DOC PREVIEW
UMD ASTR 100 - Homework #3

This preview shows page 1 out of 2 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 2 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 2 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

Homework # 3DQ#1 If liquid water is rare on the surface of the planets, then most terrestrial planets must have carbon dioxide rich atmosphere, why?Liquid water on the planet’s surface is required for the removal of atmospheric carbon dioxide. Silicate rock is transformed into carbonate rock, trapping carbon dioxide. Temperature on a planet’s surface determines whether or not water can exist as a liquid and most terrestrial planets are either too hot or too cold. Describe the surfaces of Jupiter’s moons, Io & Europa. Why are they different from the surface of the Earth’s moon?Jupiter’s moons, Io& Europa, are rocky worlds with densities similar to earth’s moon. Io’s surface has no impact caters and many of active volcanoes, while Europa is mostly rock with a thin icy crust and a watery, fluid mantle. Europa’s surface contains very few craters and has long scars suggesting cracks where liquid has flowed up through the ice crust. On the other hand Earth’s Moon has a rocky crust covered with impact craters. Earth’s moon has no watery mantle or active volcanoes as compared to the Jupiter’s moons. The moon has lost internal heat and lacks a watery mantle. The Moon has no current geological activity such as volcanoes to cover craters either. With no atmosphere, erosion cannot erase craters. Due to the tidal forces from Jupiter, Io & Europa are warm inside. Io’s volcanoes throw out material that covers craters. Europa’s crust slips, slides &fractures releasing water and destroying impact craters.Describe Pluto. Describe a typical Kupier Belt Object. How would you classify Pluto – as a planet or Kupier Belt Object? Support your answer.Kupier Belt Objects are small icy planetesimals that lie beyond the orbit of Neptune (30AU) out to 50AU from the Sun. Pluto is the outermost planet with semi major axis equal to 39.5 AU. Pluto’s low density (2 g/cm3) indicates a considerable ice component due to its temperature ( -382F). Pluto is the smallest planet ( 0.19 Earth’s radius) and has the lowest mass (0.002 Earth’s mass). When Pluto was discovered in 1930 it was classified as a planet. In 1978, Pluto’s moon Charon was discovered, strengthening its planetary status. The first Kuiper Belt Object was discovered in 1992. Pluto may be the largest Kuiper Belt Object. If it were discovered today it probably would not be classified as a planet. Astronomers voted in 2001 to keep the ninth planet.DQ#3 Pg 223From what you know now, do you think the government should spend money to locate near Earth asteroids? How serious is the risk?The Earth gets hit by small meteorites everyday and by larger objects very rarely. Still a single collision with a large asteroid or comet could cause planet wide devastation as in case of past collision, which led to dinosaur extinction. Huge mass extinction events are estimated to occur once every 100 million years. Climate altering events that would affectlife can be expected every 30 million years. Although your risk being killed in such an event is about he same as dying in a plane crash, such an event would result in the deathof a billion people. You need to evaluate personally whether or not this risk merits expenditure of government funds. Prob#9 Pg264: How long did it take radio commands to travel from Earth to Voyager 2 as it passed Neptune?This is the classic “when do we get there problem?” You drive in your car to New York City. How long does it take? You need to know how far away New York City is from your location and how fast your car travels. You must pick where you are leaving from—let’s say its 220 miles; and your average speed—60mph. So, time = distance divided by speed= 220miles/60mph=3.7 hrs. The same method applies here.To complete this problem, you must assume a distance for Neptune ( ranges from 29.8 to 30.4 AU). Pick something in that range. Average distance of Neptune from the Sun = 30.1 AUNow the Earth could be on the same side of its orbit, or the opposite meaning that you might have to add about 1 AU or subtract 1 AU. Assume Earth is on the side of its orbit nearest to Neptune, so distance from the Sun to Earth is = 1 AU That makes the distance from Neptune to the Earth = 30.1-1 = 29.1 AU = 29.1 AU x 1.496 x 108 km/AU = 4.4 x109 km Velocity of Radio wave = 3 x 105 km/ sec (Remember that all electromagnetic waves move at the speed of light.)Time taken to reach Earth from Neptune = Distance of Neptune from Earth Velocity of the Radio wave = 4.4 x109 km 3 x 105 km/sec = 1.5 x 10(9-5) secTime taken by radio commands to reach Earth from Neptune = 1.5 x 104 sec = 1.5 x 104 sec/ 60 sec/min = 2.5 x 102 mins = 2.5 x 102 min/ 60 min/hr = 4.2


View Full Document

UMD ASTR 100 - Homework #3

Download Homework #3
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Homework #3 and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Homework #3 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?