Question

A 4.0×1010kg asteroid is heading directly toward the center of the earth at a steady 32 km/s. To save the planet, astronauts strap a giant rocket to the asteroid perpendicular to its direction of travel. The rocket generates 5.0×109N of thrust. The rocket is fired when the asteroid is 4.0×106kmaway from earth. You can ignore the earth’s gravitational force on the asteroid and their rotation about the sun.

What is the actual angle of deflection if the rocket fires at full thrust for 300 s before running out of fuel?

Answer 1

The asteroid is traveling at 32 km/sec. To calculate how long it would take to reach the Earth if the mission fails, simply divide the distance it must travel by the speed (we're neglecting the radius of the Earth in this calculation because it is much smaller than the total distance):

t = D/v = (4*10^6 km)/(32 km/sec)

t = 1.25 * 10^5 sec

There are a couple of ways of calculating this. Perhaps the easiest is to note that the asteroid must acquire a velocity in the direction perpendicular to its original velocity vector that is large enough to cover 6400 km (the Earth's radius) in the 1.25*10^5 seconds before impact. (I will neglect the change in the position of the asteroid during the 300 seconds that the rocket is firing.) So the minimum transverse velocity needed is:

v_minimum = 6400 km/(1.25 * 10^5 sec)

v_minimum = 51.2 m/s

Using F = m*a, we can calculate the acceleration of the asteroid due to the rocket's thrust:

5*10^9 N = 4*10^10 kg * a

a = (5*10^9 N)/(4*10^10 kg)

a = 1.25 * 10^-1 m/s^2

The transverse velocity after 300 seconds of this acceleration is:

v_transverse = a*t = 1.25 * 10^-1 m/s^2 * 300 s = 37.5 m/s = 3.75 *10^-2 km/s

This exceeds the minimum velocity to miss the Earth by a comfortable margin.

The angle through which the asteroid is deflected is given by:

tan(deflection_angle) = v_transverse/(32 km/s)

tan(deflection_angle) = (3.75 *10^-2 km/s)/(32 km/s) = 1.172 *10^-3

deflection_angle = arctan(1.172 *10^-3)

deflection_angle = 0.067 degrees