Question

A particle is projected upward from the surface of earth(radius R) with a K.E. equal to half the minimum value needed for it to escape from earth's gravity. To what height does it rises above the surface of earth?

Answer 1

Total energy of the particle on surface of Earth
   = KE + PE = KE - G Me m / Re

  KE needed for escaping from the gravity of Earth = G Me m /Re.  Then the total energy of the particle will be zero. Hence, it will escape from the gravitational field.

  Energy given to the particle = 1/2 * G Me m / Re

   Total energy of the particle on surface of Earth = KE + PE = -1/2 G Me m / Re

Total energy of a particle in an orbit (radius R) around Earth has a total energy =
         - G Me m / R + 1/2 G Me m /R  = - 1/2 * G Me m / R

     Comparing the two above equations and as per the principle of conservation of energy,    The radius of the orbit = R = Re.

   hence, the particle stays on the surface of Earth.

Answer 2

$\huge \bold \color{lime}{ \underline \color{teal} \mathfrak{Answer}}$

$escape \: velocity ,\: v = \sqrt{ \frac{2GMe}{Re} }$

Where Me is mass of earth, Re is radius of earth.

$K.E. \: of \: particle = \frac{1}{2} escape \: K.E.$

$K.E. = \frac{1}{2} ( \frac{1}{2} m {v}^{2} )$

Where m is mass of particle

$K.E. = \frac{1}{4} \times m \times \frac{2GMe}{Re}$

$K.E. = \frac{GMem}{2Re} \implies \: eq1$

Let the height be h

$Also \: K.E. \: of \: particle = \frac{GMem}{2h } \implies \: eq2$

Equating eq 1 and eq 2

$\frac{GMem}{2Re} = \frac{GMem}{2h}$

$2 Re= 2h$

$\huge \green{h = Re}$

Hope it helps....