91Ó°ÊÓ

The given data is listed below:

• The diameter of the asteroid is $\mathrm{d}=300\mathrm{km}$.
• The density is$\mathrm{p}=2500\mathrm{kg}/{\mathrm{m}}^3$

In order to find the escape velocity, firstly, calculate the volume of the asteroid. The volume will provide the asteroid's mass, which will help to evaluate the escape velocity.

The equation of volume of asteroid can be written as,

$$\begin{gathered} \mathrm{V}=\frac{4}{3}{\mathrm{Ï€}}^3 \\ \mathrm{V}=\frac{4}{3}\mathrm{Ï€}{\left(\frac{\mathrm{d}}{2}\right)}^3 \end{gathered}$$

Here, ris the radius of the asteroid.

Substitute the values in the above equation.

$$\begin{gathered} \mathrm{V}=\frac{4}{3}\mathrm{π}{\left(\frac{300\mathrm{km}×{\displaystyle \frac{1000\mathrm{m}}{1\mathrm{km}}}}{2}\right)}^3 \\ \mathrm{V}=1.41×{10}^{16}{\mathrm{m}}^3 \end{gathered}$$

The equation of mass of asteroid can be written as,

$\mathrm{M}=\mathrm{pV}$

Substitute the values in the above equation.

$$\begin{gathered} \mathrm{M}=\left(2500\mathrm{kg}/{\mathrm{m}}^3\right)\left(1.41×{10}^{16}{\mathrm{m}}^3\right) \\ \mathrm{M}=3.525×{10}^{19}\mathrm{kg} \end{gathered}$$

The equation of escape speed can be written as,

$$\begin{gathered} \mathrm{v}=\sqrt{\frac{2\mathrm{GM}}{\mathrm{r}}} \\ \mathrm{v}=\sqrt{\frac{2\mathrm{GM}}{\left({\displaystyle \frac{\mathrm{d}}{2}}\right)}} \end{gathered}$$

Here, Gis the universal gravitational constant.

Substitute the values in the above equation.

$$\begin{gathered} \mathrm{v}=\sqrt{\frac{2\left(6.67×{10}^{-11}{\mathrm{m}}^2/\mathrm{kg}.{\mathrm{s}}^2\right)\left(3.525×{10}^{19}\mathrm{kg}\right)}{\left({\displaystyle \frac{300\mathrm{km}×{\displaystyle \frac{1000\mathrm{m}}{1\mathrm{km}}}}{2}}\right)}} \\ \mathrm{v}=177.05\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the required escape speed is 177.05 m/s.