To solve this problem we will proceed to use the equations of the centripetal Force, the force of gravity and finally the equivalence of the Period referring to the linear velocity. Mathematically this is,
[tex]F_c = F_g[/tex]
[tex]\frac{mv^2}{r} = \frac{GMm}{r^2}[/tex]
Where,
m = Mass of the object
M = Mass of the planet
r = Distance/radius
v = Velocity
Rearranging to find the velocity we have,
[tex]v^2 = \frac{GM}{r}[/tex]
Using the expression of Period for the velocity we have,
[tex]\frac{4\pi^2r^2}{T^2} = \frac{GM}{r}[/tex]
[tex]T^2 = 4\pi^2 \frac{r^3}{GM}[/tex]
Replacing we have,
[tex]T^2 = 4\pi^2 \frac{(3.4*10^6+100*10^3)^3}{(6.67*10^{-11})(6.42*10^{23})}[/tex]
[tex]T = 6287.12s = 104.7min = 1.74hours[/tex]
Therefore the period of the satellite is 1.74Hours
