A climber stands on the top of Mount Everest, \(8.85\,\text{km}\) above sea level. The Earth's pull there is weaker, because the climber is farther from the Earth's centre. The chapter Gravitational Force and Field gives \(g_h = gR^2/(R + h)^2\), with \(R = 6400\,\text{km}\). Put in the numbers and you must square \(6408.85\) to find a change in the third figure of \(g\). There is a shorter way.
Powers of a number close to one
Square a number a little larger than one: \((1 + x)^2 = 1 + 2x + x^2\). With \(x = 0.01\), this is \(1 + 0.02 + 0.0001 = 1.0201\). The last term is two hundred times smaller than the middle one. Cube it: \((1 + x)^3 = 1 + 3x + 3x^2 + x^3\), and again every term after \(3x\) is tiny. When \(x\) is small, the powers of \(x\) beyond the first can be dropped. That is the binomial approximation.
If \(x\) is much smaller than one in size, then \((1 + x)^n \approx 1 + nx\) for any real power \(n\). The line \(1 + nx\) is the tangent to the curve \((1 + x)^n\) at \(x = 0\), so the two agree closely near \(x = 0\) and part as \(x\) grows.
The rule holds for every power, not only for whole numbers. The proof uses the derivative of a power, from the chapter Differentiation. Near \(x = 0\), a smooth curve is very close to its tangent line.
\[ y = (1 + x)^n, \qquad y(0) = 1 \]
Why. the curve passes through the point \((0, 1)\) for every power \(n\).
\[ \dfrac{dy}{dx} = n(1 + x)^{n - 1} \]
Why. the power rule, applied to \(1 + x\), whose own derivative is \(1\).
\[ \left.\dfrac{dy}{dx}\right|_{x = 0} = n \]
Why. put \(x = 0\): any power of \(1\) is \(1\).
\[ y \approx 1 + nx \]
Read it. the tangent at \((0, 1)\) has slope \(n\). Close to \(x = 0\), the curve and its tangent agree.
A few forms appear so often that they are worth knowing: \(\sqrt{1 + x} \approx 1 + x/2\), \(1/(1 + x) \approx 1 - x\), \(1/(1 + x)^2 \approx 1 - 2x\) and \(1/\sqrt{1 - x} \approx 1 + x/2\). Each is the rule with \(n = \tfrac12\), \(-1\), \(-2\) and \(-\tfrac12\). For example, \(\sqrt{1.02} \approx 1.01\), and the true value is \(1.00995\).
Use the binomial approximation to find \(1/\sqrt{1 - 0.02}\).
How large is the error?
The approximation drops everything after \(nx\). The first term dropped is about \(n(n - 1)x^2/2\), and it measures the error. The error grows as \(x^2\): make \(x\) ten times smaller and the error falls a hundred times. For \(n = -2\), the error is about \(3x^2\). That is \(0.0003\) at \(x = 0.01\), \(0.03\) at \(x = 0.1\), and \(0.75\) at \(x = 0.5\).
Move the slider in the figure above and watch the gap. Near \(x = 0\) the curve and the line can hardly be told apart. Beyond about \(|x| = 0.1\) they part quickly. For \(n = \tfrac12\) the two stay close for longer, because \(n(n - 1)/2 = -0.125\) is small.
Bring the expression into shape
The rule needs the form \((1 + \text{small})^n\), with a leading one. Most expressions of physics must first be rearranged into it. Take out the large quantity as a factor. For example, \((R + h)^{-2} = R^{-2}(1 + h/R)^{-2}\), and now \(x = h/R\) is small. In the same way \((2 + 0.02)^3 = 2^3(1 + 0.01)^3 \approx 8 \times 1.03 = 8.24\). Applying the rule to \((2 + 0.02)^3\) directly, as if 2 were 1, would be wrong.
Use the binomial approximation to find by how much \(\sqrt{100.6}\) exceeds \(10\). Take out \(\sqrt{100}\) as a factor first.
Back to Everest
Now the climber's problem takes one line. Write \(g_h = gR^2/(R + h)^2 = g(1 + h/R)^{-2}\). With \(n = -2\) and \(x = h/R\), the rule gives \(g_h \approx g(1 - 2h/R)\). The fractional fall in \(g\) is simply \(2h/R\). The figure below draws the exact curve and the binomial line together.
Mount Everest is \(8.85\,\text{km}\) high. Take \(g = 9.8\,\text{m/s}^2\) at the surface and \(R = 6400\,\text{km}\). Find \(g\) at the top.
\(g_h = g(1 + h/R)^{-2}\), with \(x = h/R = 8.85/6400 = 0.00138\), far below \(0.1\).
\(g_h \approx g(1 - 2h/R)\), with \(2h/R = 0.0028\).
\(g_h \approx 9.8 \times (1 - 0.0028) = 9.8 - 0.027 = 9.77\,\text{m/s}^2\).
On the top of Everest \(g \approx 9.77\,\text{m/s}^2\), less than a third of one per cent below its value at sea level.
Expansion, pendulums and relativity
The same rule turns up wherever a quantity changes by a small fraction. A square plate whose sides each grow by the small fraction \(\alpha\Delta T\) has an area \(A(1 + \alpha\Delta T)^2 \approx A(1 + 2\alpha\Delta T)\). So area expands with \(\beta = 2\alpha\), and a volume with \(\gamma = 3\alpha\). A pendulum's period grows as \(\sqrt{l}\), so a rod warmer by \(\Delta T\) lengthens the period by the fraction \(\tfrac12\alpha\Delta T\). A brass rod \(20\,\text{K}\) too warm, with \(\alpha = 1.9 \times 10^{-5}\,\text{K}^{-1}\), makes a clock lose \(\tfrac12 \times 1.9 \times 10^{-5} \times 20 \times 86\,400 = 16.4\,\text{s}\) a day.
In relativity, the energy of a body moving at speed \(v\) is \(\gamma mc^2\), with \(\gamma = (1 - v^2/c^2)^{-1/2}\). Here \(\gamma\) is the factor of relativity, not the expansion coefficient. For a slow body, \(v^2/c^2\) is tiny, so \(\gamma \approx 1 + \tfrac12 v^2/c^2\). The kinetic energy \((\gamma - 1)mc^2\) is then \(\tfrac12 mv^2\), the familiar result of Newton's mechanics.
Take \(g = 9.8\,\text{m/s}^2\) at the surface and \(R = 6400\,\text{km}\). Use the binomial approximation to find by how much \(g\) falls at a height of \(64\,\text{km}\), in m/s².
A steel plate, \(\alpha = 1.2 \times 10^{-5}\,\text{K}^{-1}\), is heated through \(100\,\text{K}\). By what percentage does its area grow?
A student uses \(g_h \approx g(1 - 2h/R)\) for a satellite at \(h = R/2\) and gets \(g_h = 0\). What went wrong?
The binomial rule is an approximation for small \(x\) only. Past \(|x| \approx 0.1\) its error reaches a few per cent. At \(x = 0.5\) it can be worse than useless: the line for \(g\) reaches zero at \(3200\,\text{km}\). Check the size of \(x\) first. Also check the form: the rule applies to \((1 + x)^n\), never to \((a + x)^n\) until the \(a\) has been taken out as a factor.
The binomial approximation turns an exact but awkward formula into a small correction. It gives \(g\) above the surface and the energy \(mgh\) of lifting a body near the ground, from the exact law of gravitation. It gives the expansion of areas and volumes, a density that falls with temperature, and a pendulum clock that loses time in summer. Newton's law of cooling comes from Stefan's law for a small difference of temperature. In elasticity, it gives the change of density under pressure and Poisson's ratio. The field of a distant dipole uses it too. So do the kinetic energy of relativity at low speed and the recoil of an atom that emits light.
- Using the rule when \(x\) is not small.Check \(|x|\) first. Below about \(0.1\) the rule is good; at \(0.5\) use the exact formula.
- Applying the rule to \((a + x)^n\) with \(a \ne 1\).Take \(a\) out first: \((a + x)^n = a^n(1 + x/a)^n\). Then \(x/a\) is the small number.
- Losing the sign of \(n\) or of \(x\).For \((1 - x)^{-1/2}\), \(n = -\tfrac12\) and the small number is \(-x\), so \(nx = +x/2\).