(a) [tex]3.1\cdot 10^7 J[/tex]
The total mechanical energy of the space probe must be constant, so we can write:
[tex]E_i = E_f\\K_i + U_i = K_f + U_f[/tex] (1)
where
[tex]K_i[/tex] is the kinetic energy at the surface, when the probe is launched
[tex]U_i[/tex] is the gravitational potential energy at the surface
[tex]K_f[/tex] is the final kinetic energy of the probe
[tex]U_i[/tex] is the final gravitational potential energy
Here we have
[tex]K_i = 5.0 \cdot 10^7 J[/tex]
at the surface, [tex]R=3.3\cdot 10^6 m[/tex] (radius of the planet), [tex]M=5.3\cdot 10^{23}kg[/tex] (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is
[tex]U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J[/tex]
At the final point, the distance of the probe from the centre of Zero is
[tex]r=4.0\cdot 10^6 m[/tex]
so the final potential energy is
[tex]U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J[/tex]
So now we can use eq.(1) to find the final kinetic energy:
[tex]K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J[/tex]
(b) [tex]6.3\cdot 10^7 J[/tex]
The probe reaches a maximum distance of
[tex]r=8.0\cdot 10^6 m[/tex]
which means that at that point, the kinetic energy is zero: (the probe speed has become zero):
[tex]K_f = 0[/tex]
At that point, the gravitational potential energy is
[tex]U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J[/tex]
So now we can use eq.(1) to find the initial kinetic energy:
[tex]K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J[/tex]
