Universe
11th Edition
ISBN: 9781319039448
Author: Robert Geller, Roger Freedman, William J. Kaufmann
Publisher: W. H. Freeman
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Chapter 12, Problem 22Q
To determine
The data from the Galileo probe which was in agreement with the predictions made by the astronomers, and the one which was surprising.
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On February 7, 1999, NASA launched a spacecraft with the ambitious mission of making a close encounter with a comet, collecting samples from its tail, and returning the samples to Earth for analysis. This spacecraft, appropriately named Stardust, took almost five years to rendezvous with its objective-comet Wild 2 (pronounced "Vilt 2")-and another two years to return its samples. The reason for the long round trip is that the spacecraft had to make three orbits around the Sun, and also an Earth Gravity Assist (EGA) flyby, to increase its speed enough to put it in an orbit appropriate for the encounter.When Stardust finally reached comet Wild 2 on January 2, 2004, it flew within 147 miles of the comet's nucleus, snapping pictures and collecting tiny specks of dust in the glistening coma. The approach speed between the spacecraft and the comet at the encounter was a relatively "slow" 6200 m/s, so that dust particles could be collected safely without destroying the vehicle. Note that…
Galileo's telescopes were not of high quality by modern standards. He was able to see the moons of Jupiter, but he never reported seeing features on Mars. Use the small-angle formula to find the angular diameter of Mars when it is closest to Earth. How does that compare with the maximum angular diameter of Jupiter? (Assume circular orbits with radii equal to the average distance from the Sun. Using the following distances from the Sun: Mars is 228 million km, Jupiter is 778 million km, and Earth is 150 million km. The radius of Mars is 3396 km. The radius of Jupiter is 71,492 km.)
angular diameter of Mars = ( )seconds of arc
angular diameter of Jupiter =( )seconds of arc
ratio of angular diameters (Jupiter/Mars) = ( )
Chapter 12 Solutions
Universe
Ch. 12 - Prob. 1CCCh. 12 - Prob. 2CCCh. 12 - Prob. 3CCCh. 12 - Prob. 4CCCh. 12 - Prob. 5CCCh. 12 - Prob. 6CCCh. 12 - Prob. 7CCCh. 12 - Prob. 8CCCh. 12 - Prob. 9CCCh. 12 - Prob. 10CC
Ch. 12 - Prob. 11CCCh. 12 - Prob. 1QCh. 12 - Prob. 2QCh. 12 - Prob. 3QCh. 12 - Prob. 4QCh. 12 - Prob. 5QCh. 12 - Prob. 6QCh. 12 - Prob. 7QCh. 12 - Prob. 8QCh. 12 - Prob. 9QCh. 12 - Prob. 10QCh. 12 - Prob. 11QCh. 12 - Prob. 12QCh. 12 - Prob. 13QCh. 12 - Prob. 14QCh. 12 - Prob. 15QCh. 12 - Prob. 16QCh. 12 - Prob. 17QCh. 12 - Prob. 18QCh. 12 - Prob. 19QCh. 12 - Prob. 20QCh. 12 - Prob. 21QCh. 12 - Prob. 22QCh. 12 - Prob. 23QCh. 12 - Prob. 24QCh. 12 - Prob. 25QCh. 12 - Prob. 26QCh. 12 - Prob. 27QCh. 12 - Prob. 28QCh. 12 - Prob. 29QCh. 12 - Prob. 30QCh. 12 - Prob. 31QCh. 12 - Prob. 33QCh. 12 - Prob. 34QCh. 12 - Prob. 35QCh. 12 - Prob. 36QCh. 12 - Prob. 37QCh. 12 - Prob. 38QCh. 12 - Prob. 39QCh. 12 - Prob. 40QCh. 12 - Prob. 41QCh. 12 - Prob. 42QCh. 12 - Prob. 43QCh. 12 - Prob. 44QCh. 12 - Prob. 45QCh. 12 - Prob. 46QCh. 12 - Prob. 47QCh. 12 - Prob. 48QCh. 12 - Prob. 49QCh. 12 - Prob. 50QCh. 12 - Prob. 51QCh. 12 - Prob. 52QCh. 12 - Prob. 53QCh. 12 - Prob. 54QCh. 12 - Prob. 55QCh. 12 - Prob. 56QCh. 12 - Prob. 57QCh. 12 - Prob. 58QCh. 12 - Prob. 59Q
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- A spacecraft leaving Earth is deployed at burnout from its launch vehicle at 400 km altitude with flight path angle -10.0 degrees and speed 10.8 km/sec. ; Using Kepler's equation, determine how long it will take the spacecraft to cross lunar orbit (assume the Moon is on a circular orbit about Earth with 384,000 km radius, and ignore the lunar gravitational effects). Confirm your answer to part (a) using the appropriate Lambert TOF a. b. equation. If burnout speed is increased to 13 km/sec (still with flight path angle с. at -10.0 degrees and altitude 400 km), how much time can be saved for the trip to the Moon?arrow_forwardCalculate how long radio communications from the spacecraft will take when it encounters Mars. The furthest distance from Earth to Mars is 2.66 AU. Remember that 1 AU = 1.5 x 1011 m and that light travels at 3 x 108 m/s. So how long will the radio messages take to travel this greatest distance of 2.66 AU? If two way communication between the Earth and the spacecraft involve a 1 s time lapse before an acknowledging signal is sent by the spacecraft, how long a time is there between sending a command to the spacecraft and receiving a reply?arrow_forwardThe Chelyabinsk meteor entered Earth’s atmosphere at an 18° angle. What would have happened if it had entered at 90°?arrow_forward
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