Find the maximum height reached by a rocket launched…

Find the maximum height reached by a rocket launched vertically with an initial velocity equal to the first cosmic velocity. The Earth's radius $R_Z$ and mass $M$ and the gravitational constant are given; neglect air resistance and use energy conservation.

Step 1: Idea

The first cosmic velocity is the speed of a circular orbit at the surface. Launched horizontally with this speed, a body would circle the Earth; launched vertically, it goes up and comes back. We find how high it gets from energy conservation, with speed zero at the top.

Step 2: First cosmic velocity

From $mv_1^2/R_Z = \gamma Mm/R_Z^2$ (Problem 19): $v_1^2 = \frac{\gamma M}{R_Z}$.

Step 3: Energy conservation

\[ \tfrac12mv_1^2 - \frac{\gamma Mm}{R_Z} = -\frac{\gamma Mm}{R_Z + h}. \]

Insert $v_1^2$: the kinetic energy is $\tfrac12\gamma Mm/R_Z$, exactly half of the depth of the potential well at the surface: $\frac{\gamma Mm}{2R_Z} - \frac{\gamma Mm}{R_Z} = -\frac{\gamma Mm}{R_Z + h}$.

\[ \frac{1}{2R_Z} = \frac{1}{R_Z + h}. \]

\[h = R_Z \approx 6370\ \mathrm{km}\]

The rocket rises one Earth radius, to twice the distance from the centre. Note that the uniform-gravity formula $v_1^2/2g = R_Z/2$ would give only half of this.

Check

From Problem 28 with $v_0^2 = gR_Z$: $ h $ $= gR_Z^2/(2gR_Z - gR_Z) $ $= R_Z\ \checkmark $