Irodov 1.230: From Circular Orbit to Escape Speed – Amit & Deepali Sinha Foundation

Problem Irodov 1.230

A spaceship is placed in a circular orbit that grazes the Earth’s surface (orbital radius equal to the Earth’s own radius $R$ , atmospheric drag ignored). In terms of the Earth’s surface free-fall acceleration $g$ and radius $R$ , find the additional speed $\Delta v$ that must be added to the ship’s orbital speed so that it just breaks free of the Earth’s gravitational pull, reaching arbitrarily large distance with vanishing left-over speed.

The grazing circular orbit, its radius the Earth's own, carrying the ship at orbital speed v_1. The dashed extension Δ v continues along the same tangential direction, raising the speed to v_2=v_1+Δ v. — The grazing circular orbit, its radius the Earth’s own, carrying the ship at orbital speed $\vect v_1$ . The dashed extension $\Delta\vect v$ continues along the same tangential direction, raising the speed to $v_2=v_1+\Delta v$ .

Given

Find

\Delta v

Solution

Step 1Circular orbit speed at the surface

For the ship moving on a circle that grazes the surface, gravity alone supplies the centripetal force; writing the Earth’s pull as $\gamma M/R^2=g$ trades the unknown product $\gamma M$ for the two measured surface quantities.

\begin{align*}\frac{\gamma Mm_s}{R^2} &= \frac{m_s v_1^2}{R} \\ v_1^2 &= \frac{\gamma M}{R} = gR\end{align*}

v_1=\sqrt{gR}

(1)

Step 2Escape speed at the surface

\begin{align*}\frac{1}{2}m_s v_2^2 - \frac{\gamma Mm_s}{R} &= 0 \\ v_2^2 &= \frac{2\gamma M}{R} = 2gR\end{align*}

v_2=\sqrt{2gR}

(2)

Step 3The extra speed needed

$\Delta v$ is the increase from the circular speed (1) to the escape speed (2); the boost is taken parallel to $\vect v_1$ , the cheapest direction, so only magnitudes enter. Factoring out $\sqrt{gR}$ before substituting numbers keeps the algebra and the arithmetic separate.

\Delta v = v_2-v_1 = \sqrt{2gR}-\sqrt{gR} = \sqrt{gR}\left(\sqrt2-1\right)

With $g=9.8\ \text{m/s}^2$ and $R=6.37\times10^6\ \text{m}$ ,

\sqrt{gR} = \sqrt{6.24\times10^7} = 7.90\times10^3\ \text{m/s}.

\Delta v = \sqrt{gR}\left(\sqrt2-1\right) \approx 3.27\ \text{km/s}

(3)

Check

Dimensionally $[g][R]=(\text{m/s}^2)(\text{m})=\text{m}^2/\text{s}^2$ , and its square root is a speed, consistent with both $v_1$ and $v_2$ . The fraction $\Delta v/v_1=\sqrt2-1\approx0.414$ is a modest boost, well under doubling the speed, matching the everyday fact that reaching escape from a low parking orbit costs far less than launching from rest at the surface.

Remark

The ratio $v_2/v_1=\sqrt2$ holds at any circular orbit radius $r$ , not only one grazing the surface: both $v_1(r)=\sqrt{\gamma M/r}$ and $v_2(r)=\sqrt{2\gamma M/r}$ share the same $1/\sqrt r$ dependence, so their ratio cancels $r$ entirely. A spacecraft parked in a circular orbit at any altitude needs the same fractional speed increase, $\sqrt2-1$ , to escape from that altitude.