An unmanned space probe (of mass m and speed v relative to the Sun) approaches the planet Jupiter (of mass M and speed V_J relative 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 v = 10.5 km/s and V_J=13.0 km/s (the orbital speed of Jupiter).The mass of Jupiter is very much greater than the mass of the spacecraft (M >> m).
Conservation of linear momentum gives m v+M V_J=m v_f+M V_{J f}. Similarly, the total kinetic energy is conserved:
\frac{1}{2}m v^2+\frac{1}{2}M V_J^2=\frac{1}{2}m v_f^2+\frac{1}{2}M V_{J f}^2.
Solving for v_f and V_{Jf} , we obtain:
v_{1 f}=\frac{m-M}{m+M}v+\frac{2 M}{m+M}V_J, \quad V_{J f}=\frac{2 m}{m+M}v+\frac{M-m}{m+M}V_J
Since m<< M, the above expressions can be simplified to
v_{1 f}\approx-v+2 V_J, \quad V_{J f}\approx V_J
The velocity of the probe relative to the Sun is
v_{1 f}\approx-v+2 V_J=-(10.5 \,km / s )+2(-13.0\, km / s )=-36.5 \,km / s.
The speed is \left|v_{1 f}\right|=36.5 km/s.