Review. In 1963, astronaut Gordon Cooper orbited the Earth 22 times. The press stated that for each orbit, he aged two-millionths of a second less than he would have had he remained on the Earth. (a) Assuming Cooper was 160 km above the Earth in a circular orbit, determine the difference in elapsed time between someone on the Earth and the orbiting astronaut for the 22 orbits. You may use the approximation
for small x. (b) Did the press report accurate information? Explain.
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Chapter 39 Solutions
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- Kepler's third law states that the relationship between the mean distance d (in astronomical units) of a planet from the Sun and the time t (in years) it takes the planet to orbit the Sun can be given by d^3 = t^2. (A). It takes Venus 0.616 years to orbit the Sun. Find the mean distance of Venus from the Sun (in astronomical units). (B). The mean distance of Jupiter from the Sun is 5.24 astronomical units. How many years does it take Jupiter to orbit the Sun?arrow_forwardA Ferris wheel has a radius of 10 m, and the bottom of the wheel passes 1 m above the ground. If the Ferris wheel makes one complete revolution every 18 s, find an equation that gives the height above the ground of a person on the Ferris wheel as a function of time. (Let y be the height above the ground in meters and let t be the time in seconds. Assume that when t = 0 the person is 11 m above the ground and going up.) y = 10 m -1 marrow_forwardAccording to Kepler's third law of planetary motion, the mean distance D, in millions of miles, from a planet in our solar system to the sun is related to the time P, in years, that it takes for the planet to complete a revolution around the sun, and the relationship is D = 93P2/3 It takes the planet Pluto 248 years to complete a revolution around the sun. What is the mean distance from Pluto to the sun? What is the mean distance from Earth to the sun? Give your answers to the nearest million miles. from Pluto to the sun X million miles from Earth to the sun million milesarrow_forward
- In 1999, NASA lost the Mars Climate Orbiter because one group of engineers used metric units in their calculations while another group used English units. Consequently, the orbiter descended too far into the Martian atmosphere and burned up. Suppose that the orbiter was to have established orbit at 157 km and that one group of engineers specified this distance as 1.57 x 10³ m. Suppose further that a second group of engineers programmed the orbiter to go to 1.57 x 10³ ft. The $125 million Mars Climate Orbiter was lost in the Martian atmosphere in 1999 because two groups of engineers failed to communicate with each other about What was the difference in kilometers between the two altitudes? Express your answer with the appropriate units. Value Submit HA Provide Feedback Part B Complete previous part(s) Units Request Answer ?arrow_forwardTo complete this exercise, you need to know that the circumference of a circle is proportional to its radius, and that the constant of proportionality is 2π. You do not need to know either the radius of the Moon’s orbit or the radius of Earth. For purposes of this exercise, we assume that the Moon’s orbit around Earth is circular. In one trip around Earth, the Moon travels approximately 2.4 million kilometers. Another satellite orbits Earth (in a circular orbit) at a distance from Earth that is 1/4 that of the Moon. How far does this satellite travel in one trip around Earth? (Use decimal notation. Give your answer to one decimal place.) A rope is tied around the equator of Earth. A second rope circles Earth and is suspended 77 feet above the equator. How much longer is the second rope than the first?arrow_forwardA light-vear is the distance that light travels in one year. The speed of light is 3.00 x 108 m/s. How many miles are there in one light-year? (1 mi = 1609 m, 1 y =365 d) 5.88 x 1015 mi O 9.46 x 1012 mi 5.88 x 1012 mi O 9,46 x 1015 miarrow_forward
- Kepler's Third Law of planetary motion states that the square of the period T of a planet (the time it takes for the planet to make a complete revolution about the sun) is directly proportional to the cube of its average distance d from the sun. (a) Express Kepler's Third Law as an equation. (Use k for the constant of proportionality.) (c) The planet Neptune is about 2.79 × 109 mi from the sun. Find the period of Neptune. (Round your answer to the nearest whole number of years.)arrow_forwardAccording to Newton's law of universal gravitation, the attraction force between two bodies is given by: where mi and m2 are the masses of the bodies, r is the distance between the bodies, and G=6.67 × 10-" N-m²/kg² is the universal gravitational constant. Determine how many times the attraction force between the sun and the Earth is larger than the attraction force between the Earth and the moon. The distance between the sun and Earth is 149.6 x 10°m, the distance 28 between the moon and Earth is 384.4 × 10°m, mEarth = 5.98 x 10“ kg, = 2.0 x 10" kg, and mn =7.36 x 102 kg.arrow_forwardA useful and easy-to-remember approximate value for the number of seconds in a year is p * 107. Determine the percent error in this approximate value. (There are 365.24 days in one year.)arrow_forward
- Schwarzschild radius RS of a black hole is the maximum distance from the black hole’s center at which light cannot escape its gravitational field. The quantity RS (with dimensions of length) is dependent on the mass of the black hole M, the speed of light c, and the gravitational constant G. Based on the dimensions of these four parameters, predict an equation for the Schwarzschild radius. Hint: G has dimensions of [L3/MT2]arrow_forwardPolice use the formula: v = V20L to estimate the speed of a car, v, in miles per hour, based on the length, L, in feet, of its skid marks when suddenly braking on a dry, asphalt road. At the scene of an accident, a police officer measures a car's skid marks to be 158 feet long. Approximately how fast was the car traveling? Round your answer to the nearest tenth (one decimal place) of a unit. Answer: The car was traveling at approximately miles per hour. Question Help: Video D Post to forum Submit Question fa ho 144 %23 & 4. 6. 7. R. T Y 60 24arrow_forwardWhen you are at a height x above the surface of the earth, your weight can be found by the following formula: W(x) = mg (1 + x/R) −2 , where m is the mass of your body, g is the acceleration due to gravity, and R is the radius of the earth. For example, W(0) is your weight when you are on the surface of the earth and W(R) is your weight when you are at the height R above the surface of the earth (yes, you are far away from the earth’s atmosphere). In this question, we will approximate W(0) and W(R). (a) Find L0(x), the linear approximation of W(x) at x = 0. (b) Use L0(x) to approximate W(0) and W(R). Which approximation is more accurate? Why? (c) Find LR(x), the linear approximation of W(x) at x = R. (d) Use LR(x) to approximate W(0) and W(R). Which approximation is more accurate? Why?arrow_forward
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