91Ó°ÊÓ

Speed of the the meteoroid heading straight for earth =$2km/s.$

According to the law of conservation of energy,

The total energy of the meteoroid crossing the moon's orbit = The total energy of the meteoroid at the earth's surface.

$\frac{1}{2}m{v}_1^2-\frac{GMm}{r}=\frac{1}{2}m{v}_2^2-\frac{GMm}{R}.$

Where, the mass of the meteoroid = 'm'.

The initial velocity of meteoroid =$v_1.$

Mass of earth = 'M'.

The initial distance between the meteoroid and the earth (Distance of moon's orbit from Earth) = 'r'.

The radius of the earth = 'R'.

The final velocity of the meteoroid before impact $v_2.$

Simplifying the above equation and solve for $v_2$we get,

$$\begin{gathered} \frac{1}{2}m{v}_1^2-\frac{GMm}{r}=\frac{1}{2}m{v}_2^2-\frac{GMm}{R} \\ {v}_1^2-\frac{2GM}{r}=v_2^2-\frac{2GM}{R} \\ {v}_2^2=v_1^2+2GM\left(\frac{1}{R}-\frac{1}{r}\right) \\ {v}_2=\sqrt{v_1^2+2GM\left(\frac{1}{R}-\frac{1}{r}\right)} \end{gathered}$$

Where,

$$\begin{gathered} v_1=2\mathrm{km}/\mathrm{s} \\ \mathrm{G}=6.67×{10}^{-11}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2 \\ \mathrm{R}=6.637×{10}^6\mathrm{m} \\ \mathrm{M}=5.98×{10}^{24}\mathrm{kg} \\ \mathrm{r}=384×{10}^6\mathrm{m} \end{gathered}$$

Substituting all the values in the above equation we get,

$$\begin{gathered} v_2=\sqrt{v_1^2+2GM\left(\frac{1}{R}-\frac{1}{r}\right)} \\ {v}_2=\sqrt{{(2000m/s)}^2+2(6.67×{10}^{-11}N{m}^2/k{g}^2)(5.98×{10}^{24}kg)\left(\frac{1}{6.637×{10}^{6}m}-\frac{1}{384×{10}^{6}m}\right)} \\ {v}_2=11.3\mathrm{km}/\mathrm{s} \end{gathered}$$

The speed of impact of the first meteoroid is$11.3km/s.$

Speed of the the meteoroid that misses the earth by$5000km=2km/s.$

Using the law of conservation of energy we get,

$$\begin{gathered} v_1^2-\frac{2GM}{r}=v_2^2-\frac{2GM}{R} \\ {v}_2^2=v_1^2+2GM\left(\frac{1}{R}-\frac{1}{r}\right) \end{gathered}$$

Adding $5000\mathrm{km}$to 'R' value in the above equation. And simplify the equation we get,

localid="1648487563132" $$\begin{gathered} v_2=\sqrt{v_1^2+2GM\left(\frac{1}{R+6×{10}^6}-\frac{1}{r}\right)} \\ {v}_2=\sqrt{{(2000m/s)}^2+2(6.67×{10}^{-11}N{m}^2/k{g}^2)(5.98×{10}^{24}kg)\left(\frac{1}{11.637×{10}^{6}m}-\frac{1}{384×{10}^{6}m}\right)} \\ {v}_2=2.004\mathrm{km}/\mathrm{s} \end{gathered}$$

The speed of the second meteoroid at the closest point is$2km/s.$