Download Copernicus`s model did not account for the observed

Homework Chapters 2-3
Due: January 6, 2014
ASTRON 311 Introduction to Astronomy
Prof. Menningen
p. 1/3
Name: __________________________________
1.
What is retrograde motion?
a. East to west motion of the Sun.
b. Occasional west to east motion of the planets relative to the stars.
c. Occasional east to west motion of the planets relative to the stars.
d. East to west motion of the Moon relative to the stars.
2.
The purpose of describing planetary orbits in terms of epicycles and deferents in Ptolemy’s model was to account for the
a. generally eastward motion of a planet compared to background stars while the whole sky appeared to
move westward.
b. variation of brightness of a planet with time.
c. difference between the sidereal period and the synodic period of a planet.
d. pattern of direct and retrograde motion of a planet as it moved slowly against the background of stars.
3.
Which of the following statements correctly states the significance of Galileo's observation that Venus shows phases?
a. Since the phases were NOT correlated with angular size, they actually provided more support for the
geocentric theory than for the heliocentric theory.
b. The phases were interesting but did not have any particular significance.
c. The phases showed that, like the Moon, Venus is always much closer to Earth than the Sun.
d. The phases were correlated with angular size in a way that supported the heliocentric theory.
4.
Give at least two (nonreligious) reasons why Copernicus’s heliocentric model of the solar system was not immediately
accepted by the scientific community.
Copernicus’s model did not account for the observed positions of the planets any more
accurately than did Ptolemy’s geocentric model. Stellar parallax could not be measured and
therefore could not verify that the Earth orbits the sun. Finally, his views ran counter to the
“common sense” understanding that the Earth is at rest and the heavenly objects move.
5.
Tycho Brahe
a. built improved refracting telescopes that confirmed and extended Galileo's observations of the sky.
b. made accurate measurements of planetary positions, which Kepler later used to find the shapes of
planetary orbits.
c. developed the first detailed heliocentric model for the solar system.
d. developed a reflecting telescope, which used a curved mirror to focus the light.
6.
What is the difference between the synodic orbital period and the sidereal orbital period of a planet?
The synodic orbital period is the time it takes for a planet to go from some configuration to
that same configuration again from the vantage point of an observer on Earth. The sidereal
orbital period is the time it takes for a planet to complete one orbit around the Sun with
respect to the "fixed" stars.
Homework Chapters 2-3
7.
ASTRON 311 Introduction to Astronomy
Prof. Menningen
p. 2/3
The orbit of an asteroid about the Sun has a perihelion distance of 0.10 AU and an aphelion distance of 0.40 AU. What is
the semimajor axis of the asteroid’s orbit? What is its orbital period?
The sum of the perihelion distance and the aphelion distance is twice the semimajor axis, so
a  12  0.10  0.40 AU   0.25 AU .
The orbital period is: P  a3 
8.
 0.25 AU 
3
 0.125 y
A comet with a period of 125 years moves in a highly elongated orbit about the Sun. At perihelion, the comet comes very
close to the Sun's surface. What is the comet's average distance from the Sun? What is its maximum distance from the
Sun?
If the period is 125 years then the average distance is: a 
3
P 2  3 125 y   25 AU .
2
The farthest it can get from the Sun is almost twice the average distance, or 50 AU.
9.
A comet orbits the Sun with a sidereal period of 27.0 years. (a) Find the semimajor axis of the orbit. (b) At aphelion, the
comet is 17.5 AU from the Sun. What is its perihelion distance?
(a) a  3 P 2  3  27.0 y   9.0 AU .
2
(b) The perihelion distance is the major axis (2a) minus the aphelion distance:
Perihelion distance = 2×9.0 AU – 17.5 AU = 0.5 AU
10. Suppose that you traveled to a planet with 3 times the mass and 3 times the diameter of the Earth. Would you weigh more
or less on that planet than on Earth? By what factor?
If Fold 
m  3M E 
GmM E
is your weight on Earth, and Fnew  G
is your weight on the
2
2
RE
 3RE 
G
m  3M E 
 3RE 
F
F
planet, then you would weigh new times as much there: new 
mM
Fold
Fold
G 2E
RE
would weigh less by a factor of 3.
2

3 1
 . You
9 3
Homework Chapters 2-3
ASTRON 311 Introduction to Astronomy
Prof. Menningen
p. 3/3
11. On Earth, a 50 kg astronaut weighs 490.0 newtons. What would she weigh if she landed on Jupiter's moon Europa? What
fraction is this of her weight on Earth? See Appendix Table 5 for relevant data about Europa.
Europa’s mass is 4.797×1022 kg and its radius is 1565 km = 1.565×106 m. The astronaut
would therefore weigh:
 50 kg   4.797 1022 kg 
mM
11
2
2
F  G 2   6.67  10 N  m /kg 
 65.3 N
2
d
1.565 106 m 
This is less than 490 newtons by a factor of 490/65.3 = 7.50
12. The average distance from center of the Moon to the center of the Earth is 3.844×108 m, the radius of the Earth is
6.378×106 m, and the mass of the Moon is 7.35×1022 kg. Calculate the gravitational force that the Moon exerts (a) on a
1.0 kg rock at the point on the Earth's surface closest to the Moon and (b) on a 1.0 kg rock at the point on the Earth's
surface farthest from the Moon. (c) Find the difference between the two forces you calculated in parts (a) and (b). This
difference is the tidal force pulling these two rocks away from each other. (d) Explain why tidal forces cause only a very
small deformation of the Earth.
(a) For the closest side: d = 3.844×108 m – 6.378×106 m = 3.780×108 m. Then calculate:
11
2
2
22
m1 m2  6.67  10 N·m /kg  7.35  10 kg  1 kg 
F G 2 
 3.431  105 N
2
8
d
3.780 10 m 
(b) For the farthest side: d = 3.844×108 m + 6.378×106 m = 3.908×108 m. Then,
 6.67 10
F
11
N·m 2 /kg 2  7.35 1022 kg  1 kg 
3.908 10 m 
8
2
= 3.210 105 N
(c) The difference is 2.21×10−6 N.
(d) There is only a very small deformation of the Earth’s surface due to tidal forces because
the tidal force is very small compared to the Earth’s gravity, which tends to maintain a
spherical surface.