrow_forwardMars is 1.5 times as far away from the Sun as Earth. Earth’s axis is tilted at 23.5o compared to the ecliptic. The axis of Mars is tilted at 25o compared to the ecliptic. The atmosphere on Earth is 100 times as thick as the atmosphere on Mars. Which of the following statements is true? 1.)Mars is so cold that the water there is ice, while Earth does not have any ice 2.)When it is summer in Earth’s northern hemisphere, it is winter on Mars’ southern hemisphere 3.) Earth has seasons, Mars does not 4.) All of the water on Mars is frozen, while Earth has water in solid, liquid and gas formarrow_forward
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