Question

A dwarf planet with mass M = 2.43 x 1016 kg and radius R = 2360 km is crashed into by a large comet. The dwarf planet initially has a day with length 12 hours. The comet hits the planet at its equator and at an angle of 0 = 80° with the radius of the planet, as shown in the view of the planet looking above from the North Pole. The speed of the comet is V comet = 9000 m/s, and its mass is m = 1.99 x 10+1 kg. The comet fuses with the planet after the collision. Take the moment of inertia of the planet to be 0.33MR2 (a) What is the length of day on the dwarf planet after the collision? (b) If the collision take place in 40 ms, then how much torque is delivered to the planet during the collision? What is the angular acceleration of the planet during these 40 ms? comet (c) If the comet hit at 8 = 90°, how fast would it have to be traveling in order for the length of the day of the dwarf planet to become one half its former length?

Answer 1

Peg O Given hat 243 M= kg R 236 0Km Comel 900om/s of inestia =033MR. momeht P Wahgula Veloc:ty of PIont - 12X60X60 =1454メ 8od/sec morehtum of the system obout an Ih.tiol angula8 that axi's th8ough the Nes Pole Passes アメマ)(of comet) Iw Cof Planct) + m 45 + m(9000 XRメ Sin B0°) MR 2 the value s ode NOw Substibute

Peg = 2.43 10 x (2360x 163)j.45 X10 + .99 b"x q00D X2360 X1 Sng80 24 Length K9 m /sec Finol momentum aflev impac angulas = Iw'+ mwlR2 he angula veloily of Planel whede impoct after Citenal fo Sina thede is ho ahgula momehlum MR2 w MiwR 2 3x 243 X1y (23602 451 Y1.45 W 1451 Yio dadI sei 516 102 2TT 1-45 211 12.03 houS 2x6DX6D

Pes (b) Change monetum of Planct TOddue deliVeded to Planet Lime -40xi0Se MR2 che Substi utng values of M, R W 3.92313NM Tosdue (T) = IXd MR2X 3 dk MR 2 3.923 XID X3 2.4310 360102 whese the aceladaoh

Peg Th:tal lehgh of doy - 12.ho s w454 d/se lehglh of day hs w Final 2908x1adscc Cohvextion Df ahgulo mmomentum usng as done in whede w = 2W R X2.431 ¥ }.454 XI, oh solvihg 9ct We