91Ó°ÊÓ

Mass of the bullet$4.0\text{ }\mathrm{g}$

Muzzle speed of$950\text{ }\text{m/s}$

Satellite in a circular orbit at$500\text{ k³¾}$above the earth's surface

Using the formula of kinetic energy, wecan find thekinetic energy of the pellet in the reference frame of the satellite just before the collision and the ratio of this kinetic energy to the kinetic energy of a $\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{g}$bullet from a modern army rifle with a muzzle speed of$\mathbf{950}\mathbf{\text{″¾/²õ}}$

Formula:

$\mathbf{\text{K}}\mathbf{=}\frac{\mathbf{\text{1}}}{\mathbf{\text{2}}}{\mathbf{\text{mv}}}^{\mathbf{\text{2}}}$

$\mathbf{\text{v=}}\sqrt{\frac{\mathbf{\text{GM}}}{\mathbf{\text{R}}}}$

As $\text{v}=\sqrt{\frac{\text{GM}}{\text{R}}}$and pellets have the same speed but in the opposite direction of motion, so the relative speed between the pellet and satellite is ${\text{v}}_{\text{rel}}=\text{2v}$

$\begin{array}{c}{\text{K}}_{\text{rel}}=\frac{\text{1}}{\text{2}}\text{m}{\left(\text{2}\sqrt{\frac{\text{GM}}{\text{R}}}\right)}^{\text{2}}\\ =\frac{\text{2GMm}}{\text{R}}\end{array}$

As$\mathrm{R}=(6370×{10}^3\text{ }\mathrm{m}+500×{10}^3\text{ }\mathrm{m})=6870×{10}^3\text{ }\mathrm{m}$

$\begin{array}{c}{\text{K}}_{\text{rel}}=\frac{2\left(6.67×{10}^{−11}\frac{{\mathrm{m}}^3}{\mathrm{kg}â‹\dots {\mathrm{s}}^2}\right)(5.98×{10}^{24}\text{ }\mathrm{kg})(0.0040\text{ }\mathrm{kg})}{6870×{10}^3\text{ }\mathrm{m}}\\ =4.6×{10}^5\text{ }\mathrm{J}\end{array}$

Therefore, the kinetic energy of the pellet in the reference frame of the satellite just before the collision is $4.6×{10}^5\mathrm{J}$

$\text{K}=\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}$

Therefore,

$\begin{array}{c}\frac{{\text{K}}_{\text{rel}}}{{\text{K}}_{\text{bullet}}}=\frac{4.6×{10}^5\text{ }\mathrm{J}}{\frac{1}{2}\left(0.0040\text{ }\mathrm{kg}\right){\left(950\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)}^2}\\ =\frac{4.6×{10}^5\mathrm{J}}{1805\mathrm{J}}\\ =2.6×{10}^2\end{array}$

Therefore, theratio of this kinetic energy to the kinetic energy of a $4.0\text{g}$bullet from a modern army rifle with a muzzle speed of$950\text{ }\text{m/s}$is$2.6×{10}^2$.