The escape speed isthe speed when the escape kinetic energy equals the magnitude

Question

The escape speed isthe speed when the escape kinetic energy equals the magnitude ofthe initial gravitational potential energy. Did you conserve themechanical energy (the sum of the kinetic and potential energies)?Did you use GMm/r for the potentialenergy (not mgh)?
A projectile is shot directly away from Earth^'s surface. Neglectthe rotation of Earth. (a) As a multiple of Earth^'s radius R_E,what is the radial distance a projectile reaches if its initialspeed is four-fifths of theescape speed from Earth? x R_E (b) As a multiple of Earth^'s radius R_E,what is the radial distance a projectile reaches if its initialkinetic energy is three-fifths of the kinetic energy required toescape Earth? x R_E (c) What is the least initial mechanical energy required at launchif the projectile is to escape Earth? J The escape speed isthe speed when the escape kinetic energy equals the magnitude ofthe initial gravitational potential energy. Did you conserve themechanical energy (the sum of the kinetic and potential energies)?Did you use GMm/r for the potentialenergy (not mgh)?

Explanation / Answer

Hey, I think I may actually be in your class at Penn State withFoster, but for part C: since mechanical energy = KE + PE = final KE + final PE = 0 + 0 =0, the net mechanical energy = 0 J for the projectile to just escape the earth's gravitationalfield. I still don't know how to do part B but that answer seems to be wayoff. Should be in the range of 0.1-5 it seems.

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