21st Century Astronomy 6E
6th Edition
ISBN: 9780393690675
Author: Laura Kay, Stacy Palen, George Blumenthal
Publisher: W. W. Norton
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Chapter 4, Problem 16QP
To determine
The fundamental difference between Kepler’s laws and Newton’s laws.
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Newton’s law of gravitation and the formula for centripetal acceleration can be used to show that:
T^2=(4π^2/Gms)R^3 where G is the universal constant of gravitation and MS is the mass of the Sun. Take logarithms to base 10 of both sides of the equation to complete the expression for 2 lg T.2 lg T = ……………… × lg R + ……………………
What is Kepler's first law of
planetary motion?
The period of a planet's
orbit is proportional to
its distance from the
sun.
Planets have circular
orbits.
Planets have elliptical
orbits.
The eccentricity of a
planet's orbit is
proportional to its
distance from the sun.
Which of the following statements is supported by Kepler's laws of planetary motion?
Earth orbits the Sun at a constant speed, never speeding up or slowing down.
Earth's orbit is a perfect circle, with the Sun located at the center of the circle.
Earth orbits the Sun at a slightly faster speed every year.
Earth has an elliptical orbit, with the Sun located at one focus of the ellipse.
Chapter 4 Solutions
21st Century Astronomy 6E
Ch. 4.1 - Prob. 4.1ACYUCh. 4.1 - Prob. 4.1BCYUCh. 4.2 - Prob. 4.2CYUCh. 4.3 - Prob. 4.3CYUCh. 4.4 - Prob. 4.4CYUCh. 4 - Prob. 1QPCh. 4 - Prob. 2QPCh. 4 - Prob. 3QPCh. 4 - Prob. 4QPCh. 4 - Prob. 5QP
Ch. 4 - Prob. 6QPCh. 4 - Prob. 7QPCh. 4 - Prob. 8QPCh. 4 - Prob. 9QPCh. 4 - Prob. 10QPCh. 4 - Prob. 11QPCh. 4 - Prob. 12QPCh. 4 - Prob. 13QPCh. 4 - Prob. 14QPCh. 4 - Prob. 15QPCh. 4 - Prob. 16QPCh. 4 - Prob. 17QPCh. 4 - Prob. 18QPCh. 4 - Prob. 19QPCh. 4 - Prob. 20QPCh. 4 - Prob. 21QPCh. 4 - Prob. 22QPCh. 4 - Prob. 23QPCh. 4 - Prob. 24QPCh. 4 - Prob. 25QPCh. 4 - Prob. 26QPCh. 4 - Prob. 27QPCh. 4 - Prob. 28QPCh. 4 - Prob. 29QPCh. 4 - Prob. 30QPCh. 4 - Prob. 31QPCh. 4 - Prob. 32QPCh. 4 - Prob. 33QPCh. 4 - Prob. 34QPCh. 4 - Prob. 35QPCh. 4 - Prob. 36QPCh. 4 - Prob. 37QPCh. 4 - Prob. 38QPCh. 4 - Prob. 39QPCh. 4 - Prob. 40QPCh. 4 - Prob. 41QPCh. 4 - Prob. 42QPCh. 4 - Prob. 43QPCh. 4 - Prob. 44QPCh. 4 - Prob. 45QP
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- Kepler's 1st law says that our Solar System's planets orbit in ellipses around the Sun where the closest distance to the Sun is called perihelion. Suppose I tell you that there is a planet with a perihelion distance of 2 AU and a semi-major axis of 1.5 AU. Does this make physical sense? Explain why or why not.arrow_forwardMars' period (its "year") was noted by Kepler to be about 687 days (Earth days), which is (687d / 365 d) = 1.88 yr. Determine the distance of Mars from the Sun using the Earth as reference. (The distance of Earth from the Sun is 1.50 x 10 m) !3! Thu TMS IMS TE TES TES TE 2 28 x 10 m TES yr After reading and understanding the concept Gravity, please do the following problems: 1. What keeps a satellite up in its orbit around the Earth?arrow_forwardIn a distant star system there are many inhabitable planets. One of these planets is named Qomar. Qomar is 3.2 AU's from its star and takes 6.5 Earth years to go around its star once. There is another planet in the same star system called Ferenginar. Ferenginar is 0.9 AUs from the star. What is the length of a Ferengi year (on Ferenginar) in terms of Earth years?arrow_forward
- Two exoplanets, UCF1.01 and UCF1.02 are found revolving around the same star. The period of planet UCF1.01 is 92.4 days, and that of planet UCF1.02 is 7.1 days. If the average distance of UCF1.01 to the sun is 5,828.0 km, what is the average distance of UCF1.02 to the sun in km? Please keep four digits after decimal points.arrow_forwardMars has an orbital radius of 1.523 AU and an orbital period of 687.0 days. What is its average speed v in SI units? (1 AU is the astronomical unit, the mean distance between the Sun and the Earth, which is 1.496×1011 m) a. 0.00221 AU/day b. 3838 m/s c. 0 d. 1.28×10−9 m/sarrow_forwardA new planet is discovered orbiting a distant star. Observations have confirmed that the planet has a circular orbit with a radius of 12 AU and takes 117 days to orbit the star. Determine the mass of the star. State your answer with appropriate mks units. [NOTE: AU ..stands.for...astronomical unit". It is the average distance between Earth & the Sun. 1 AU≈ 1.496 x 1011 m.] Enter a number with units. I be quite large and your calculator will display the answer as a power of 10. If, as an example, your answer was 8.54 x 1056, you would type "8.54e56" into the answer box (remember to state your units with your answer).]arrow_forward
- You may attempt this question 3 more times for credit. In this problem, we will directly calculate the surface gravity and your weight on another planet. In metric, your weight is measured in "Newtons", and 1 Newton = 1 kg m / s². Newton's constant G = 6.67 x 10-11 m³/(kg s²). Earth has a mass = 5.97 x 1024 kg and a radius of 6378 km. You should be able to verify that g = 9.8 m/s² on Earth using the formula for surface gravity. If your mass is 64 kg, you should also be able to verify you should weigh 626 Newtons. If you can do that you should be OK for what's next. The mass of Venus is 4.87E+24 kg, and it's radius is 6.05E+3 km. What is the surface gravity of this planet? (Watch your units!). m/s² If your mass is 64 kg, what would you weigh on Venus? Newtons. Note: Remember if your answer requires scientific notation to use the "e" notation: "1.1 x 105" is "1.1e5" to OWL.arrow_forwardSolve the Equator. Thanks!arrow_forwardQuestion 4: Use Kepler's 3rd law to find the orbital periods (assume circular orbits) for the inner planets given that their orbital radii are: Mercury: 5.8 x 107 km Venus: 1.08 x 108 kmarrow_forward
- Many people mistakenly believe that the astronauts who orbit Earth are "above gravity." Earth's mass is 6×1024kg , and its radius is 6.38×106m (6380 km ). Use the inverse-square law to find a height above Earth's surface at that the force of gravity on a shuttle is about 94% that at Earth's surface. Express your answer to two significant figures and include the appropriate units.arrow_forwardDescribe three propositions, now known as Kepler’s laws of planetary motion?arrow_forwardUse Kepler's Law, which states that the square of the time, T, required for a planet to orbit the Sun varies directly with the cube of the mean distance, a, that the planet is from the Sun.Using Earth's time of 1 year and a mean distance of 93 million miles, the equation relating T (in years) and a (in million miles) is 804375T2=a3.Use that relation equation to determine the time required for a planet with mean distance of 206 million miles to orbit the Sun. Round to 2 decimal places. yearsarrow_forward
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