Answer
$Less, By\space about\space 6802\space km/h$
Work Step by Step
Here we use the principle of conservation of mechanical energy.
Final mechanical energy = Initial mechanical energy
$K+U=K_{0}+U_{0}$
$-\frac{GM_{S}m}{r}+\frac{1}{2}mV^{2}=0+0$
Where, m - is the mass of the spacecraft, $M_{S}$- the mass of Saturn, V - the speed at the atmosphere of Saturn, and r - The distance to the end of the atmosphere from the center of Saturn.
Let's plug known values into the above equation.
$V^{2}= \frac{2GM_{S}}{r}= \frac{2\times6.67\times10^{-11}Nm^{2}/kg^{2}\times568\times10^{24}kg}{58291\times10^{3}m}$
$V^{2}=\frac{7577.12\times10^{13}}{58291\times10^{3}}m^{2}/s^{2}$
$V= \sqrt {13\times10^{8}}m/s= 36056\space m/s= 129802\space km/h$
The value is larger than the 123000 km/h. Therefore 123000 km/h is less & it less by 6802 km/h
