A meteor headed straight towards the earth has an approach speed of 2.00e3 m/s at a distance of 7 earth radii from the center of the earth. What will be its speed when it reaches earth a 6.38e6 m(ignore the effects of atmosphere)?

Question

A meteor headed straight towards the earth has an approach speed of 2.00e3 m/s at a distance of 7 earth radii from the center of the earth.  What will be its speed when it reaches earth a 6.38e6 m(ignore the effects of atmosphere)?

bobpursley · Answer

CAlculate the potential energy at each point:

PE= GMe*m/r

Then, set that change of energy into
original KE+ change of PE=final ke
Vf^2=Vi^2+ 2*change in PE/m

vf^2=vi^2+2GMe/r1-2GMe/r2
r1 original distance, r2 final