# Escape speed

1. Dec 13, 2008

### mariahkraft

A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE gives the radial distance (from the Earth's center) the projectile reaches if (a) its initial speed is 0.758 of the escape speed from Earth and (b) its initial kinetic energy is 0.758 of the kinetic energy required to escape Earth?

So basically what i got from lecture is that K+U=1/2mv2+-GmM/R=0
v=$$\sqrt{2GM/R}$$

for part a would I just multiply v=$$\sqrt{2GM/R}$$ by .758 then plug that in for v in the 1/2 mv2?

then for part b would i multiply the entire 1/2mv2 by .758? Also one more thing for part b am i using my answer from part a for the speed or am i using standard escape speed?

I have my final on monday and I am trying to review for the test any help would be awesome Thanks!.

2. Dec 13, 2008

### LowlyPion

Welcome to PF.

Your first equation is for escape velocity. But for escape velocity it is assumed that potential is 0 and kinetic energy is 0.

So what they are really giving you is 1/2m(.758v)2 + (-μ*m/Re) = PE at max height which is given by -μ*m/Rmax

The second part wants you to recognize the difference between initial velocity and initial KE which is v:v2

(Note I used μ as GM the standard gravitational parameter for earth which numerically is 398K in SI units)

3. Dec 13, 2008

### mariahkraft

thank you!!!!