11. A projectile is shot directly away from the Earth?s surface. Neglect the rot

Question

11. A projectile is shot directly away from the Earth?s surface. Neglect the rotation of the Earth. a) What multiple of the Earth?s radius RE gives the radial distance a projectile reaches if its initial speed is 0.500 of the escape speed of the Earth? b) What multiple of the Earth?s radius RE gives the radial distance a projectile reaches if its initial kinetic energy is 0.500 of the kinetic energy required to escape the Earth? c) What is the least initial mechanical energy required at launch if the projectile is to escape the Earth

Explanation / Answer

1/2mv^2=mMG/R=mgR [escape kinetic energy]
ve=(2MG/R)^1/2=(2gR)^1/2 [escape speed]

A>maximum height is
h=v^2/(2g)
if v=kve
h=k^2ve^2/(2g)
h=k^2R
h=0.25R [R=Earth radius]

B>if 1/2mv^2=kmgR
v^2=2kgR
then the maximum height is
h=v^2/(2g)=kR
h=0.500 R [R=Earth radius]

C>E =0.5m(v^2) = 0.5=0.5m*2GM/R= GmM/R = mgR (Initial least energy)

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