Irodov 1.229: Braking a Grazing Parabola into a Circle

Problem Irodov 1.229

A spaceship coasts in toward the Moon (mass $M$ , radius $R$ ) along a parabolic trajectory whose point of closest approach lies essentially at the Moon’s own surface. At that instant of closest approach, a brief rocket burn transfers the ship onto a circular orbit of radius equal to the Moon’s radius. Find the change $\Delta v$ in the ship’s speed produced by that burn.

The parabolic approach (solid) grazes the Moon M at closest approach, where the burn transfers the ship to a circular orbit of the Moon's own radius, drawn on the rim. v_1 is the speed before the burn, v_2 (dashed) after. — The parabolic approach (solid) grazes the Moon $M$ at closest approach, where the burn transfers the ship to a circular orbit of the Moon’s own radius, drawn on the rim. $\vect v_1$ is the speed before the burn, $\vect v_2$ (dashed) after.

Given

parabolic approach with closest-approach distance $\approx R$
transfer to a circular orbit of radius $R$

Find

the change

\Delta v

in speed during the braking burn

Solution

Step 1Speed on the parabola at closest approach

A parabolic trajectory has exactly zero total mechanical energy, so at any distance $r$ from the Moon’s center the kinetic energy equals the magnitude of the potential energy; at the point of closest approach $r=R$ .

\begin{align*} \frac12 m v_1^2 - \frac{\gamma Mm}{R} &= 0 \\ v_1^2 &= \frac{2\gamma M}{R}. \tag{1}\end{align*}

v_1 = \sqrt{\dfrac{2\gamma M}{R}}

(2)

Step 2Speed on the new circular orbit

Immediately after the burn, gravity alone supplies the centripetal force for the circular orbit of radius $R$ .

\frac{\gamma Mm}{R^2} = \frac{mv_2^2}{R} \;\Longrightarrow\; v_2^2 = \frac{\gamma M}{R}. \tag{3}

v_2 = \sqrt{\dfrac{\gamma M}{R}}

(4)

Step 3The change in speed

The burn happens at one instant at the same point, so $\Delta v$ is simply the new speed (4) minus the old one (2); factoring out the common $\sqrt{\gamma M/R}$ isolates the numerical factor.

\begin{align*} \Delta v &= v_2 - v_1 \\ &= \sqrt{\frac{\gamma M}{R}} - \sqrt{\frac{2\gamma M}{R}} \\ &= \sqrt{\frac{\gamma M}{R}}\left(1-\sqrt2\right). \tag{5}\end{align*}

Substituting the Moon’s mass $M$ and radius $R$ gives the common factor:

\gamma M = 4.90\times10^{12}\ \mathrm{m^3/s^2}, \qquad \sqrt{\frac{\gamma M}{R}} = 1.68\ \mathrm{km/s}.

\Delta v = \sqrt{\dfrac{\gamma M}{R}}\left(1-\sqrt2\right) \approx -0.70\ \mathrm{km/s}

(6)

Check

The boxed $1.68$ km/s is exactly the Moon’s own circular (first cosmic) speed, and $v_1=\sqrt2\,v_2\approx2.37$ km/s its escape speed at $R$ ; the burn must remove the fraction $1-1/\sqrt2\approx0.293$ of that escape speed, $0.293\times2.37\ \mathrm{km/s}\approx0.70\ \mathrm{km/s}$ , matching the boxed magnitude by a route independent of the algebra that derived it.

Remark

The ratio $v_2/v_1 = 1/\sqrt2$ holds at any $M$ and $R$ whatsoever, since both speeds are the same $\sqrt{\gamma M/R}$ times a pure number: the circular speed at a body’s surface is always a factor $\sqrt2$ below the speed of a trajectory that just grazes that surface with zero total energy. A braking burn that circularises a grazing parabolic approach therefore always removes the same fraction, $1-1/\sqrt2\approx29\%$ , of the incoming speed, whichever world is being orbited.