An unmanned space probe (of mass mand speed vrelative to the Sun) approaches the planet Jupiter (of mass Mand speed VJrelative to the Sun) as shown in Fig. 9-84. The spacecraft rounds the planet and departs in the opposite direction. What is its speed (in kilometers per second), relative to the Sun, after this slingshot encounter, which can be analyzed as a collision? Assume $\mathit{v}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{5}\frac{\mathbf{k}\mathbf{m}}{\mathbf{s}}$and $\mathit{V}\mathit{J}\mathbf{=}\mathbf{13}\mathbf{.}\mathbf{0}\frac{\mathbf{km}}{\mathbf{s}}$(the orbital speed of Jupiter).The mass of Jupiter is very much greater than the mass of the spacecraft (M $\mathbf{\gg }$m).

1. Mass of the space probe is m.
2. Massof Jupiteris M.
3. Initial speed of the spaceprobe relativeto the sun is$v=10.5\frac{\mathrm{km}}{s}·$
4. Initial speed of the Jupiter relative to the sun is$V_J=13\frac{\mathrm{km}}{s}·$

Two equations can be found from conservation of momentum and energy. Solving them, the expression for speed of the space probe relative to the sun can be found. Then, applying the condition for masses, approximate the speed of the space probe relative to the sun. Inserting values in it, its magnitude can be found.

Formulae are as follow:

$P_i=P_f$

$E_i=E_f$

Here, P_i,and P_fare initial and final momentum, E_i,andE_fare initial and final energies.

Gravitational slingshot corresponds to the elastic collision. Therefore, momentum is conserved in it. Hence,

$P_i=P_f$

Let, v’ and v'_f are the final speeds of the space probe and Jupiter respectively. Then,

$mv+M{V}_J=mv\text{'}+MV{\text{'}}_J$ …… (1)

According to the law conservation of energy,

$E_i=E_f$

$\frac{1}{2}m{v}^2+\frac{1}{2}M{V}_J^2=\frac{1}{2}mv{\text{'}}_J^2$ …… (2)

Solving equations (1) and (2).

$v\text{'}=\frac{m-M}{m+M}+\frac{2M}{m+M}V{\text{'}}_J$

But, m << M, Then,$m+M~Mandm-M~-M$.Hence,

$$\begin{gathered} v\text{'}~-\frac{M}{M}v+\frac{2M}{M}V{\text{'}}_J \\ \therefore v\text{'}=-v+2V{\text{'}}_J \end{gathered}$$

Let’s consider the direction of the space probe after collision as negative. Then final speed of it relative to sun is,

$$\begin{gathered} -v\text{'}=10.5+2\left(13\right) \\ -v\text{'}=36.5 \\ \left|v\text{'}\right|=36.5 \end{gathered}$$

Hence, speed of the space probe relative to the sun after collision is$|v\text{'}|=36.5\frac{\mathrm{km}}{s}·$

Therefore,the speed of the space probe relative to the sun can be found after the slingshot using the law of conservation momentum and energy.