Question

What is the escape speed on a spherical asteroid whose radius is 700 km and whose gravitational acceleration at the surface is 0.685 m/s2? (b) How far from the surface will a particle go if it leaves the asteroid's surface with a radial speed of 694 m/s? (c) With what speed will an object hit the asteroid if it is dropped from 1066 km above the surface? (SHOW ALL WORK)

Answer 1

a) escape speed is Ve = sqrt(2*g*R) = sqrt(2*0.685*700*1000) = 979.3 m/s

b)using law of conservation of energy

energy at the surface = energy at the maximum height

-G*m*M/r + (0.5*m*v^2) = (-G*m*M/r1)

G is the universal gravitational constant

m is the mass of the particle

M is the mass of the asteroid

v = 694 m/s

m cancels on both sides

but g = G*M/R^2

0.685 = (6.67*10^-11*M)/(700*1000)^2

M = 5*10^21 Kg

then

[(-6.67*10^-11*m*5*10^21)/(700*1000)]+(0.5*m*694^2) = (-6.67*10^-11*m*5*10^21)/r1

m cancels on both sides

then

-476428.57 + 240818 = -333500000000/r1

r1 = 1.415*10^6 m = 1415 km

so required height from the surface is r2 = 1415-700 = 715 km

c) again using law of conservation of energy

energy at the surface = energy at the maximum height

-G*m*M/r + (0.5*m*v^2) = (-G*m*M/r1)

[(-6.67*10^-11*m*5*10^21)/(700*1000)]+(0.5*m*v^2) = (-6.67*10^-11*m*5*10^21)/[(1066+700)*1000]

m cancels on both sides

-476428.57+(0.5*v^2) = -188844.85

v = 758.4 m/s