Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

#34? weigh on the te same mass as our sun and a lbit and return to you so yo act

ID: 1775578 • Letter: #

Question

#34?

weigh on the te same mass as our sun and a lbit and return to you so yo actually throw it at that spee or 20.0 km? dTh asteroid 243 Ida has a mass ofabout 40) ×1016 kg and an aver 12 age (a) Ca astron drop us of about 16 km (it's not spherical, but you can assume it is) ulate the acceleration of gravity on 243 Ida. (b) What would an t whose earth weight is 650 N weigh on 243 Ida? (c) If you a rock from a height of 1.0 m on 243 Ida, how long would r the rock to reach the ground? (d) If you can jump 60 cm General Problems 41. I International Space S launched in November 19 each day in a circular or it take the asterld 's gravity doesn't weaken significantly over the distance of your jum. lite. (by Apply Newton height (in kilometers) a on earth, how high could you jump on 243 Ida? (Assume 42. 1 Artificial gravity. station is to spin it. is to spin about its c rpm) must it turn s 6.5 Satellie Motion 33. equal to g? I An of 6200 mS. (a) Find the time of one revolution of the satellite. (b) Find the radial acceleration of the satellite in its orbit. (c) Make a free-body diagram of the satellite. earth satellite moves in a circular orbit with an orbital speed 43. Il Shortest possib the equator is und of the earth. (a) I is placed at the become shorter maximum spe still remain or 34. I What is the period of revolution of a satellite with mass im that orbits the eath in a circular path of radius 7880 km (about 1500 km 44. 11 Volcanoes it is the mos above the surface of the earth)? what must be the orbital speed of a satellite that is in a circular 6. Il Planets beyond the solar system. On October 15, 2001, a planet llian kilometers from the center of material asl orbit 200 km above the surface of the moon? (Consult Appendix E.) in gravity discovered orbiting around the star HD68988. Its orbital dis- mass of

Explanation / Answer

34) period T = sqrt [4 pi2 R3 / G M]

= sqrt [4 pi2 * (7880 * 103)3 / (6.67 * 10-11 * 5.98 * 1024)]

period T = 6959 s