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I'm specifically asked to compute the charge in the Earth-Moon system knowing that the gravitational force between the two bodies is much greater ($\gg$) than the electrostatic one. However, I don't know how many orders of magnitude should I take in order to make a good estimation.

Does anybody know, not just in this specific case, but perhaps in a more general one, how can one quantify these comparisons?

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The most common case where this comes up is when you're dealing with a problem where it's helpful to linearize using a Taylor expansion. For example, a decaying exponential $$ e^{-t/t_0} = 1 - \frac{t}{t_0} + \frac12\left(\frac{t}{t_0}\right)^2 - \frac1{3!}\left(\frac{t}{t_0}\right)^3 + \cdots $$ or a smallish logarithm $$ \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots $$ or the sine of an angle $$ \sin\theta = \theta - \frac1{3!}\theta^3 + \frac{1}{5!} \theta^5 - \frac{1}{7!} \theta^7 + \cdots $$ or an actual binomial, like the acceleration due to gravity near Earth's surface $$ g(h) = \frac{GM_\oplus}{(R_\oplus + h)^2} = \frac{GM_\oplus}{R_\oplus^2} \left( 1 - 2 \frac{h}{R_\oplus} + \frac{(-2)(-3)}{2!} \left( \frac{h}{R_\oplus} \right)^2 + \frac{(-2)(-3)(-4)}{3!} \left( \frac{h}{R_\oplus} \right)^3 + \cdots \right) $$ Now, when we compute a physically interesting quantity, like $$ g\approx 9.806\,65\,\rm m/s^2 \approx 9.81\, m/s^2 \approx 9.8\,m/s^2 \approx 10\,m/s^2, $$ we have a built-in way to indicate the precision that we intend: we truncate the "insignificant" digits from the end of the decimal representation. It's pretty common to limit a result to two or three significant figures, which corresponds to an implied precision of somewhere under 1%.

In the Taylor expansions, there is always some dimensionless parameter which is raised to a different power in every term ($h/R_\oplus$ in the last example). If that dimensionless parameter is smaller than 0.1, then each term in the Taylor expansion corresponds, roughly, to a single significant digit in the final result. And if the dimensionless parameter is smaller than 0.01, you can expect to get your three-ish significant digits from the first two terms in the series! This is why you hear people, when pressed, say that "$\gg$" means something like "different by a factor of ten or more."

For your specific example: we have data on the moon's orbit from lunar laser ranging that tells us the orbit is described by general relativity to a precision of about a centimeter. A one-centimeter shift in the orbit of the moon would change the moon-earth gravitational potential energy by $$ \delta U = GM_\oplus M_\text{moon} \left( \frac{1}{d} - \frac{1}{d+1\,\rm cm} \right) = \frac{GM_\oplus M_\text{moon}}{d} \left( \frac{1\,\rm cm}{d} + \cdots %\mathcal O\left( %\frac{1\,\rm cm}{d} %\right)^2 \right) $$ where $d$ is the average Earth-Moon distance. I think your numerical intuition should confirm that $d\gg 1\,\rm cm$.

In jabirali's answer he shows that the electrostatic and gravitational interactions would have roughly equal strength if the earth-moon system had $\sqrt{Qq}\approx 10^{14}\,\rm C$. By specifying how well we know the earth-moon distance we can put up a better limit: \begin{align} \sqrt{Qq} &\lesssim 10^{14}\,\rm C \frac{1\,cm}{384,400\,km} \\&\lesssim 10^3\,\rm C \end{align} The feebleness of the gravitational force is pretty well-discussed, but I'm actually surprised at how small that is.

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$\gg$ doesn't mean anything about a specific order of magnitude. If you are told to assume $Y \gg x$, what that means is if you have a series expansion of the form $$\sum C_n \biggl(\frac{x}{Y}\biggr)^n\tag{1}$$ you can neglect all subleading terms (i.e. all except the first one with a nonzero coefficient).

For comparison, a simple statement $Y > x$ does not imply that you can neglect all subleading terms in series (1). You would have to evaluate the coefficients, and it may be the case that the series converges so slowly that several consecutive terms have similar magnitudes, like if $\frac{x}{Y} = 0.9$.

It should be clear from this that the precise meaning of $\gg$, in terms of the relative size of $x$ and $Y$, is context-dependent.

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  • $\begingroup$ I guess $x$ and $y$ are interchanged here. I tried to edit, but it was not big enough. $\endgroup$ – pfnuesel Feb 14 '15 at 10:37
  • $\begingroup$ it's not formal, rigorous math, is it? i'd say $x\gg y \Rightarrow x>y$ and nothing more. is it e.g. "false" that $2\gg1$? $\endgroup$ – innisfree Feb 14 '15 at 12:34
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    $\begingroup$ A faculty once told my classmates & I that $x\gg y$ should be read as"$x$ is at least 10 times bigger than $y$." I don't think that's a hard rule written down anywhere, but I would think that some people do interpret $\gg$ to say something about the order of magnitude of the two values. $\endgroup$ – Kyle Kanos Feb 14 '15 at 13:52
  • $\begingroup$ @kyle so arbitrary though. he probably picked 10 because it's one order of magnitude in base 10, but if i like base 2, i might say that $2\gg1$ by one order of magnitude. $\endgroup$ – innisfree Feb 14 '15 at 14:17
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    $\begingroup$ @innisfree: $\gg$ means much greater than. I would not consider 2 to be much greater than 1, even if it's an order of magnitude in base 2. But yes, I would say it's arbitrary. $\endgroup$ – Kyle Kanos Feb 14 '15 at 14:25
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I rather like the accepted answer, but I feel it necessary to offer a basic rule of thumb for using the $\gg$ and $\ll$ signs; something short that you don't have to wade through lots of interesting information to get to. Note that there are situations where what I say here is not exactly true, since there is no specific rules for using these notations other than "One value should be reasonably much greater than the other in the context". But on to the rule of thumb:

When one says $a\gg b$, it is usually safe to take that to mean that $a^2+b^2\approx a^2$ or in other words, $\left(\frac{b}{a}\right)^2\approx0$

It also usually means that when determining the expected order of magnitude of a result, one can safely assume that $a+b\sim a$

There may be cases when $a\gg b$ means $\frac{b}{a}\approx0$, but as a general guideline, what I stated above is the safest bet.

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Given the context of your question, I think the point is to use the observational fact that the gravitational attraction between the Earth and the Moon dominates over the electrostatic interactions to create an upper limit on the net charges of the objects.

Newtons law of gravitation can be written as follows, where capital $M$ is the mass of the Earth, lowercase $m$ is the mass of the moon, and $r$ is the distance inbetween: $$ F_g = G\frac{Mm}{r^2} $$ Coulombs law of electrostatics can be written like this, where capital $Q$ is the net charge of the Earth, and lowercase $q$ is the charge of the Moon: $$ F_e = K\frac{Qq}{r^2} $$ From observations, we know that $F_e \ll F_g$, which using the equations above and solving for $Qq$ becomes: $$ Qq \ll \frac{G}{K} Mm $$ Using the approximate values for Newtons gravitational constant $G \approx 10^{-10}~\text{Nm}^2/\text{kg}^2$, the Coulomb constant $K \approx 10^{10}~\text{Nm}^2/\text{C}^2$, the Earth mass $M \approx 10^{25}~\text{kg}$, and Lunar mass $m \approx 10^{23}~\text{kg}$, we get: $$ Qq \ll 10^{28}~\text{C}^2$$ So the geometric mean of the charges of the Earth and the Moon has the upper limit: $$ \sqrt{Qq} \ll 10^{14}~\text{C}$$ So we have now obtained an upper bound on the possible charges that the Earth and Moon can have, which is what I believe your assignment wanted you to do. I have no idea exactly how much lower than this limit the net charges of the Earth and Moon are, but I suspect that Google might reveal it if you're interested :).

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