Answer:
a. t = 69.4 hr = 2.89 days
b. theta = 91.67*10^-3 degrees
c. deflection_angle = 0.134 degrees
Explanation:
a).
The asteroid impacts the earth in t
t = x/v = (4.0*10^6 km)/(16 km/sec)
t = 2.5 * 10^5 sec
t = 69.4 hr = 2.89 days
b).
tan(theta) = 6400 km/(4.0*10^6 km)
tan(theta) = 1.6*10^-3
theta = arctan(1.6*10^-3)
theta = 1.6*10^-3 radians (for small angles, tan(theta) ~= theta)
theta = 91.67*10^-3 degrees
c).
v_minimum = 6400 km/(2.5 * 10^5 sec)
v_minimum = 25.6 m/s
Using F = m*a, we can calculate the acceleration of the asteroid due to the rocket's thrust:
5.0*10^9 N = 4.0*10^10 kg * a
a = (5.0*10^9 N)/(4.0*10^10 kg)
a = 0.125 m/s^2
The transverse velocity after 300 seconds of this acceleration is:
v_transverse = a*t = 0.125 m/s^2 * 300 s
v_transverse = 37.5 m/s = 37.5*10^-3 km/s
tan(deflection_angle) = v_transverse/(20 km/s)
tan(deflection_angle) = (37.5*10^-3 km/s)/(16 km/s) = 2.34^-3
deflection_angle = arctan(2.34*10^-3)
deflection_angle = 2.34*10^-3 radians = 0.134 degrees
v_transverse/(16 km/s) > (6400km)/(5.0*10^6 km)
(note that the right hand side if this inequality is tan(theta) calculated above)
v_transverse > 23.704 m/
