The following table gives the mean velocity of planets in their orbits versus their mean distance from the sun. Note that 1 AU (astronomical unit) is the mean distance from Earth to the sun, about 93 million miles. Planet d = distance (AU) V = velocity (km/sec) Mercury Venus 0.39 47.4 0.72 35.0 Earth 1 29.8 Mars 1.52 24.1 Jupiter Saturn 5.20 13.1 9.58 9.7 Uranus 19.20 6.8 Neptune 30.05 5.4 Astronomers tell us that it is reasonable to model these data with a power function. (a) Use power regression to express velocity as a power function of distance from the sun. (Round regression parameters to two decimal places.) v= 21.41 x d-0.50 V = 21.41 x d-0.29 ov = 29.73 x d-0.50 v = 29.73 x d0.73 v = 29.73 x d0.50 (b) Plot the data along with the regression equation. 50 50 50 50 40 40 40 40 30 30 30 30 20 20 20 20 10 10 10 10 d 5 10 15 20 25 30 5 10 15 20 25 30 5 10 15 20 25 30 5 10 15 20 25 30 (c) An asteroid orbits at a mean distance of 3 AU from the sun. According to the power model you found in part (a), what is the mean orbital velocity of the asteroid? (Round your answer to two decimal places.) | km/sec
The following table gives the mean velocity of planets in their orbits versus their mean distance from the sun. Note that 1 AU (astronomical unit) is the mean distance from Earth to the sun, about 93 million miles. Planet d = distance (AU) V = velocity (km/sec) Mercury Venus 0.39 47.4 0.72 35.0 Earth 1 29.8 Mars 1.52 24.1 Jupiter Saturn 5.20 13.1 9.58 9.7 Uranus 19.20 6.8 Neptune 30.05 5.4 Astronomers tell us that it is reasonable to model these data with a power function. (a) Use power regression to express velocity as a power function of distance from the sun. (Round regression parameters to two decimal places.) v= 21.41 x d-0.50 V = 21.41 x d-0.29 ov = 29.73 x d-0.50 v = 29.73 x d0.73 v = 29.73 x d0.50 (b) Plot the data along with the regression equation. 50 50 50 50 40 40 40 40 30 30 30 30 20 20 20 20 10 10 10 10 d 5 10 15 20 25 30 5 10 15 20 25 30 5 10 15 20 25 30 5 10 15 20 25 30 (c) An asteroid orbits at a mean distance of 3 AU from the sun. According to the power model you found in part (a), what is the mean orbital velocity of the asteroid? (Round your answer to two decimal places.) | km/sec
Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
6th Edition
ISBN:9781337111348
Author:Bruce Crauder, Benny Evans, Alan Noell
Publisher:Bruce Crauder, Benny Evans, Alan Noell
Chapter5: A Survey Of Other Common Functions
Section5.3: Modeling Data With Power Functions
Problem 2E: Planetary Velocity The following table gives the mean velocity of planets in their orbits versus...
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Transcribed Image Text:The following table gives the mean velocity of planets in their orbits versus their mean distance from the sun. Note that 1 AU (astronomical unit) is the mean distance from Earth to the sun, about 93 million miles.
Planet
d = distance (AU)
v = velocity (km/sec)
Mercury
0.39
47.4
Venus
0.72
35.0
Earth
1
29.8
Mars
1.52
24.1
Jupiter
5.20
13.1
Saturn
9.58
9.7
Uranus
19.20
6.8
Neptune
30.05
5.4
Astronomers tell us that it is reasonable to model these data with a power function.
(a) Use power regression to express velocity as a power function of distance from the sun. (Round regression parameters to two decimal places.)
V = 21.41 x d
-0.50
V = 21.41 x d-0.29
° v = 29.73 x d¬0.50
V = 29.73 x d-0.73
v = 29.73 x d0.50
(b) Plot the data along with the regression equation.
v
V
50
50 H
50
50H
40
40
40
40
30
30
30
30
20
20
20
20-
10
10
10
10
d
10
15
20
25
30
d
d
5
10
15
20
25
30
5
10
15
20
25
30
5
10
15
20
25
30
(c) An asteroid orbits at a mean distance of 3 AU from the sun. According to the power model you found in part (a), what is the mean orbital velocity of the asteroid? (Round your answer to two decimal places.)
km/sec
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