Find approximately the third cosmic velocity `v_3`, i.e. the minimum velocity that has to be imparted to a body relative to the Earth's surface to drive it out of the Solar system. The rotation of the Earth about its own axis is to be neglected.

Explanation

Assume first that the attraction of the earth can be neglected. Then the minimum velocity, that must be imparted to the body to escape from the Sun's pull, is, as in `1*230`, equal to
`(sqrt2-1)v_1`
where `v_1^2=gammaM_s//r`, r=radius of the earth's orbit, `M_s=` mass of the Sun.
In the actual case near the earth, the pull of the Sun is small and does not change much over distances, which are several times the radius of the Earth. The velocity `v_3` in question is that which overcomes the earth's pull with sufficient velocity to escape the Sun's pull.
Thus
`1/2mv_3^2-(gammaM_E)/(R)=1/2m(sqrt2-1)^2v_1^2`
where R=radius of the earth, `M_E`=mass of the earth.
Writing `v_1^2=gammaM_E//R`, we get
`v_3=sqrt(2v_2^2+(sqrt2-1)^2v_1^2)=16*6km//s`