NASA has taken an increased interest in near Earth asteroids. These objects, popularized in recent blockbuster movies, can pass very close to Earth on a cosmic scale, sometimes as close as 1 million miles. Most are small-less than \(500 \mathrm{~m}\) across \(-\) and while an impact with one of the smaller ones could be dangerous, experts believe that it may not be catastrophic to the human race. One possible defense system against near Earth asteroids involves hitting an incoming asteroid with a rocket to divert its course. Assume a relatively small asteroid with a mass of \(2.10 \cdot 10^{10} \mathrm{~kg}\) is traveling toward the Earth at a modest speed of \(12.0 \mathrm{~km} / \mathrm{s}\). a) How fast would a large rocket with a mass of \(8.00 \cdot 10^{4} \mathrm{~kg}\) have to be moving when it hit the asteroid head on in order to stop the asteroid? b) An alternative approach would be to divert the asteroid from its path by a small amount to cause it to miss Earth. How fast would the rocket of part (a) have to be traveling at impact to divert the asteroid's path by \(1.00^{\circ}\) ? In this case, assume that the rocket hits the asteroid while traveling along a line perpendicular to the asteroid's path.

To stop the asteroid, we will use the conservation of linear momentum principle. The momentum before and after the impact will be the same: \(p_{before} = p_{after}\) The momentum of the asteroid before impact is given by: \(p_{asteroid,initial} = m_{asteroid} \cdot v_{asteroid}\) The momentum of the rocket before impact is: \(p_{rocket,initial} = m_{rocket} \cdot v_{rocket}\) After the impact, we want the asteroid to stop; hence its final velocity will be zero. Therefore, the final momentum of the asteroid is zero. The final momentum of the rocket is given by: \(p_{final} = (m_{asteroid} + m_{rocket}) \cdot v_{final}\) Using the conservation of momentum, we get: \(m_{asteroid} \cdot v_{asteroid} + m_{rocket} \cdot v_{rocket} = (m_{asteroid} + m_{rocket}) \cdot v_{final}\) We know the masses of the asteroid and the rocket, as well as the velocity of the asteroid. We can solve for the velocity of the rocket \(v_{rocket}\): \(v_{rocket} = \frac{(m_{asteroid} + m_{rocket}) \cdot v_{final} - m_{asteroid} \cdot v_{asteroid}}{m_{rocket}}\)

For this part of the question, we need to analyze the situation using conservation of momentum in two dimensions. Since the rocket impacts the asteroid perpendicular to its path, we'll have: \(p_{y_{before}} = p_{y_{after}}\) The initial vertical momentum is only due to the rocket: \(p_{y_{before}} = m_{rocket} \cdot v_{y_{rocket}}\) The final vertical momentum is shared between the asteroid and the rocket: \(p_{y_{after}} = m_{asteroid} \cdot v_{y_{asteroid,final}} + m_{rocket} \cdot v_{y_{rocket,final}}\) By conserving vertical momentum, we have: \(m_{rocket} \cdot v_{y_{rocket}} = m_{asteroid} \cdot v_{y_{asteroid,final}} + m_{rocket} \cdot v_{y_{rocket,final}}\) The change in the asteroid's path is given as 1 degree: \(\Delta \theta = 1^{\circ}\) Let's convert this angle to radians: \(\Delta \theta_{rad} = \frac{\pi}{180}\) Using the relation: \(\tan (\Delta \theta) = \frac{v_{y_{asteroid,final}}}{v_{x_{asteroid,final}}}\) We can solve for the final vertical velocity of the asteroid: \(v_{y_{asteroid,final}} = v_{x_{asteroid,final}} \tan (\Delta \theta_{rad})\) Since the asteroid keeps moving horizontally, it means that the final horizontal velocity of the asteroid remains the same: \(v_{x_{asteroid,final}} = v_{asteroid} = 12.0 \space km/s\) Plugging back into our conservation of vertical momentum equation: \(m_{rocket} \cdot v_{y_{rocket}} = m_{asteroid} \cdot (v_{x_{asteroid,final}} \tan (\Delta \theta_{rad})) + m_{rocket} \cdot v_{y_{rocket,final}}\) And finally, we can solve for the requisite vertical velocity of the rocket: \(v_{y_{rocket}} = \frac{m_{asteroid} \cdot (v_{x_{asteroid,final}} \tan (\Delta \theta_{rad}))}{m_{rocket}}\)