Radial Distance of Projectile Shot from Earth's Surface

AI Thread Summary
A projectile shot directly away from Earth's surface can reach a radial distance from the Earth's center based on its initial speed and kinetic energy. For part (a), the initial speed should be calculated as 0.758 times the escape speed, and then plugged into the kinetic energy equation. In part (b), the initial kinetic energy is 0.758 of the escape kinetic energy, and it is important to use the standard escape speed for calculations. The discussion emphasizes the distinction between initial velocity and kinetic energy in determining the projectile's maximum height. Understanding these concepts is crucial for solving the problem effectively.
mariahkraft
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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!.
 
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mariahkraft said:
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!.

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)
 
thank you!
 
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