+ (38 yr) 1 yr. = 742 mo. 1 yr The number of days in 742 months is

Size: px

Start display at page:

Download "+ (38 yr) 1 yr. = 742 mo. 1 yr The number of days in 742 months is"

Brandon Hunter
5 years ago
Views:

1 ASTR 101 Homework 2 Solutions 3-44 Chinese Calendar The traditional Chinese lunar calendar has 12 months in most years but adds a thirteenth month to 22 of every 60 years. How many days does this give the Chinese calendar in each 60-year period? How does this compare to the number of days in 60 years on a solar calendar? Based on your answers, explain how this scheme is similar to the scheme used by lunar calendars that follow the Metonic cycle. (5pt) If there are 22 years that have 13 months, then there are 38 years that have the normal 12 months in this 60-year cycle. Therefore, the total number of months in the Chinese lunar calendar are (22 yr) ( ) 13 mo + (38 yr) 1 yr ( ) 12 mo = 742 mo. 1 yr The number of days in 742 months is ( ) d (742 mo) = d. 1 mo For the solar calendar, there are only 12 months in every year. Therefore, the number of days is ( ) d (60 yr) = d. 1 yr The solar calendar has d more days than the Chinese lunar calendar. This is similar to the lunar calendars that follow the Metonic cycle in that it will overshoot the length of a solar year, then undershoot, eventually matching the length of a solar year before overshooting again Method of Eratosthenes I You are an astronomer on planet Nearth, which orbits a distant star. It has recently been accepted that Nearth is spherical in shape, though no one knows its size. One day, while studying in the library of Alectown, you learn that on the equinox your sun is directly overhead in the city of Nyene, located 1000 kilometers due north of you. On the equinox, you go outside and observe that the altitude of your sun is 80. What is the circumference of Nearth? (5pt) Since the altitude of this sun is 80, the angle it makes with the zenith is = 10. Therefore, the circumference of Nearth is 10 C = 1000 km km C = = km. It turns out that Nearth has about 1.5 times the circumference of Earth! 1

2 3-48 Eris Orbit The dwarf planet Eris orbits the Sun every 557 years. What is its average distance (semimajor axis) from the Sun? How does its average distance compare to that of Pluto? (5pt) We can use Kepler s Third Law here: p 2 = a 3 a = 3 p 2 = 3 (557 yr) 2 = AU. The semimajor axis of Pluto is AU (see Appendix E). Eris obviously has a larger average distance, almost about twice that of Pluto. Does It Make Sense? Decide whether the statement makes sense (or is clearly true) or does not make sense (or is clearly false). Explain clearly; not all of these have definitive answers, so your explanation is more important than your chosen answer Suppose you could enter a vacuum chamber (a chamber with no air in it) on Earth. Inside this chamber, a feather would fall at the same rate as a rock. This statement is true (0.5pt). Since there is no air, there is no air resistance to slow the descent of the feather. Both the rock and the feather would fall at the same rate (0.5pt) If an astronaut goes on a space walk outside the Space Station, she will quickly float away from the station unless she has a tether holding her to the station. This statement is false (0.5pt). Since the astronaut maintains the same orbit as the Space Station, she ll continue to remain next to the station. The only way she would float away is if she makes an effort to, i.e. pushing off from the Space Station when she leaves (0.5pt) I used Newton s version of Kepler s third law to calculate Saturn s mass from orbital characteristics of its moon Titan. This statement is true (0.5pt), because Newton s version of Kepler s third law has the mass of the central body in it (0.5pt). If you were to actually do this calculation, you would get exactly Saturn s mass of kg! 4-19 If the Sun were magically replaced with a giant rock that had precisely the same mass, Earth s orbit would not change. This statement is true (0.5pt). Because the giant rock and the Sun have the same mass, the force of gravity would still be the same between it and the other bodies in the solar system. They d remain on their orbit, including Earth (0.5pt). If the Sun was replaced by a giant rock, we d lose the heat and sunlight that powers life on Earth, but it d be easier to do astronomy! 4-24 Someday soon, scientists are likely to build an engine that produces more energy than it consumes. This statement is false (0.5pt). You can t produce more energy than you consume because that would violate the law of conservation of energy (0.5pt). 2

3 4-52 Understanding Newton s Version of Kepler s Third Law Find the orbital period for the planet in each case. (a) A planet with twice Earth s mass orbiting at a distance of 1 AU from a star with the same mass as the Sun. (b) A planet with the same mass as Earth orbiting at a distance of 1 AU from a star with four times the Sun s mass. (5pt) In both cases, the semimajor axis of the planet is 1 AU. Note that the mass of the orbiting planet does not matter compared to the mass of the star, so you can safely ignore any changes to the mass of the planet. See Mathematical Insight 4.3 on page 126 for more details. (a) Since the planet is orbiting a star of the same mass as the Sun at Earth s distance, the orbital period is still one year. The planet s mass does not enter into Newton s version of Kepler s third law. (b) Now we have a star with quadruple the Sun s mass, 4M (the symbol means sun ). We can write Newton s version of Kepler s third law: p 2 = 4π2 G(4M ) a3. We can factor out the 1 4 from the equation and bring it outside the fraction: p 2 = 1 4π 2 a 3 4 GM 1 4π p = 2 a 3. 4 GM The stuff under the second square root is just Kepler s third law for the Earth, which we know the period of: 1 yr. Now, we just need the square root of onefourth, which is one-half. Therefore, the period of this planet orbiting a star with four times the Sun s mass is 0.5 yr Weights on Other Worlds Calculate the acceleration of gravity on the surface of each of the following worlds. How much would you weigh, in pounds, on each of these worlds? (a) Mars (mass = 0.11M Earth, radius = 0.53R Earth ). (b) Venus (mass = 0.82M Earth, radius = 0.95R Earth ). Jupiter (mass = 317.8M Earth, radius = 11.2R Earth ) Bonus: Given that Jupiter has no solid surface, how could you weigh yourself on Jupiter? (d) Jupiter s moon Europa (mass = 0.008M Earth, radius = 0.25R Earth ) (e) Mars s moon Phobos (mass = kg, radius = 12 km) (5pt) One of your TAs, Ray, weighs 195 lbs or 88 kg. We ll use his weight in these problems, but obviously you should use yours. The gravitational acceleration on Earth is 9.8 m s 2. 3

4 (a) The mass of Mars is M Mars = 0.11M Earth and its radius is R Mars = 0.53R Earth. a = G(0.11M Earth) (0.53R Earth ) 2 = 0.11 (0.53) 2 GM Earth = 0.392(9.8 m s 2 ) = m s 2. Ray s weight can be determined from Newton s Second Law:, where m is Ray s mass and a is the gravitational acceleration. = (88 kg)(3.838 m s 2 ) = N = lbs. (b) The mass of Venus is M Venus = 0.82M Earth and its radius is R Venus = 0.95R Earth. a = G(0.82M Earth) (0.95R Earth ) 2 = 0.82 (0.95) 2 GM Earth = 0.909(9.8 m s 2 ) = m s 2. = (88 kg)(8.904 m s 2 ) = N = N. (c) The mass of Jupiter is M Jupiter = 317.8M Earth and its radius is R Jupiter = 11.2R Earth. a = G(317.8M Earth) (11.2R Earth ) 2 = (11.2) 2 GM Earth = 2.533(9.8 m s 2 ) = m s 2. 4

5 = (88 kg)( m s 2 ) = N = lbs An example of how you might weigh yourself on Jupiter without a scale would be to measure how much gas you displace and convert that volume into a mass, assuming you have densities of yourself and the gas around you. (d) The mass of Europa is M Europa = 0.008M Earth and its radius is R Europa = 0.25R Earth. a = G(0.008M Earth) (0.25R Earth ) 2 = (0.25) 2 GM Earth = 0.128(9.8 m s 2 ) = m s 2. = (88 kg)(1.254 m s 2 ) = N = lbs. Fun fact: if we discovered life on Europa, they would be called Europeans! (e) The mass of Phobos is M Phobos = kg and its radius is R Phobos = m. a = GM Phobos R 2 Phobos = ( N m 2 kg 2 )( kg) ( m) 2 = m s 2. = (88 kg)(0.005 m s 2 ) = 0.44 N = lbs. 5

6 Extra Credit 3-50 Halley Orbit Halley s Comet orbits the Sun every 76.0 years and has an orbital eccentricity of (a) Find its average distance from the Sun (semimajor axis). (b) Find its perihelion and aphelion distances. (2pt) Halley s Comet s period is 76 yr with an orbital eccentricity e = (a) We can use Kepler s Third Law to determine the semimajor axis. p 2 = a 3 a = 3 p 2 = 3 (76 yr) 2 = AU. (b) The perihelion distance and aphelion distance are d p = a(1 e) = ( AU)(1 0.97) = AU d a = a(1 + e) = ( AU)( ) = AU. 6

A = 6561 times greater. B. 81 times greater. C. equally strong. D. 1/81 as great. E. (1/81) 2 = 1/6561 as great Pearson Education, Inc.

A = 6561 times greater. B. 81 times greater. C. equally strong. D. 1/81 as great. E. (1/81) 2 = 1/6561 as great Pearson Education, Inc. Q13.1 The mass of the Moon is 1/81 of the mass of the Earth. Compared to the gravitational force that the Earth exerts on the Moon, the gravitational force that the Moon exerts on the Earth is A. 81 2