The escape speed of a projectile on the Earth 's surface i.e `11.2kms^(-1)` .A body is project with thrice this speed .what is the speed of the body far away from the Earth ? Ignore the pressure of sun and other places.

Explanation

`v_(e)=11.2xx10^(3)ms^(-1),v_(e)^(1)=3xx11.2xx10^(3)ms^(-1)=33.6xx10^(3)ms^(-1)`
applying `1/2mv_(e)^(2)-(GMm)/(R)=1/2mv^(2)` where PE = 0 at a
distance far away from earth.
`(v_(e)^(2))/(2)-gR=(v^(2))/(2)`
i.e., `v^(2)=v_(e)^(2)-2gR"where " sqrt(2gR)=v_(e)^(1),v^(2)=(3v_(e)^(1))^(2)-(v_(e)^(1))^(2)`
i.e., `v^(2)=8(v_(e)^(1))^(2)`
`:.v=(2sqrt(2))v_(e)^(1)=2xx1.414xx11.2kms^(-1)`
`v = 31 ksm^(-1)`