The orbital speed ($v_o$) of a satellite at a height $h$ above the Earth's surface is given by the formula $v_o = \sqrt{\frac{GM}{R+h}}$. Since the acceleration due to gravity on the surface is $g = \frac{GM}{R^2}$, we can substitute $GM = gR^2$ into the velocity equation: $v_o = \sqrt{\frac{gR^2}{R+h}} = R \sqrt{\frac{g}{R+h}}$. Given values: $R = 6.38 \times 10^6$ m $h = 0.25 \times 10^6$ m $g = 9.8$ m s$^{-2}$ The orbital radius is $r = R+h = (6.38 + 0.25) \times 10^6 = 6.63 \times 10^6$ m. Substituting these into the formula: $v_o = 6.38 \times 10^6 \sqrt{\frac{9.8}{6.63 \times 10^6}}$ $v_o = 6.38 \times 10^6 \times \sqrt{1.478 \times 10^{-6}}$ $v_o \approx 6.38 \times 10^6 \times 1.216 \times 10^{-3} \approx 7.76 \times 10^3$ m/s = 7.76 km/s.