Q.
An asteroid of mass m is approaching earth, initially at a distance of 10Re with speed vi. It hits the earth with a speed vf(Re and Me are radius and mass of earth), then
Applying law of conservation of energy for asteroid at a distance 10Re and at earth's surface, Ki+Ui=Kf+Uf (i)
Now, Ki=21mvi2 and Ui=10ReGMem Kf=21mvf2 and Uf=−ReGMem
Substituting these values in Eq. (i), we get 21mvi2−10ReGMem=21mvf2−ReGMem ⇒21mvf2=21mvi2+ReGMem−10ReGMem ⇒vf2=vi2+Re2GMe−10Re2GMe ∴vf2=vi2+Re2GMe(1−101)