Escaping the Solar System

Problem

Escaping the Solar System. To send a probe out of the Solar System it must climb out of two gravitational wells: the Earth’s and the Sun’s. Estimate the minimum speed the probe must have on leaving the Earth’s surface, neglecting Earth’s rotation and planetary perturbations. Compare it with the “Earth-only” escape speed ($\sim 11{,}2\ \text{km/s}$): why is it larger than one might naively expect from simply adding the two?

Solution

The two wells. The probe must first climb out of Earth’s well and then out of the Sun’s. Earth’s escape speed at the surface is $v_{f,T} \approx 11{,}2\ \text{km/s}$.

Escape from the Sun at Earth’s distance. With $GM_\odot \approx 1{,}33\cdot 10^{20}\ \text{m}^3/\text{s}^2$ and $r_T = 1{,}5\cdot 10^{11}\ \text{m}$: $v_{f,\odot} = \sqrt{\frac{2GM_\odot}{r_T}} = \sqrt{\frac{2\cdot 1{,}33\cdot 10^{20}}{1{,}5\cdot 10^{11}}} \approx \sqrt{1{,}77\cdot 10^{9}} \approx 42{,}1\ \text{km/s}$

Exploiting Earth’s orbital motion. Earth already moves at $v_\oplus \approx 29{,}8\ \text{km/s}$ around the Sun. Launching the probe in the direction of Earth’s motion, relative to the Sun only the excess is needed: $v_\infty = v_{f,\odot} - v_\oplus \approx 42{,}1 - 29{,}8 = 12{,}3\ \text{km/s}$

This $v_\infty$ is the residual speed the probe must retain, relative to Earth, after freeing itself from Earth’s well.

Combining the two wells (energies, not speeds). Conservation of energy at launch requires that the initial kinetic energy cover both the escape from Earth and the residual energy $v_\infty$: $\tfrac{1}{2}v_\text{launch}^2 = \tfrac{1}{2}v_{f,T}^2 + \tfrac{1}{2}v_\infty^2 \quad\Longrightarrow\quad v_\text{launch} = \sqrt{v_{f,T}^2 + v_\infty^2}$

$v_\text{launch} = \sqrt{11{,}2^2 + 12{,}3^2} = \sqrt{125{,}4 + 151{,}3} = \sqrt{276{,}7}$

$\ev{\Delta v \sim 16{,}6\ \text{km/s} \ \text{from the Earth's surface}}$

Why greater than the simple $11{,}2\ \text{km/s}$? Because it’s not enough to leave Earth: enough extra energy must also be supplied to climb the Sun’s well. Speeds do not add linearly: kinetic energies ($v^2$) add, hence $v_\text{launch} = \sqrt{v_{f,T}^2 + v_\infty^2}$. The result $16{,}6\ \text{km/s}$ is larger than Earth escape alone, but is still far smaller than $11{,}2 + 42{,}1$ km/s: by exploiting the $29{,}8\ \text{km/s}$ of Earth’s orbital motion, most of the energy needed to escape the Sun is “gifted” to the probe (Povey 2015, §10.10).

Links

Topics: Gravitation Concepts: Escape velocity · Conservation of mechanical energy · Orbits Skills: Fermi estimates · Conservation of energy