The orbital speed of the satellite is

${v_o} = R\sqrt {{g \over {\left( {R + h} \right)}}}$

where R is the earth’s radius, g is the acceleration due to gravity on earth’s surface and h is the height above the surface of earth.

Here, R = 6.38 × 10⁶m, g = 9.8 m s^–2 and h = 0.25 × 10⁶ m

${v_o} = \left( {6.38 \times {{10}^6}m} \right)\sqrt {{{9.8\,m{s^{ - 2}}} \over {\left( {6.38 \times {{10}^6}m + 0.25 \times {{10}^6}m} \right)}}}$

= 7.76 × 10³ m s^–1 = 7.76 km s^–1