Projectile fired from earth. Conservation of energy

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SUMMARY

The discussion focuses on the mechanics of a projectile launched from Earth's surface, specifically addressing calculations related to its escape velocity and mechanical energy. The initial speed of the projectile is set at 0.5 times the escape speed, and the kinetic energy is half of the required energy for escape. Participants clarify the equations needed to solve for the radial distance the projectile reaches, including the total mechanical energy formula and the escape velocity equation, v = √(2GM/R). Key corrections were made regarding the equations used in parts a and b of the problem.

PREREQUISITES
  • Understanding of gravitational potential energy and kinetic energy concepts
  • Familiarity with escape velocity calculations
  • Knowledge of the formula for total mechanical energy
  • Basic algebra for solving equations
NEXT STEPS
  • Study the derivation of escape velocity using the formula v = √(2GM/R)
  • Learn about conservation of mechanical energy in gravitational fields
  • Explore the implications of initial kinetic energy on projectile motion
  • Investigate the effects of varying initial speeds on the trajectory of projectiles
USEFUL FOR

Students in physics, educators teaching mechanics, and anyone interested in understanding projectile motion and energy conservation principles in gravitational fields.

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Homework Statement


A projectile is shot directly away from Earth's surface. Neglect the rotation of Earth.
What multiple of Earth's radius R_e gives the radial distance a projectile reaches if its (a) initial speed is .5 the escape speed from earth, (b) its initial kinetic energy is .5 of the kinetic energy required to escape Earth? (c) What is the least initial mechanical energy required at launch if the projectile is to escape Earth?

Homework Equations



v =.5\sqrt{2GM/R}

The Attempt at a Solution



Total mechanical energy = 1/2 mv^2 - GMm/R =?

I know that I'm going to need to plug in the escape velocity for v^2. However, that is all I know.
(As for b and c, I don't even know what the relevant equations are. )
 
Last edited:
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I'm not sure why it states mechanical energy in part of the problem but I'm assuming this is kinetic energy.
K = \frac{1}{2} mv^2
v = \sqrt{\frac{2GM}{R}}

a) v = 0.5\sqrt{\frac{2GM}{R}} Solve for R
b) \frac{1}{2}mv^2 = \frac{1}{2}\frac{GMm}{R} Solve for R
c) Start with your given formula for v (without the 0.5 that you added) and get to \frac{1}{2}mv^2 on the LHS
 
Last edited:
So does anyone know how to answer this cause the person before doesn't know how to answer it.

For one thing, part a is just wrong. That is not true equation.

... Unless I'm just totally confused.
 
You're correct that part a) is wrong. I should have had the 0.5 on the RHS of the equation. And the same for part b). I'll edit my original post to reflect this.
 

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