Math Detours · Grade 12 · Connections · 1

Escape velocity

Integrate the inverse-square law from the surface to infinity and the energy needed is finite. 11.2 km/s for Earth, and why a black hole follows.

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Tonight's idea

A mass \(m\) at distance \(r\) from the Earth's centre is pulled with force \(F = \dfrac{GMm}{r^2}\), where \(M\) is the Earth's mass and \(G\) the gravitational constant. Lifting it a short distance \(dr\) against that pull takes work \(F\,dr\). To lift it from the surface \(r = R\) all the way out, add up the slices:

R2R3R4Rr → ∞ F = GMm/r² area = GMm/R
Work is area under the force curve. The curve falls like \(1/r^2\), fast enough that the area out to infinity is finite.

\[W = \int_R^{\infty} \frac{GMm}{r^2}\,dr = \left[-\frac{GMm}{r}\right]_R^{\infty} = 0 - \left(-\frac{GMm}{R}\right) = \frac{GMm}{R}.\]

The journey is infinite but the cost is not, because the force dies off fast enough. Had gravity fallen like \(1/r\), the integral would be a logarithm, unbounded, and nothing could ever leave.

The energy comes from speed. Launch at \(v\), ignore air. The kinetic energy \(\tfrac12 mv^2\) is spent against gravity, and to just reach infinity it must equal \(W\):

\(\tfrac12 mv^2 = \dfrac{GMm}{R}\), so \(v_{\text{esc}} = \sqrt{\dfrac{2GM}{R}}\).

The \(m\) cancels: a pebble and a spacecraft need the same speed. Direction does not appear either; energy does not care which way you point. Compare the circular orbit speed \(\sqrt{GM/R}\) from Kepler's laws: escaping costs exactly \(\sqrt2\) times orbiting.

Watch one: the Earth

\(G = 6.674 \times 10^{-11}\) in SI units, \(M = 5.972 \times 10^{24}\) kg, \(R = 6.371 \times 10^{6}\) m. First \(GM = 3.986 \times 10^{14}\). Then \(v_{\text{esc}} = \sqrt{2 \times 3.986 \times 10^{14} / (6.371 \times 10^{6})} = \sqrt{1.251 \times 10^{8}} = 11{,}190\) m/s \(\approx 11.2\) km/s, about 33 times the speed of sound.

A check without \(G\): surface gravity is \(g = GM/R^2 = 9.82\) m/s², so \(2GM/R = 2gR\) and \(v_{\text{esc}} = \sqrt{2 \times 9.82 \times 6.371 \times 10^6} = 11{,}190\) m/s. Same answer, because \(g\) already contains \(GM\).

Your turn

  1. The Moon: \(M = 7.35 \times 10^{22}\) kg, \(R = 1.74 \times 10^{6}\) m. Find its escape velocity in km/s. Gas molecules in a warm atmosphere move at a few hundred m/s, with a tail of much faster ones. Why does this make it hard for the Moon to keep an atmosphere?
  2. Show algebraically that \(v_{\text{esc}} = \sqrt{2gR}\) where \(g = GM/R^2\). Then: a planet has the same surface gravity as Earth but twice the radius. What is its escape velocity?
  3. A projectile leaves the surface straight up at 5.0 km/s, no air. Use \(\tfrac12 v^2 - \dfrac{GM}{R} = -\dfrac{GM}{r_{\max}}\) (energy per kilogram at launch equals energy at the top, where it is momentarily still) to find \(r_{\max}\) and the height above the surface. Compare with the constant-\(g\) estimate \(v^2/2g\). Which is bigger, and why?
  4. (a) Set \(v_{\text{esc}} = c = 3.00 \times 10^{8}\) m/s and solve for \(R\) in terms of \(M\). Compute it for the Sun (\(M = 1.989 \times 10^{30}\) kg) and for the Earth. (b) \(R\) is proportional to \(M\). How does the average density needed, \(M / \tfrac43 \pi R^3\), depend on \(M\)? Which is easier to turn into a black hole, something heavy or something light?

Stretch: leaving the Sun

Escape velocity from the Sun starting at the Earth's orbit (\(r = 1.496 \times 10^{11}\) m) is \(\sqrt{2GM_{\odot}/r}\). Compute it. The Earth already moves at \(\sqrt{GM_{\odot}/r}\) around the Sun. A probe launched in the direction of the Earth's motion keeps that speed for free. How much extra does it need? (Ignore the Earth's own pull.)

Answers

  1. \(v = \sqrt{2 \times 6.674 \times 10^{-11} \times 7.35 \times 10^{22} / 1.74 \times 10^{6}} = 2{,}370\) m/s, about 2.4 km/s, a fifth of Earth's. The fastest molecules in the tail of the speed distribution exceed 2.4 km/s far more often than they exceed 11.2 km/s, so over billions of years the Moon's gas leaks away while Earth's stays.
  2. \(g = GM/R^2\) gives \(GM = gR^2\), so \(2GM/R = 2gR\) and \(v_{\text{esc}} = \sqrt{2gR}\). Doubling \(R\) at fixed \(g\) multiplies \(v_{\text{esc}}\) by \(\sqrt2\): \(11.2 \times 1.414 = 15.8\) km/s.
  3. Per kilogram: \(\tfrac12 (5000)^2 = 1.25 \times 10^{7}\) and \(GM/R = 6.256 \times 10^{7}\). So \(GM/r_{\max} = 6.256 \times 10^{7} - 1.25 \times 10^{7} = 5.006 \times 10^{7}\) and \(r_{\max} = 3.986 \times 10^{14} / 5.006 \times 10^{7} = 7.96 \times 10^{6}\) m. Height \(= 7.96 \times 10^{6} - 6.371 \times 10^{6} = 1{,}590\) km. Constant \(g\) gives \(v^2/2g = 2.5 \times 10^{7}/19.62 = 1{,}270\) km. The true height is bigger because gravity weakens on the way up; the constant-\(g\) formula keeps pulling at full strength.
  4. \(c^2 = 2GM/R\) gives \(R = 2GM/c^2\). Sun: \(2 \times 6.674 \times 10^{-11} \times 1.989 \times 10^{30} / (9.00 \times 10^{16}) = 2{,}950\) m, about 3 km. Earth: \(2 \times 3.986 \times 10^{14} / 9.00 \times 10^{16} = 8.9 \times 10^{-3}\) m, under a centimetre. (b) With \(R \propto M\), the density \(M/R^3 \propto M/M^3 = 1/M^2\). A heavier body needs a lower density to trap light, so very massive objects are the easy ones: a black hole of a hundred million solar masses has roughly the density of water. This Newtonian argument is not the right physics for light, but by a happy accident it gives exactly the radius that Einstein's theory does, the Schwarzschild radius.

Stretch. \(GM_{\odot} = 6.674 \times 10^{-11} \times 1.989 \times 10^{30} = 1.327 \times 10^{20}\). Escape speed at Earth's orbit: \(\sqrt{2 \times 1.327 \times 10^{20} / 1.496 \times 10^{11}} = 42{,}100\) m/s \(= 42.1\) km/s. Earth's orbital speed: \(42.1/\sqrt2 = 29.8\) km/s. Extra needed: \(42.1 - 29.8 = 12.3\) km/s, about the same as leaving Earth itself. That is why probes to the outer planets launch in the direction the Earth is already going.

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