Question: Distance from the sun (au) Orbital Period (years)

Answer: .....

Question: Mercury

Answer: 0.39 0.242

Question: Venus

Answer: 0.72 0.616

Question: Earth

Answer: 44927

Question: Mars

Answer: 1.52 1.88

Question: Jupiter

Answer: 5.20 N/A

Question: Saturn

Answer: N/A 29.5

Question: Uranus

Answer: 19.18 84.0

Question: Neptune

Answer: 30.06 165

Question: Mercury and Venus are the sun than Earth is.

Answer: closer to

Question: So, their orbital periods are than Earth's orbit.

Answer: shorter

Question: The further a planet is from the sun, the its orbit is.

Answer: longer

Question: A power regression equation for this data is:T = A^

Answer: 1.5

Question: Use the fact that 1.5 is the same as 3/2 to select all of the equations below that are equivalent toT = A1.5.

Answer: T = A3/2T = (A1/2)3(all four options)

Question: The equation T = A3/2 is great if you know A and want to find T.Now, identify the correct inverse equation, which can be used to find A when you are given T.

Answer: A = T2/3

Question: You used a telescope and other mathematics to discover that Jupiter is 5.20 au from the sun. Use the equation to find its orbital period. Round to the nearest tenth of a year.Jupiter's Orbit: years

Answer: 11.9

Question: Suppose that your data shows that Saturn orbits every 29.5 years. To the nearest hundredth of an au, how far is Saturn from the Sun?Saturn's Distance from the Sun: au

Answer: 9.55

Question: Convert the equation you found, T = A3/2, into a form without any rational exponents.

Answer: C T2=A3

Question: Translate the answer, T2 = A3, into words:The squaresquare rootcube of the orbital period, T, of aplanet is equal to the squaresquare rootcube of the average distance, A, of the planet from the Sun.

Answer: squarecube