In Fig. 13-50, two satellites, A and B, both of mass $\mathbf{m}\mathbf{\text{=125kg}}$ , move in the same circular orbit of radius $\mathbf{r}{\mathbf{\text{=7.87×10}}}^{\mathbf{\text{6}}}\mathbf{\text{m}}$around Earth but in opposite senses of rotation and therefore on a collision course.

(a) Find the total mechanical energy role="math" localid="1661161625366" ${\mathbf{E}}_A\mathbf{+}{\mathbf{E}}_B$of the$\mathbf{\text{twosatellites+Earth}}$ system before the collision.

(b) If the collision is completely inelastic so that the wreckage remains as one piece of tangled material ( $\mathbf{\text{mass=2m}}$), find the total mechanical energy immediately after the collision.

(c) Just after the collision, is the wreckage falling directly toward Earth’s center or orbiting around Earth?

The mass of each satellite A and B is$\text{M}=125\text{\hspace{0.17em}kg}$

The radius of the orbit of satellites around the Earth is$\text{R}=7.87\times {10}^6\text{\hspace{0.17em}m}$

Using the formula for the total mechanical energy of an orbiting satellite around the Earth, findthe total mechanical energy of the two satellites + Earth system before the collision and after the collision. From the velocity of the wreckage, interpret its direction of motion.

The formula is as follows:

$\mathrm{E}=\frac{-{\mathrm{GM}}_{\mathrm{E}}\mathrm{m}}{2\mathrm{r}}$

where E is total mechanical energy, G is gravitational constant, ${\text{M}}_{\text{E}}$, m are masses and r is the radius.

The total mechanical energy of an orbiting satellite around the Earth is,

$\mathrm{E}=\frac{-{\mathrm{GM}}_{\mathrm{E}}\mathrm{m}}{2\mathrm{r}}$

The total mechanical energy of the two satellites + Earth system before the collision is,

${\mathrm{E}}_{\mathrm{A}}+{\mathrm{E}}_{\mathrm{B}}=\frac{-{\mathrm{GM}}_{\mathrm{E}}\mathrm{m}}{2\mathrm{r}}+\frac{-{\mathrm{GM}}_{\mathrm{E}}\mathrm{m}}{2\mathrm{r}}$

${\mathrm{E}}_{\mathrm{A}}+{\mathrm{E}}_{\mathrm{B}}=\frac{-{\mathrm{GM}}_{\mathrm{E}}\mathrm{m}}{\mathrm{r}}$

$\begin{array}{c}{\mathrm{E}}_{\mathrm{A}}+{\mathrm{E}}_{\mathrm{B}}=\frac{-(6.67\times {10}^{-11}\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{kg}}^2)(5.98\times {10}^{24}\mathrm{kg})\left(125\mathrm{kg}\right)}{7.87\times {10}^6\mathrm{m}}\\ =-6.33\times {10}^9\mathrm{J}\end{array}$

Hence,the total mechanical energy E_A +E_B of the two satellites + Earth system before the collision is $-6.33\times {10}^{\text{9}}\text{J.}$

The total mechanical energy immediately after the collision is,

$\mathrm{E}=\frac{-{\mathrm{GM}}_{\mathrm{E}}\left(2\mathrm{m}\right)}{2\mathrm{r}}$

$\mathrm{E}=\frac{-{\mathrm{GM}}_{\mathrm{E}}\mathrm{m}}{\mathrm{r}}$

$\begin{array}{c}\mathrm{E}=\frac{-(6.67\times {10}^{-11}\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{kg}}^2)(5.98\times {10}^{24}\mathrm{kg})\left(125\mathrm{kg}\right)}{7.87\times {10}^6\mathrm{m}}\\ =-6.33\times {10}^9\mathrm{J}\end{array}$

Hence, the total mechanical energy immediately after the collision is $-\text{6.33}\times {\text{10}}^{\text{9}}\text{\hspace{0.17em}J.}$

Just after the collision, the wreckage has zero velocity. So, it will fall towards the Earth’s center.

Therefore, using the formula for the total mechanical energy of an orbiting satellite around the Earth, the total mechanical energy before and after the collision of satellites can be found.