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I'm mostly having a problem incorporating the approximation. My trouble is towards the bottom of the post, but I wrote down all of my steps just in case I made an error somewhere.

So in this question 3 balls start off with zero velocity and begin to fall towards the earth. They're vertically aligned, with ball A in the middle, ball B 22m directly above A, and ball C 22m directly below A.

The question asks me to find the magnitude of the acceleration of ball B and C with respect to A. The question also says to use the binomial approximation because otherwise when you numerically compute the magnitude your calculator will not store enough sig figs to give you an accurate result.

So far I did:

$F = ma $

The acceleration for ball A is $ \vec{a}_A = \frac{F_A}{m_A} $, and ditto for balls B and C

Force due to gravity is $ F_B = \frac{G M_e m_B}{r_B^2}$ where $M_e$ is the mass of the earth, and $r_B$ is the radius between the center of the earth and the center of ball B.

Now the acceleration due to the earth on ball B is $ \vec{a}_B = \frac{G M_e}{r_B^2}$

We want the acceleration of B with respect to A so we can write

$ \vec{a}_B - \vec{a}_A = \frac{G M_e}{r_B^2} - \frac{G M_e}{r_A^2} $

We can rewrite $r_B = r_A + 22m$

So $ \vec{a}_B - \vec{a}_A = \frac{G M_e}{(r_A + 22m)^2} - \frac{G M_e}{r_A^2} $

$ \vec{a}_B - \vec{a}_A = G M_e \left[\frac{1}{(r_A + 22m)^2} - \frac{1}{r_A^2}\right] $


Now this is where I'm lost. I don't know how to incorporate the approximation $ (1+x)^n \approx 1+nx $

$ (r_A + 22m)^{-2}$ doesn't follow the $(1+x)^n$ format

I put the expression $\frac{1}{(r + x)^2}$ into wolfram to see if there are any alternate forms of the expression that I can apply to the approximation and I couldn't find any. I also tried to factor out $\frac{1}{r_A^2}$ to get

$ \frac{G M_e}{r_A^2} \left[\frac{r_A^2}{(r_A + 22m)^2} - 1 \right] $ But this doesn't work with the approximation either.

If anyone could point me in the right direction or give me a hint it would be very much appreciated

Edit: The question is P1.5 from Moore's General Relativity Workbook

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When you have an expression of the form $(a + b)^n$, where $|a| \gg |b|$, you can rewrite it as $$(a + b)^n = \left(a\left(1+\frac ba\right)\right)^n = a^n \left(1+\frac ba\right)^n$$ and then, given that $\left|\frac{nb}a\right| \ll 1$ (and $a + b > 0$), apply the binomial approximation to obtain $$a^n \left(1+\frac ba\right)^n ≈ a^n \left(1 + n\frac ba\right).$$


In your example $r_A ≈ 6371\,{\rm km} \gg 22\,{\rm m}$, so we can write $$\frac1{(r_A + 22\,{\rm m})^2} = (r_A + 22\,{\rm m})^{-2} = r_A^{-2}\left(1 + \frac{22\,{\rm m}}{r_A}\right)^{-2} ≈ r_A^{-2}\left(1 - 2 \cdot \frac{22\,{\rm m}}{r_A}\right)$$ and then subtract $r_A^{-2}$ to obtain $$\frac1{(r_A + 22\,{\rm m})^2} - \frac1{r_A^2} ≈ r_A^{-2}\left(1 - 2 \cdot \frac{22\,{\rm m}}{r_A}\right) - r_A^{-2} = - 2 \cdot 22\,{\rm m} \cdot r_A^{-3}.$$

Multiply this by $GM_e$, and you'll have your answer.

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    $\begingroup$ Thanks for the info about when the binomial approximation is appropriate! I really appreciate you going in depth about that. I kept trying to take out the 22m, when I should have been trying to take out the $r_a$, like you said. Thanks man! $\endgroup$
    – M-B
    Nov 18 '21 at 20:29
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Just do $$\frac{1}{(r_A + 22)^2}=\frac{1}{ (22)^2 (\frac{r_A}{22} + 1)^2}$$ Now you have a term $$(1+\frac{r_A}{22})^{-2}\approx 1-\frac{r_A}{11}$$

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    $\begingroup$ In the first line, $\sqrt{22}$ should be replaced by $22^2$. $\endgroup$
    – Mark H
    Nov 18 '21 at 7:50
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    $\begingroup$ Thanks for spotting that. $\endgroup$
    – joseph h
    Nov 18 '21 at 7:58
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    $\begingroup$ Isn't this also backwards (and also missing a unit)? $r_A \gg 22\,{\rm m}$, so the binomial approximation isn't valid as you've written it. $\endgroup$ Nov 18 '21 at 15:57
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For Binomial approximation, you need take out the $22m$. So:

$$(r_A + 22m)^{-2} = \frac{1}{(r_A + 22m)^2} = \frac{1}{(22m)^2(\frac{r_A}{22m}+1)^2}$$

$$\therefore \ \ \ \text{binomial expansion:} \ \ \ (1+x)^n \approx 1 + nx$$ Also:

$$(1 + a x)^n \approx 1 + n a x \ \ \ \ \ \text{where} \ \ \text{a} \ \ \text{is} \ \ \text{constant}$$

We have:

$$\text{Constant} \times (1 + ax)^n = \frac{1}{(22m)^2}\left(\frac{r_A}{22m}+1\right)^{-2} \approx \frac{1}{(22m)^2} \left( 1 - 2 \frac{r_A}{22m} \right) \ \ , \ \text{where} \ \ n = -2$$

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    $\begingroup$ Thank you so much! I can't believe it didn't occur to me to take out the 22m. Kudos $\endgroup$
    – M-B
    Nov 18 '21 at 1:25
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    $\begingroup$ Wait, I think there is a typo in your solution. Instead of $\frac{1}{\sqrt{22}}$ shouldn't it be $\frac{1}{22^2}$ ? $\endgroup$
    – M-B
    Nov 18 '21 at 5:13
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    $\begingroup$ @M-B You are right about $22^2$. $\endgroup$
    – Mark H
    Nov 18 '21 at 7:51
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    $\begingroup$ To make this approach more general if you have some condition like $r\ll R$ you can rewrite it as $\frac{r}{R}\equiv x\ll 1$. If you have a function $f(r,R)$ that you want to approximate you can replace each $r$ with $xR$ and finally you can Taylor expand in $x$ because $x$ is a small parameter (while taking $R$ constant) $\endgroup$ Nov 18 '21 at 12:02
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    $\begingroup$ Isn't this backwards? The binomial approximation is valid for $x$ close to $0$, but in this case $\frac{r_A}{22\,{\rm m}} \gg 1$. $\endgroup$ Nov 18 '21 at 15:56

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