Irodov 1.221: Approaching Escape Speed from the Pole

Problem Irodov 1.221

A small body is launched straight upward from the Earth’s pole with initial speed $v_0$ , and thereafter moves under the Earth’s gravitational pull alone, air resistance neglected. Given the Earth’s radius $R$ and the free-fall acceleration $g$ at its surface, find the maximum height $h$ above the surface that the body reaches, the point where its upward speed momentarily falls to zero.

The body leaves the pole radially, along the local vertical, with speed v_0; launching from the pole keeps the whole flight on this one radial line, with no sideways deflection to track. — The body leaves the pole radially, along the local vertical, with speed $v_0$ ; launching from the pole keeps the whole flight on this one radial line, with no sideways deflection to track.

Given

$v_0$
$R$
$g$

Find

h

Solution

Step 1The turning point from energy conservation

Once the body leaves the pole the Earth’s attraction is the only force on it, air drag being neglected, and gravity is conservative, so the mechanical energy of the body is the same at every instant of the flight. Launching from the pole puts the whole path on the rotation axis, where the centrifugal term vanishes and the Coriolis term vanishes too because the velocity is parallel to the rotation axis, so the Earth’s rotation never enters. Equating the energy at launch (distance $R$ from the center, speed $v_0$ ) to the energy at the highest point (distance $R+h$ , speed zero), with the gravitational potential energy $U=-\gamma M m/r$ , and then dividing through by the mass:

\begin{align*}\frac{1}{2}mv_0^2-\frac{\gamma M m}{R} &= 0-\frac{\gamma M m}{R+h} \\ \frac{1}{2}v_0^2-\frac{\gamma M}{R} &= -\frac{\gamma M}{R+h} \\ \frac{v_0^2}{2} &= \frac{\gamma M}{R}-\frac{\gamma M}{R+h}\end{align*}

\frac{v_0^2}{2} = \gamma M\left(\frac1R-\frac1{R+h}\right)

(1)

Step 2Write

\gamma M

through the surface value of

g

The surface value of $g$ fixes the product $\gamma M$ , so the energy relation can be written in the two quantities the problem supplies.

g = \frac{\gamma M}{R^2} \;\Longrightarrow\; \gamma M = gR^2

Step 3Solve for the height

Substituting $\gamma M=gR^2$ into (1) and combining the two fractions over the common denominator $R(R+h)$ isolates $h$ .

\begin{align*}\frac{v_0^2}{2} &= gR^2\left(\frac1R-\frac1{R+h}\right) \\ &= gR^2\cdot\frac{h}{R(R+h)} \\ &= \frac{gRh}{R+h}\end{align*}

Clearing the denominator and collecting the terms in $h$ ,

\begin{align*}v_0^2(R+h) &= 2gRh \\ v_0^2 R &= h(2gR-v_0^2)\end{align*}

h = \frac{v_0^2 R}{2gR-v_0^2} = \frac{R}{2gR/v_0^2-1}

(2)

Check

As $v_0\to\sqrt{2gR}$ the denominator $2gR-v_0^2\to0$ and $h\to\infty$ : the body never comes back down, exactly the surface escape speed $v_{\mathrm{esc}}=\sqrt{2gR}$ found independently from $v_0^2/2=\gamma M/R$ . For $v_0>\sqrt{2gR}$ the formula gives a negative $h$ , correctly signaling that no finite turning point exists.

Remark

For $v_0^2\ll2gR$ , a launch speed far below escape speed, dividing the boxed result through by $2gR$ and expanding $(1-x)^{-1}\approx1+x$ gives $h\approx(v_0^2/2g)\,(1+v_0^2/2gR)$ : the familiar constant-gravity result $h=v_0^2/2g$ is the leading term, and the correction term is the first sign that gravity has weakened by the time the body has climbed a height comparable to $R$ itself.