Explanation Needed for Equation for Gravitation/Conservation of Energy equation

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SUMMARY

The discussion centers on the conservation of energy in the context of a projectile launched from Earth's surface at half the escape velocity. The key equations involved are the conservation of mechanical energy, represented as Ki + Ui = Kf + Uf, and the initial velocity equation v0 = √(Gm/2R). The correct formulation for the energy conservation equation is identified as (1/2)mv0² - (GMm/R_e) = (GMm/r), where r is the radial distance from Earth. The confusion arises from misapplying the total mechanical energy equation for satellites, which is not applicable to projectiles launched from the surface.

PREREQUISITES
  • Understanding of gravitational potential energy and kinetic energy
  • Familiarity with the conservation of energy principle
  • Knowledge of escape velocity concepts
  • Basic grasp of orbital mechanics
NEXT STEPS
  • Study the derivation of escape velocity from gravitational potential energy
  • Learn about the conservation of mechanical energy in different gravitational contexts
  • Explore the differences between projectiles and satellites in orbital mechanics
  • Investigate the implications of energy conservation in multi-body gravitational systems
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Students studying physics, particularly those focusing on mechanics and gravitational forces, as well as educators seeking to clarify concepts related to energy conservation and projectile motion.

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Homework Statement


A projectile is shot directly away from Earth's surface. Neglect rotation of Earth. What multiple of Earth's radius R_e gives the radial distance a projectile reaches if its initial speed is 1/2 of the escape speed from Earth.


Homework Equations



Ki + Ui = Kf + Uf,
where K is the kinetic energy and u is the gravitational potential energy.

v0 =[tex]\sqrt{Gm/2R}[/tex] ... this is the initial velocity when its initial speed is 1/2 of the escape speed.

The Attempt at a Solution



We know that energy is conserved. Then what I do is:

[tex]\frac{1}{2}[/tex]mv02 - [tex]\frac{GMm}{R_e}[/tex] = 0

I plug in v0 =[tex]\sqrt{Gm/2R}[/tex] for v0 in the previous equation. I do not get the right answer. What is wrong with this equation?

Apparently, this is the right equation:

[tex]\frac{1}{2}[/tex]mv02 - [tex]\frac{GMm}{R_e}[/tex] = [tex]\frac{GMm}{r}[/tex], where r is just some radial distance from earth.

Now, I do not understand where [tex]\frac{GMm}{r}[/tex] comes from.

Also, why doesn't the [tex]\frac{1}{2}[/tex]mv02 - [tex]\frac{GMm}{R_e} =\frac{-GMm}{2r}[/tex], since that is what my textbook says is what the total mechanic energy is for satellites. Isn't this thing going to be become a satellite? So why couldn't I use the aforementioned equation?
 
Last edited:
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Initially on the surface of the earth, the projectile has both kinetic and potential energy. Now at the peak of it's motion, the projectile has lost all its kinetic energy which has been converted totally into potential energy. That's where the GMm/r on the right hand side of the equation comes from.

For your last question, that equation is for an object which is already in orbit, not for something which is shot from the Earth into orbit. That equation enables us to find the speed at which the satellite is moving around the Earth if we know the altitude of the satellite. You wrote the equation wrongly too. Its

[tex]\frac{1}{2}mv^2-\frac{GMm}{r}=-\frac{GMm}{2r}[/tex]

in which both r's are the distance of the satellite from the center of the earth.
 
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