Explanation Needed for Equation for Gravitation/Conservation of Energy equation

[auk411]

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

K_i + U_i = K_f + U_f,
where K is the kinetic energy and u is the gravitational potential energy.

v₀ =\sqrt{Gm/2R} ... 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:

\frac{1}{2}mv₀² - \frac{GMm}{R_e} = 0

I plug in v₀ =\sqrt{Gm/2R} 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:

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

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

Also, why doesn't the \frac{1}{2}mv₀² - \frac{GMm}{R_e} =\frac{-GMm}{2r}, 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?

 

[kudoushinichi88]

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

\frac{1}{2}mv^2-\frac{GMm}{r}=-\frac{GMm}{2r}

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