In the early morning hours of June 14, 2002, the Earth had a remarkably close encounter with an...

Question:

In the early morning hours of June 14, 2002, the Earth had a remarkably close encounter with an asteroid the size of a small city. The previously unknown asteroid, now designated 2002 MN, remained undetected until three days after it had passed the Earth. At its closest approach, the asteroid was 73,600 miles from the center of the Earth-about a third of the distance to the Moon.

Part A

Find the speed of the asteroid at closest approach, assuming its speed at infinite distance to be zero and considering only its interaction with the Earth.

Part B

Observations indicate the asteroid to have a diameter of about 2.0 km. Estimate the kinetic energy of the asteroid at closest approach, assuming it has an average density of 3.35 g/cm{eq}^{3} {/eq}. (For comparison, a 1-megaton nuclear weapon releases about 5.6 x 10{eq}^{15} {/eq} J of energy.) Express your answer in joules using two significant figures.

Conservation of Energy:

The law of conservation of energy says the total energy of a closed system is conserved; no energy is created nor destroyed. The energy is transferred from one form to another, but the total will remain constant.

Part A

The mechanical energy of the asteroid and the Earth system is conserved from the law of conservation of energy.

Let i and f subscripts denote the initial (when the asteroid is very far away) and the final (when it is its closest approach) stages, respectively.

Let KE and PE be the kinetic energy and the potential energy of the system. Then;

{eq}KE_i = 0 {/eq} since the asteroid is at rest at infinity, and {eq}PE_i = 0 {/eq} since it is very far way (at infinity).

Thus,

{eq}\begin{align} PE_i + KE_i &= PE_f + KE_f\\ 0 + 0 &= -\frac{GM_eM_a}{R_f} + \frac{1}{2}M_av_f^2\\ \therefore v_f &= \sqrt{\frac{2GM_e}{R_f}} \end{align} {/eq}

Here,

• {eq}G = 6.67\times 10^{-11} \ \rm N\cdot m^2/kg^2 {/eq} is the gravitational constant,
• {eq}M_e = 5.972 \times 10^{24} \ \textrm{kg} {/eq} is the mass of Earth,
• {eq}M_a {/eq} is the mass of the asteroid,
• {eq}R_f = 73600 \ \textrm{mi} = 1.185 \times 10^8 \ \textrm m {/eq} is the distance of the asteroid at the final stage (closest to Earth),
• {eq}v_f {/eq} is the velocity of the asteroid at this distance.

Thus,

{eq}\begin{align} v_f &= \sqrt{\frac{2GM_e}{R_f}}\\ &= \sqrt{\frac{2\times 6.67\times 10^{-11} \ \rm N\cdot m^2/kg^2 \times 5.972 \times 10^{24} \ \textrm{kg}}{1.185 \times 10^8 \ \textrm m}}\\ &=\boxed{ 2.593 \ \textrm10^3 \ \textrm{m/s}} \end{align} {/eq}

Part B

The kinetic energy of the asteroid is;

{eq}\begin{align} KE_f &= \frac{1}{2}M_av_f^2\\ &= \frac{1}{2}\rho_aV_av_f^2 \end{align} {/eq}

Here,

• {eq}\rho_a = 3.35 \ \textrm{g/cm}^3= 3.35 \times 10^{3} \ \textrm{kg/m}^3 {/eq} is the density of the asteroid,
• {eq}V_a = \frac{4\pi}{3}R_a^3 = \frac{4\pi}{3}\times (1000 \ \textrm m)^3 = 4.19 \times 10^9 \ \textrm m^3 {/eq} is the volume of the asteroid,
• {eq}R_a = \frac{2.0 \ \textrm{km}}{2} = 1000 \ \textrm m {/eq} is the radius of the asteroid.

Thus,

{eq}\begin{align} KE_f &= \frac{1}{2}\rho_aV_av_f^2\\ &= \frac{1}{2} \times 3.35 \times 10^{3} \ \textrm{kg/m}^3 \times 4.19 \times 10^9 \ \textrm m^3 \times (2.593 \ \textrm10^3 \ \textrm{m/s})^2\\ &= \boxed{4.7 \times 10^{19} \ \textrm J} \end{align} {/eq}

This is ~8400 times more than 1-megaton nuclear weapon releases. 