COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Question
Chapter 25, Problem 38QAP
To determine
The speed of the satellite relative to the spaceship using the Galilean transformation.
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Check out a sample textbook solutionChapter 25 Solutions
COLLEGE PHYSICS
Ch. 25 - Prob. 1QAPCh. 25 - Prob. 2QAPCh. 25 - Prob. 3QAPCh. 25 - Prob. 4QAPCh. 25 - Prob. 5QAPCh. 25 - Prob. 6QAPCh. 25 - Prob. 7QAPCh. 25 - Prob. 8QAPCh. 25 - Prob. 9QAPCh. 25 - Prob. 10QAP
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- Unreasonable Results Suppose you wish to shoot supplies straight up to astronauts in an orbit 36,000 km above the surface of the Earth. (a) At what velocity must the supplies be launched? (b) What is unreasonable about this velocity? (c) Is there a problem with the relative velocity between the supplies and the astronauts when the supplies reach their maximum height? (d) Is the premise unreasonable or is the available equation inapplicable? Explain your answer.arrow_forwardSpace debris left from old satellites and their launchers is becoming a hazard to other satellites. (a) Calculate the speed of a satellite in an orbit 900 km above Earth’s surface. (b) Suppose a loose rivet is in an orbit of the same radius that intersects the satellite’s orbit at an angle of 90 . What is the velocity of the rivet relative to the satellite just before striking it? (c) If its mass is 0.500 g, and it comes to rest inside the satellite, how much energy in joules is generated by the collision? (Assume the satellite’s velocity does not change appreciably, because it mass is much greater than the rivets’s.)arrow_forwardOur solar system orbits the center of the Milky Way Galaxy. Assuming a circular orbit 30,000 ly in radius and an orbital speed of 250 km/s, how many years does it take for one revolution? Note that this is approximate, assuming constant speed and circular orbit, but it is representative of the time for our system and local stars to make one revolution around the galaxy.arrow_forward
- (a) Use the distance and velocity data in Figure 3.64 to find the rate of expansion as a function of distance. (b) If you extrapolate back in time, how long ago would all of the galaxies have been at approximately the same position? The two parts of this problem give you some idea of how the Hubble constant for universal expansion and the time back to the Big Bang are determined, respectively. Figure 3.64 Five galaxies on a straight line, showing their distances and velocities relative to the Milky Way (MW) Galaxy. The distances are in millions of light years (Mly), where a light year is the distance light travels in one year. The velocities are nearly proportional to the distances. The sizes of the galaxies are greatly exaggerated; an average galaxy is about 0.1 MlY across.arrow_forwardAstronomical observations of our Milky Way galaxy indicate that it has a mass of about 8.1011 solar masses. A star orbiting on the galaxy's periphery is about 6.0104 light years from its center. (a) What should the orbital period of that star be? (b) If its period is 6.0107 instead, what is the mass of the galaxy? Such calculations are used to imply the existence of "dark matter" in the universe and have indicated, for example, the existence of very massive black holes at the centers of some galaxies.arrow_forwardAstronomical observatrions of our Milky Way galaxy indicate that it has a mass of about 8.01011 solar masses. A star orbiting on the galaxy’s periphery is about 6.0104 light-years from its center. (a) What should the orbital period of that star be? (b) If its period is 6.0107 years instead, what is the mass of the galaxy? Such calculations are used to imply the existence of other matter, such as a very massive black hole at the center of the Milky Way.arrow_forward
- Find the speed needed to escape from the solar system starting from the surface of Earth. Assume there are no other bodies involved and do not account for the fact that Earth is moving in its orbit. [Hint: Equation 13.6 does not apply. Use Equation 13.5 and include the potential energy of both Earth and the Sun. Substituting the values for Earth’s mass and radius directly into Equation 13.6, we obtain vesc=2GMR=2(6.67 10 11Nm2/kg2)(5.96 10 24kg)(6.37 106m)=1.12104m/s That is about 11 km/s or 25,000 mph. To escape the Sun, starting from Earth’s orbit, we use R=RES=1.501011m and MSum=1.991030kg . The result is vesc=4.21104m/s or about 42 km/s. We have 12mvesc2GMmR=12m02GMm=0 Solving for the escape velocity,arrow_forwardA massive black hole is believed to exist at the center of our galaxy (and most other spiral galaxies). Since the 1990s, astronomers have been tracking the motions of several dozen stars in rapid motion around the center. Their motions give a clue to the size of this black hole. a. One of these stars is believed to be in an approximately circular orbit with a radius of about 1.50 103 AU and a period of approximately 30 yr. Use these numbers to determine the mass of the black hole around which this star is orbiting, b. What is the speed of this star, and how does it compare with the speed of the Earth in its orbit? How does it compare with the speed of light?arrow_forward
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