Question

Gravitational Fields: a) What is the escape velocity for a stationary object (of mass 100 kg) from a planet if an object dropped on its surface falls 5m in 1.4 seconds. (diameter of planet is 12,000 km) b) What is the objects velocity at a distance of 60, 000 km if the velocity is 1.2 times the escape velocity? Please show and explain as much work as possible. Thank you!

Answer 1

5m As it is given that an object in covering in 1-4 oce whow it is chopped. by and ean of motion Szut + fat2 ; Uz nitial velocity zo a= we have tocalailate 5=<xax(1.4)2 s- distance =5 lo t=1-48ce = g (1-4)2 gast on the given planet value of g=51MER Now eo cabe velocity to defried Oro the minimum velocity with colich the particle should be projected so that ct leaves the gravitatimal field of the planet. and reaches to D. By energy conservation we can say the hat we have to provide the particle at least mininun KE with which particle to bounded to the planet @ PE epotential energy Ve=eskape speed kom Potential benitivo object iogiron as - . GMm PE Re ki=lmue2 finally particle is stopped at a Kg By energy Conservation Kit Pi= Rjt Pf Live 2 gumzo Re arra to Ro 2 P=0 /

fror 2 gue² = Gmina ve fan g=GM ao we know the formula of gravitantial acceleration at surface R=60ookm Ve=N29R Drameter=12 cookm 245.14 Gorokm will be = R =7-82 km/o Ano Now the particle is given speed 1-2 times the escape opeed 80 know oto kmetic energy. am 12ve) initially posto Ki=1mX144x2GM = 1:49MM / 20 R = 6000kn Pi=-qum Ineque nitially Kt=£ mor -GMm Clors Gorrykn) IOR By energy Conpervation "Kit Ri= Pf thy > 1.44 GMm GMm Imozzamm ₂ OTAGMA +GMA TOR arve = 5.4 am = uz 2 ) ² sty 2 GM o=1054 I 2 = 5.75km Reboooking Pf = TOR - R TO

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