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Unformatted text preview: Question (1170006)
Position (update) of Earth
At some instant in time, Earth is located at the position this instant is 1.5 × 1011 , 0, 0 m relative to Sun. Its velocity at 3.0 × 10 4 m/s in the +y direction. (a) Where will it be 1 day later? (b) What assumption did you make in doing this calculation, and is your calculation exact or approximate? Solution
(a)First,
sketch a picture of the situation. Figure 1: The position of Earth and Sun (not drawn to scale). Assume that the velocity of Earth is constant for the given time interval. The smaller the time interval, the better the approximation. vavg vavg = = ∆r ∆t rf − ri ∆t ∆ t. Solve for the nal position of Earth at the end of the time interval ∆r rf − ri rf = vavg ∆t = vavg ∆t = ri + vavg ∆t Substitute the given quantities. Be sure to convert a day to seconds. rf = = = ( 1.5 × 1011 , 0, 0 m) + ( 0, 3.0 × 104 , 0 m/s)(1 ( 1.5 × 10 , 0, 0 m) + ( 0, 2.6 × 10 , 0 m) 1.5 × 1011 , 2.6 × 109 , 0 m
11 9 day)(24 h/day)(3600 s/h) (b) According to our calculation, Earth traveled straight upward" in the +y direction, as shown in Figure 1. However, we know from experience that Earth travels approximately in a circle around Sun, thus it should have moved slightly to the left. Our calculation assumed a constant velocity, which occurs if the net force on Earth is zero. Yet, in reality there is a gravitational force on Earth by Sun, and Earth's velocity is not constant. Therefore, our calculation is an approximation. ...
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This note was uploaded on 02/04/2011 for the course PHYS 303K taught by Professor Niu during the Spring '08 term at University of Texas at Austin.
 Spring '08
 NIU
 Physics

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