\({v_0} = R\sqrt {\frac{g}{{\left( {R + h} \right)}}} \)

Where \(R\) is the eath's radius, \(g\) is the acceleration due to gravity on earth's surface and \(h\) is the height above the surface of earth.

Here, \(R = 6.38 \times {10^{ 6}}m,g\, = 9.8m\,{s^{ - 2}}\,and\)

\(h = 0.25 \times {10^6}\,m\)

\(\therefore {v_0} = \left( {6.38 \times {{10}^6}m} \right)\sqrt {\frac{{\left( {9.8\,m\,{s^{ - 2}}} \right)}}{{\left( {6.38 \times {{10}^6}\,m + 0.25 \times {{10}^6}m} \right)}}} \)

\( = 7.76 \times {10^3}\,m\,{s^{ - 1}} = 7.76\,km\,{s^{ - 1}}\)