Question

t = r^{3/2}
or
t = \sqrt{r^3}
for example, if a planet is 4 au from the sun, its period would be:
t = 4^{3/2} = 8
so, this planet would take 8 earth years to complete one orbit around the sun. scientists use mathematical models to predict the motion of planets and other objects in the solar system.

fill in the blank: fill in the blank with the correct words.

1. the shape of a planet’s orbit is an ______, and its “stretchiness” is measured by ______.
2. when a planet is at its closest point to the sun, it is at the ________.
3. kepler’s third law can be written as t = ________^{3/2}, where t is the orbital period and r is the average distance from the sun in au.
4. planets with ________ eccentricity have orbits that are nearly circular.
5. when a planet is farther from the sun, its speed ________ according to kepler’s second law.

word bank:
ellipse, lower, eccentricity, perihelion, r, decreases

multiple choice questions: highlight the correct response.

1. what does the eccentricity value of 0 mean for a planet’s orbit?

a) the orbit is a perfect circle
b) the orbit is a straight line
c) the orbit is a triangle
d) the orbit is highly stretched out

1. if a planet’s orbit has high eccentricity, what does its path look like?

a) a perfect circle
b) a very stretched ellipse
c) a straight line
d) a spiral

1. at which point in its orbit does a planet move fastest, based on kepler’s second law?

a) perihelion (closest to the sun)
b) aphelion (farthest from the sun)
c) the middle of the orbit
d) always at the same speed

1. according to kepler’s third law, if a planet is 9 au from the sun, what is its orbital period?

a) 3 years
b) 9 years
c) 27 years
d) 1 year

1. which law relates a planet’s orbital period to its distance from the sun?

a) first law
b) second law

Explanation:

Fill in the Blank Answers:

1. The shape of a planet’s orbit is an ellipse, and its “stretchiness” is measured by eccentricity.
2. When a planet is at its closest point to the Sun, it is at the perihelion.
3. Kepler’s Third Law can be written as \( T = \boldsymbol{R}^{3/2} \), where \( T \) is the orbital period and \( R \) is the average distance from the Sun in AU.
4. Planets with lower eccentricity have orbits that are nearly circular.
5. When a planet is farther from the Sun, its speed decreases according to Kepler’s Second Law.

Multiple Choice Answers:

1. What does the eccentricity value of 0 mean for a planet’s orbit?

• a) The orbit is a perfect circle (Eccentricity of 0 means no “stretch”—a perfect circle).

1. If a planet’s orbit has high eccentricity, what does its path look like?

• b) A very stretched ellipse (High eccentricity means the orbit is more elongated/stretched).

1. At which point in its orbit does a planet move fastest, based on Kepler’s Second Law?

• a) Perihelion (closest to the Sun) (Kepler’s Second Law: Planets move fastest when closest to the Sun).

1. According to Kepler’s Third Law, if a planet is 9 AU from the Sun, what is its orbital period?

• Using \( T = R^{3/2} \): \( T = 9^{3/2} = (\sqrt{9})^3 = 3^3 = 27 \). Thus, c) 27 years.

1. Which law relates a planet’s orbital period to its distance from the Sun?

• Kepler’s Third Law (not listed as an option here? Wait, the options are “a) First Law” and “b) Second Law”—this may be a typo, but if forced to choose from given options, none are correct. However, if assuming a formatting error, the correct law is Kepler’s Third Law, but based on the provided options, there’s a mistake. If we proceed with the given options, this question has an error.)

Note for Question 5 (Multiple Choice):

The provided options (a) First Law, (b) Second Law) do not include Kepler’s Third Law (which relates orbital period to distance). This is likely a typo in the question. If we assume a missing option (e.g., “c) Third Law”), that would be correct.

Answer:

Fill in the Blank Answers:

1. The shape of a planet’s orbit is an ellipse, and its “stretchiness” is measured by eccentricity.
2. When a planet is at its closest point to the Sun, it is at the perihelion.
3. Kepler’s Third Law can be written as \( T = \boldsymbol{R}^{3/2} \), where \( T \) is the orbital period and \( R \) is the average distance from the Sun in AU.
4. Planets with lower eccentricity have orbits that are nearly circular.
5. When a planet is farther from the Sun, its speed decreases according to Kepler’s Second Law.

Multiple Choice Answers:

1. What does the eccentricity value of 0 mean for a planet’s orbit?

• a) The orbit is a perfect circle (Eccentricity of 0 means no “stretch”—a perfect circle).

1. If a planet’s orbit has high eccentricity, what does its path look like?

• b) A very stretched ellipse (High eccentricity means the orbit is more elongated/stretched).

1. At which point in its orbit does a planet move fastest, based on Kepler’s Second Law?

• a) Perihelion (closest to the Sun) (Kepler’s Second Law: Planets move fastest when closest to the Sun).

1. According to Kepler’s Third Law, if a planet is 9 AU from the Sun, what is its orbital period?

• Using \( T = R^{3/2} \): \( T = 9^{3/2} = (\sqrt{9})^3 = 3^3 = 27 \). Thus, c) 27 years.

1. Which law relates a planet’s orbital period to its distance from the Sun?

• Kepler’s Third Law (not listed as an option here? Wait, the options are “a) First Law” and “b) Second Law”—this may be a typo, but if forced to choose from given options, none are correct. However, if assuming a formatting error, the correct law is Kepler’s Third Law, but based on the provided options, there’s a mistake. If we proceed with the given options, this question has an error.)

Note for Question 5 (Multiple Choice):

The provided options (a) First Law, (b) Second Law) do not include Kepler’s Third Law (which relates orbital period to distance). This is likely a typo in the question. If we assume a missing option (e.g., “c) Third Law”), that would be correct.