Question

(a) What is the escape speed on a spherical asteroid whose radius is 550 km and whose gravitational acceleration at the surface is 3.3m/s²

m/s

(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?
  m

(c) With what speed will an object hit the asteroid if it is dropped from 1000 km above the surface?
  m/s

Answer 1

escape speed ve = √2 R* g
R = 550*10^3 m, g (R) = 3.3 m/s^2
ve = 1905.26 m/s
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b) Total energy at surface = Total energy at H(max)
KE(R) + PE(R) = KE(H-max) + PE(H-max)
when particle at H-max, v =0 so KE (H-max) =0
----------------------------
0.5 mu^2 - GMm/R = 0 - GMm/(R+H-max)
GM/(R+H-max) = GM/R - 0.5 u^2
-----------------
g(R) = GM/R^2 or >>>>>>> GM = g(R) * R^2
-----------
g(R) * R^2/(R+H-max) = g(R) * R - 0.5 u^2
(R+H-max) = g R^2/ [g * R - 0.5 u^2]
H-max) = - R + 1/ [(1/R) - 0.5 (u^2/gR^2)] ---- (1)
calculation in KM>>>
g = 3.3 m/s^2 = 3.3*10^-3 km/s^2
u = 1 km/s, R = 550 km >>>
0.5 u^2/gR^2 =
H-max = - 550 + 1/ [0.001818 - 0.0005]
H-max = - 550 + 759.126 = 209 km
particle will reach max height of 209 km above surface
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c) at t=o, H = 1000 km, KE(H) = 0 from rest dropped
0 - GMm/(R+H) = 0.5 mv-hit^2 - GMm/R
0.5 v-hit^2 = GM/R - GM/(R+H)
0.5 v-hit^2 = gR^2/R - gR^2/(R+H)
v-hit^2 = 2gR - 2gR^2/(R+H)
v-hit^2 = 2gR [ 1 - R/(R+H)] = 2gRH/(R+H)
v-hit^2 = 2*3.3*550*10^3*1000*10^3/(1550*10^3)
v-hit^2 = 2341935.484 (all used in m/s)
v-hit = 1530.34 m/s