Life in the Universe (4th Edition)
4th Edition
ISBN: 9780134089089
Author: Jeffrey O. Bennett, Seth Shostak
Publisher: PEARSON
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Chapter 7, Problem 45IF
Earth Mass. The Moon orbits Earth in an average time of 27.3 days at an average distance of 384,000 kilometers. Use these facts to determine the mass of Earth. (Hint: You may neglect the mass of the Moon, since its mass is only about
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Chapter 7 Solutions
Life in the Universe (4th Edition)
Ch. 7 - Why do we expect the elements of life to be widely...Ch. 7 - How does the strength of sunlight vary with...Ch. 7 - Under what conditions does it seem reasonable to...Ch. 7 - Why is a liquid medium important for life? Why...Ch. 7 - Summarize the three major environmental...Ch. 7 - Why do the Moon and Mercury seem unlikely to be...Ch. 7 - Why is Venus so much hotter than Earth? How does...Ch. 7 - Why does Mars seem such a good candidate for life?Ch. 7 - Briefly discuss the possibility of life on Jupiter...Ch. 7 - With regard to habitability, how do the cases of...
Ch. 7 - What characteristics make some of the large moons...Ch. 7 - Briefly describe the prospects for habitability of...Ch. 7 - Describe and distinguish between space missions...Ch. 7 - For a few of the most important past, present, or...Ch. 7 - On the smallest moon of Uranus, my team discovered...Ch. 7 - New spacecraft images show lakes of liquid water...Ch. 7 - We are pumping water for our new Moon colony from...Ch. 7 - I was part of the first group of people to land on...Ch. 7 - We sent a robotic airplane into the atmosphere of...Ch. 7 - On a moon of Neptune, we discovered photosynthetic...Ch. 7 - We deposited bacteria (from Earth) that get energy...Ch. 7 - The drilled sample showed no signs of life on...Ch. 7 - We cut holes in the frozen surface of a methane...Ch. 7 - The drilled sample from Mars brought up rock that...Ch. 7 - Oxygen and carbon are (a) rarer than almost all...Ch. 7 - On an asteroid that is twice as far as Earth from...Ch. 7 - Compared to liquid water, liquid methane is (a)...Ch. 7 - Frozen lakes often have liquid water beneath their...Ch. 7 - Temperatures on Mercury are (a) always very hot;...Ch. 7 - On Venus, liquid water (a) does not exist...Ch. 7 - The reason Venus is so much hotter than Earth is...Ch. 7 - Life is probably not possible in Jupiters...Ch. 7 - Which of the following are you most likely to find...Ch. 7 - The Cassini spacecraft (a) flew past Pluto; (b)...Ch. 7 - Bizarre Forms of Life. Discuss some forms of life...Ch. 7 - Making a Living. Consider various methods by which...Ch. 7 - Solar System Tour. Based on the brief tour in this...Ch. 7 - Galileo Spacecraft. In 2003, scientists...Ch. 7 - Greenhouse Effect. The text (in Chapter 4) makes...Ch. 7 - Prob. 41IFCh. 7 - Understanding Newtons Version of Keplers Third Law...Ch. 7 - Understanding Newtons Version of Keplers Third Law...Ch. 7 - Earth Mass. The Moon orbits Earth in an average...Ch. 7 - Jupiter Mass. Jupiters moon Io orbits Jupiter...Ch. 7 - Pluto/Charon Mass. Plutos moon Charon orbits Pluto...Ch. 7 - Mission to Pluto. The New Horizons spacecraft took...Ch. 7 - Planetary Missions. Visit the web page for one of...
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- Calculate how many seconds there are in 1.0 Mars year (678 Earth days). Report your answer in scientific notation with the correct number of significant digits.arrow_forward2. On August 27, 2003, the planet Mars was at a distance of 0.373 AU from Earth. The diameter of Mars is 6788 km. a) Calculate the angular diameter of Mars, as seen from Earth on August 27, 2003. Give your answer in arcminutes.arrow_forwardA planet orbits the Sun every 539.4 years. What is its distance (semi-major axis of the orbit) from the Sun? (Give your answer in SI units and include the unit.)arrow_forward
- The moons Prometheus and Pandora orbit Saturn at 139,350 and 141,700 kilometers, respectively. a. Using Newton's version of Kepler's third law, find the orbital periods of the two moons. b. Find the percent difference in their.distances and in their orbital periods. c. Consider the two in a race around Saturn: In one Prometheus orbit, how far behind is Pandora (in units of time)? In how many Prometheus orbits will Pandora have fallen behind by one of its own orbital periods? Convert this number of periods back into units of time. This is how often the satellites pass by each other.arrow_forwardSuppose you were given a 3 in diameter ball to represent the Earth and a 1 in diameter ball to represent the Moon. (The actual ratio of Earth diameter to Moon diameter is 3.7 to 1.) The actual average Earth–Moon distance is about 384,000 kilometers, and Earth’s diameter is about 12,800 kilometers. How many “Earth diameters” is the distance from Earth to the Moon? Based on your answer to Question 2, what is the correct scaled distance of the Moon, using the 3-inch ball as Earth? The Sun’s actual diameter is about 1,400,000 kilometers. How many “Earth diameters” is this? Given your 3-inch Earth, how large (i.e what diameter) of a ball would you need to represent the Sun? Give your answer in feet. The average Earth–Sun distance is about 149,600,000 km. To represent this distance to scale, how far away would you have to place your 3-inch Earth from your Sun? Give your answer in feet. Could we use this scale to visualize the solar system instead of just the Earth and Moon? Why or Why…arrow_forwardIt is important to have an idea about the distances between and relative sizes of celestial objects in the solar system. In Part 1 we will pretend to shrink the solar system until its center piece, the Sun, is 67.3 cm in diameter. This will represent the Sun which is 1,390,000 km in diameter. The scale of our model is thus: 67.3 cm = 4.84 x 10-5 cm km Scale 1, 390, 000 km To find the size or distance between objects in centimeters for the model, simply multiply the actual size or distance in kilometers by the scale factor above. 1. Fill in following table: Quantity Actual Distance (km) Model Distance (cm) Diameter of Sun 1,390,000 Diameter of Earth 12,760 Diameter of Moon 3,480 Distance Between Earth and Sun 1.5 x 108 Distance Between Earth and Moon 384,000 Distance to Proxima Centauri 3.97 x 1013arrow_forward
- A planet's speed in orbit is given by V = (30 km/s)[(2/r)-(1/a)]0.5 where V is the planet's velocity, r is the distance in AU's from the Sun at that instant, and a is the semimajor axis of its orbit. Calculate the Earth's velocity in its orbit (assume it is circular): What is the velocity of Mars at a distance of 1.41 AU from the Sun? What is the spacecraft's velocity when it is 1 AU from the Sun (after launch from the Earth)? What additional velocity does the launch burn have to give to the spacecraft? (i.e. What is the difference between the Earth's velocity and the velocity the spacecraft needs to have?) How fast will the spacecraft be traveling when it reaches Mars? Does the spacecraft need to gain or lose velocity to go into the same orbit as Mars?arrow_forwardIf the moon is 238,000 miles from earth, convert this distance to meters and leave your answer in scientific notation. ( Recall 5280 ft = 1 mi , 1 m = 3.3 ft) b) In a school zone, the speed limit is 25 miles/hour. Convert this speed to meters/second (m/s). Round your answer to the nearest tenth.arrow_forwardAt present there are 8 planets in the solar system. In the early models, there were only 6 planets. What is the reason behind this? Describe a model of the modern solar system in terms of the number of planets, their arrangement and the model’s center.arrow_forward
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