Question

(a) What is the escape speed on a spherical asteroid whose radius is 376 km and whose gravitational acceleration at the surface is 0.683 m/s²? (b) How far from the surface will a particle go if it leaves the asteroid's surface with a radial speed of 445 m/s? (c) With what speed will an object hit the asteroid if it is dropped from 592.7 km above the surface?

Answer 1

escape speed v = sqrt(2gR)

escape speed v = sqrt(2*0.683*376*10^3) = 716.67 m/s

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(b)

Vescape = sqrt(2G*M/R) = 716.67

sqrt(2*6.67*10^-11*M/376000) = 716.67

M = 1.45*10^21 kg

initial energy on the surface Ei = Ki + Ui = (1/)*m*vi^2 - G*M*m/R

final energy at maximum height h, Ef = Kf + Uf = (1/2)*m*vf^2 - G*M*m/(R+h)

at maximum height vf = 0

from energy conservation

Ef = Ei

- G*M*m/(R+h) = (1/2)*m*vi^2 - G*M*m/R

- G*M/(R+h) = (1/2)*vi^2 - G*M/R

-6.67*10^-11*1.45*10^21/(376000+h) = (1/2)*445^2 - 6.67*10^-11*1.45*10^21/376000

h = 235.3 km

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(c)

initial energy Ei = -G*M*m/(R+h)

after reaching the surface

final energy Ef = (1/2)*m*v^2 - G*M*m/R

Ef = Ei

(1/2)*m*v^2 - G*M*m/R = -G*M*m/(R+h)

(1/2)*v^2 - G*M/R = -G*M/(R+h)

(1/2)*v^2- 6.67*10^-11*1.45*10^21/376000 = -6.67*10^-11*1.45*10^21/(376000+592700)

v = 561 m/s <<<------ANSWER