You launch a rocket with an initial speed of $5\times 10^3$ m/s from Earth's surface. At what height above the Earth will it have ¼ of its initial launch speed? Assume the rocket's engines shut off after launch.

The initial kinetic energy is given by \( KE_i = \frac{1}{2} m v_0^2 \) and the initial potential energy is \( PE_i = -\frac{G M m}{R} \), where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

The final kinetic energy is \( KE_f = \frac{1}{2} m v_f^2 \) and the final potential energy is \( PE_f = -\frac{G M m}{R + h} \), where \( h \) is the height above the Earth's surface.