Question

A $4.0*1010kg$ asteroid is heading directly toward the center of the earth at a steady $20km/s$. 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 $4.0*106km$ away from earth. You can ignore the earth’s gravitational force on the asteroid and their rotation about the sun.

a. If the mission fails, how many hours is it until the asteroid impacts the earth?

b. The radius of the earth is $6400km$. By what minimum angle must the asteroid be deflected to just miss the earth?

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

Short answer

a). If the mission fails, it takes $56h$by the asteroid to hit the Earth.

b). Minimum angle must the asteroid be deflected to just miss the earth is $0.{092}^o$.

c). This angle is greater than the angle of minimum deviation, Earth is saved.

Step-by-step solution

Given Information (Part a)

The asteroid travels with constant speed of $20\mathrm{km}/\mathrm{s}$.

Its distance from the Earth is $4.0\times {10}^6\mathrm{km}$.

Explanation (Part a)

If the mission fails, the time taken by the asteroid to hit the Earth is given by

$t=\frac{\text{Distance}}{\text{Speed}}=\frac{4.0\times {10}^6\mathrm{km}}{20\frac{\mathrm{km}}{\mathrm{s}}}=2.0\times {10}^5\mathrm{s}$

$t=2.0\times {10}^5\S \times \frac{1h}{3600\varphi }=55.56h\approx 56h$

Final Answer (Part a)

If the mission fails, it takes $56h$by the asteroid to hit the Earth.

Given Information (Part b)

The distance of the asteroid from Earth is $4.0\times {10}^6\mathrm{km}$.

The radius of the Earth is $6400\mathrm{km}$.

Explanation (Part b)

The minimum angle of deflection to miss the Earth is given by

$\theta ={\tan }^{-1}\left(\frac{6400\mathrm{km}}{4.0\times {10}^6\mathrm{km}}\right)=0.092°$

Final Answer (Part b)

The minimum angle must the asteroid be deflected to just miss the earth is $0.{092}^o$.

Given Information (Part c)

Initially there is no horizontal motion. After the rocket is fired, there is an acceleration along the horizontal direction

The initial horizontal velocity is zero.

Explanation (Part c)

The horizontal velocity of the asteroid after $300\mathrm{s}$ is given by

$v_x=\frac{F_x}{m}\times \Delta t=\frac{5\times {10}^9}{4\times {10}^{10}}\times 300=37.5\mathrm{m}.{\mathrm{s}}^{-1}$

The vertical velocity of the asteroid is constant and given by $v_y=20000\mathrm{m}.{\mathrm{s}}^{-1}$.

The angle of deflection due to horizontal velocity is given by

$\theta ={\tan }^{-1}\left(\frac{v_x}{v_y}\right)={\tan }^{-1}\left(\frac{37.5}{20000}\right)=0.107°$

Since this angle is greater than the angle of minimum deviation, Earth is saved.

Final Answer (Part c)

This angle is greater than the angle of minimum deviation, Earth is saved.