How did the approximation $\left(1+\frac{h}{R}\right)^{-2} \approx 1-\frac{2h}{R}$ come about?

Question

In physics there is a formula calculate the variation in the acceleration due to gravity at a height, it is given by:

$$g=g_0\left(1+\frac{h}{R}\right)^{-2}$$

But it is approximated as $$g=g_0\left(1-\frac{2h}{R}\right)$$

Some how the terms in the brackets are almost equal could anyone show why ?

Answer 1

For small $\frac hR$, ignore the terms of $(\frac hR)^2$ below

$$\left(1+\frac{h}{R}\right)^{-2} = \frac1{1+\frac{2h}{R}+( \frac{h}{R})^2} \approx \frac1{1+\frac{2h}{R}} = \frac{1-\frac{2h}{R}}{1-(\frac{2h}{R})^2} \approx 1-\frac{2h}{R} $$

Answer 2

Even without knowing the binomial series or approximations, this can be shown via geometric series: recall $$f(z) = \frac{1}{1-z} = 1 + z + z^2 + \cdots, \quad |z| < 1,$$ so that $$\frac{1}{(1-z)^2} = (1 + z + z^2 + \cdots )^2 = 1 + 2z + 3z^2 + \cdots.$$ Then with the choice $z = -h/R$, we have $$\left(1 + \frac{h}{R}\right)^{-2} = 1 - \frac{2h}{R} + \frac{3h^2}{R^2} - \cdots.$$ The approximation improves as $h/R \to 0$.

Answer 3

Reacall binomial expansion for negative index as $$(1+x)^{-2}=1-2x+\frac{-2(-3)}{2!}x^2+\frac{-2(-3)(-4)}{3!}x^3+...$$ When $x<<1$, then by retaining first two terms we wet the approximation: $$(1+x)^{-2}\approx 1-2x.$$

Answer 4

I suppose this is not inherently different from the other answers posted, but you can get this from the simple Taylor expansion: $$ f(x) \approx f(0) + x f'(0),\qquad\text{where}\qquad f(x)= (1 + x)^{-2}. $$ This gives $f(x) \approx 1 - 2x$ or, with $x = h/R$, $$ \left(1 + \frac h R\right)^{-2} \approx 1 - \frac{2h}{R}. $$ By Taylor's theorem with remainder, the error is at most $3h^2 / R^2$ (i.e., the quadratic term in the series expressions give by other users).

Answer 5

We want to show that for small $|x|$, $$(1+x)^{-2}\approx 1-2x$$ Equivalently, $$(1+x)^2(1-2x)\approx 1$$ Now, $$\begin{align} &(1+x)^2(1-2x)\\ &=(1+2x+x^2)(1-2x)\\ &=1+2x+x^2-2x-4x^2-2x^3\\ &=1-3x^2-2x^3\\ \end{align}$$

When $x$ is sufficiently small, we can ignore the $x^2$ and $x^3$ terms.

Also, $$\frac1{(1+x)^2} = \frac{1-2x}{1-3x^2-2x^3}$$ Once again, for small $x$ we can ignore the higher powers of $x$, so the denominator of the RHS is very close to $1$.

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If we look at the graphs of these functions, we can see that the line $y=1-2x$ is tangent to the curve $y=(1+x)^{-2}$ at $(0, 1)$.

So the slope of the curve at $(0, 1)$ is $-2$. When you learn calculus, you will learn how to find the slope of a tangent at a point on a curve. But here's a "sneak preview". Let's find the slope $m$ of the line through $(0, 1)$ and $(x, 1/(1+x)^2)$ for some small $x$. $$m=\frac{1-1/(1+x)^2}{x}$$ $$m=\frac{-2-x}{(1+x)^2}$$ As $x$ gets close to $0$, $m$ gets close to $-2$.

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FWIW, we can find the series

$$\frac1{(1+x)^2}=1-2x+3x^2-4x^3+5x^4-6x^5+\dots$$ which is valid for $|x|<1$, using polynomial long division.

$$\begin{align} &\underline{1-2x+3x^2-\dots}\\ 1+2x+x^2)&1\\ &\underline{1+2x+x^2}\\ &-2x-x^2\\ &\underline{-2x-4x^2-2x^3}\\ &3x^2+2x^3\\ &\underline{3x^2+6x^3+3x^4}\\ &-4x^3-3x^4\\ \end{align}$$ Etc.

And we can verify it by multiplication:

$$\begin{array}{r|rrrrrrr} 1 & 1 & -2x & +3x^2 & -4x^3 & +5x^4 & -6x^5 & \dots\\ 2x & & 2x & -4x^2 & +6x^3 & -8x^4 & +10x^5 & \dots\\ x^2 & & & x^2 & -2x^3 & +3x^4 & -4x^5 & \dots\\ \end{array}$$