Recall that the magnitude of the acceleration [tex]a[/tex] of a particle moving with speed [tex]v[/tex] in a circular path around a point at a distance [tex]R[/tex] away from the particle is given by
[tex]a=\dfrac{v^2}R[/tex]
So, the satellite has velocity
[tex]6.2\,\dfrac{\rm m}{\mathrm s^2}=\dfrac{v^2}{800\,\rm m}\implies v=70.4\,\dfrac{\rm m}{\rm s}[/tex]
pointing in the direction tangent to the circular path.
