A body is launched vertically from the Earth's surface with…

A body is launched vertically from the Earth's surface with initial velocity $v_0$. Given the Earth's radius $R_Z$ and the surface gravity $g$, find the maximum height the body reaches (no air resistance).

Explanation of Method

Height reached in a vertical launch.

Step 1: Idea

For small heights we would use $h = v_0^2/2g$, but that assumes a constant gravity. When the body rises high, gravity weakens, so it rises higher than this formula says. We therefore use energy conservation with the full potential energy $-\gamma Mm/r$.

Step 2: Energy conservation

At launch the body is at $r = R_Z$ with speed $v_0$; at the highest point it is at $r = R_Z + h$ with speed zero: $\tfrac12mv_0^2 - \frac{\gamma Mm}{R_Z} = -\frac{\gamma Mm}{R_Z + h}$.

Replace $\gamma M$ by $gR_Z^2$ and divide by $mgR_Z$: $\frac{v_0^2}{2gR_Z} - 1 = -\frac{R_Z}{R_Z + h}$.

\[ \frac{R_Z}{R_Z + h} = 1 - \frac{v_0^2}{2gR_Z} = \frac{2gR_Z - v_0^2}{2gR_Z}. \]

Inverting both sides gives $1 + h/R_Z = 2gR_Z/(2gR_Z - v_0^2)$, so

\[h = \frac{v_0^2R_Z}{2gR_Z - v_0^2}\]

Step 3: Discussion

If $v_0$ is small compared with $\sqrt{2gR_Z}$, the denominator is almost $2gR_Z$ and $h \approx v_0^2/2g$, the familiar result. As $v_0$ approaches $\sqrt{2gR_Z} = 11.2\ \mathrm{km/s}$ the denominator goes to zero and the height becomes infinite: the body escapes (see the graph).

Check

Small $v_0$: $ h \approx v_0^2/2g $, the uniform-field result $\checkmark$

$v_0 = \sqrt{gR_Z}$: $h = R_Z$ (Problem 29) $\checkmark$

$h/R$ $v_0/v_2$ $1/\sqrt2$ $1$ $1$ $v_0 = v_1 \Rightarrow h = R$ $h = \frac{v_0^2R}{2gR - v_0^2}$

Maximum height against launch speed (in units of the escape speed $v_2 = \sqrt{2gR}$). For small speeds $h \approx v_0^2/2g$; as $v_0 \to v_2$ the height grows without limit.

Result

\[\boxed{\,h = \frac{v_0^2R_Z}{2gR_Z - v_0^2}\,}\]