(a) What is the escape speed on a spherical asteroid whose radius is 510 km and
ID: 1904444 • Letter: #
Question
(a) What is the escape speed on a spherical asteroid whose radius is 510 km and whose gravitational acceleration at the surface is 3.3 m/s2? (b) How far from the surface will a particle go if it leaves the asteroid's surface with a radial speed of 1000 m/s? (c) With what speed will an object hit the asteroid if it is dropped from 1000 km above the surface?Explanation / Answer
a) escape speed is speed to have v=0 and r-> infinity 1/2 mv^2 - G M m/R = 0 v = sqrt( 2 G M/R) now we use that g = G M/R^2 so GM/R = gR and GM = gR^2 v = sqrt( 2 g R ) = sqrt(2*3.3*510E3)=1835 m/s b) 1/2 m v^2 - G M m/R = - G M m/r 1/2*1000^2 - gR = - g R^2/r 1/2* 1000^2 - 3.3*510E3 = - 3.3 *(510E3)^2/r solve for r r=725554 m but this is front center of planet h= 725 km - 510 km = 215 km c)similar to before 1/2*v^2 - gR = - g R^2/r 1/2 v^2 - 3.3*510E3 = -3.3*(510E3)^2/(1000E3) solve for v v= 1284 m/s
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