Calculate the escape velocity for an atmospheric particle 1600km above the earth's surface,Given;The radius of earth is 6400km and acceleration due to gravity on the surface of earth is 9.8.

We know that bindin energy of particle on the surface of earth is GMm/R+h.If this much energy is supplied in form of kinetic energy the body will escape out of the gravitaional field of earth.v = 2gRputting the valuesv = 2*9.8*(6400+1600)*10-3 km/s= 12.52 km/sec

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Given, Re=6400km= 6.4*106 m r=1600km= 1.6*106 m

g= 9.8m/s2

Hence Escape Velocity =√2*gR

= Ve=√2*g*(Re+r)

= Ve=√2*9.8*(6.4+1.6)*106

= Ve= 12.5 *103 m/s= 12.5 km/s

Hence the Escape Velocity in the given condition is 12.5 km/s

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