College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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- 2. A comet is traveling in outer space with a velocity of 2000 m/s. How long will it take to travel 300,000 meters? 2000 s 0.0006 s 6,000,000 s 150 sarrow_forward7. Use Kepler's third law to evaluate the distance of Saturn from the Sun, using the following facts: (1) The period of Saturn is 29.5 years. (2) The Earth's distance to the Sun is 1 AU. (3) The period of the Earth is one year. (The Astronomical Unit (AU) is a measure of distance used in astron- omy.)arrow_forwardKepler'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_forward
- According to Lunar Laser Ranging experiments the average distance LM from the Earth to the Moon is approximately 3.85 × 10° km. The Moon orbits the Earth and completes one revolution in approximately 27.5 days (a sidereal month). How can the mass of the Earth be calculated using the information above? Select the correct statements. Select one or more: O a. Use Newton's third law. O b. Use Newton's first law. c. Use Coulombs law. O d. Use Newton's second law.arrow_forward1. Moon is at the distance 384400 km from Earth and orbits the Earth every ~28 days. If the radius of the Moon is 1737 km (consider it to be spherical), what is the area of the moon as measured by the observer on Earth? (Hint: Length contraction)arrow_forwardLeave the moon's amplitude (semi-major axis) constant. How does the changing of the moon's orbital period change the calculated value of the planet's mass? a. Increasing the orbital period results in a higher calculated value for the mass of the planet. Decreasing the orbital period results in a lower calculated value for the mass of the planet. b. Increasing the orbital period results in a lower calculated value for the mass of the planet. Decreasing the orbital period results in a higher calculated value for the mass of the planet. c. Increasing the orbital period results in a higher calculates value for the mass of the planet. Decreasing the orbital period results in a higher calculated value for the mass of the planet.arrow_forward
- 6. True or false, if Earth were twice as massive but it revolved at the same distance from the sun, it's orbital period would be 2.88 yearsarrow_forward3. What is the official definition of a planet and why is Pluto not a planet?arrow_forward5. A spherical planet has a mass of 5 X 1024 kg. Its diameter is 10,000 km. First calculate its vol- ume in meters (1 km = 1000 m). Then calculate its density (Mass/Volume) in kg/m³.arrow_forward
- A satellite encircles Mars at a distance above its surface equal to 3 times the radius of Mars. If gm the acceleration due to gravity at the surface of Mars, what is the acceleration due to gravity at the location of the satellite? A. gm/16 B. gm/9 C. gm/4 D. gm/3 O B O Aarrow_forward19. Saturn’s moon Mimas has an orbital period of T= 82,800 s at a distance of d= 1.87 × 10^8 m from Saturn. Using Kepler's 3rd law listed below for mass to determine Saturn’s mass. You must show your calculations for credit.arrow_forwardKepler's third law of planetary motion describes a relationship between the__ * a, shape of orbit and the location of the Sun. b. orbital velocity and position in orbit c. distance from the Sun and length of year. d. path of epicycle and position.arrow_forward
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