Question

Problem 3.29 A star of mass 6 x 1030 kg is located at <6 x 1012, 4 x 1012, 0> m. A planet of mass 2 x 1024 kg is located at <5 x 1012, 9 x 1012, 0> m and is moving with a velocity of <0.5 x 104, 1.4 x 104, 0> m/s. (a) During a time interval of 1 x 10° seconds, what is the change in the planet's velocity? > m/s (b) During this time interval of 1 x 106 seconds, what is the change in the planet's position? (c) Why are the procedures you followed in parts (a) and (b) able to produce usable results but wouldn't work if you used a value of At of 1 x 109 seconds instead of 1 x 106 seconds? Check all that apply: The momentum change in the short time interval is small compared to the momentum, so the velocity in the position update equation is nearly constant. With the larger time interval it would be important to consider relativistic effects. The distance moved in the short time interval is small compared to the distance from the star to the planet, so the force is nearly constant in the momentum update equation.

Answer 1

Given data The distance between star and planet is, d-,(4x1012-6x1012 +(9x101-5x1012 4.47 x1012 m The force between star and planet is, Gmm2 6.67x10-11(6 x1030 (2x1024 4.47x1012 -4x1019 N The acceleration is, 4 x1019 2x1024 -2x10-5 m/s a) The change in velocity is, -(2x10) (1x109) - 20 m/s b) The change in position is, (20)(1x10 -2x10' m For t-1x10 s The change in velocity is, - at (2x10)1x109) -20x103 m/s The change in position is, (20x10' (10) -2x100 m