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My problem:

How many orders of magnitude larger is the sun $(1.99$ x $10^{30} kg)$ than the moon $(7.32$ x $10^{22} kg)$.

Justification 1:

So according to my teacher, we can do this question by looking at which one is closer to ten and which one is closer to zero. In this case, he said $7.32$ is closer to ten so, he rounded the order of magnitude up to $10^{23} kg$. And then use $30 - 23 = 7$.

But what happens if the number is like $5$ x $10^{22} kg$, when its equally close to zero and ten?

Justification 2:

However, my previous teacher said otherwise. He said if anything that is greater than $3.16$ we round the order of magnitude up, anything that is less than $3.16$ the order of magnitude remains the same.

So my question is which approach is better ( i say #2) or which one is more correct than other? Also, please answer what happens if the number is like $5$ x $10^{22} kg$, when its equally close to zero and ten?

Thank you.

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  • $\begingroup$ "Closer" means "relative to 3.16", so both methods are the same. Have you considered logarithms? $\endgroup$ – safesphere Feb 1 at 8:31
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There isn't a hard and fast answer to this because order of magnitude estimates are inevitably approximations. Given this it isn't a productive use of a physicist's time to worry about the exact value of an estimate.

Having said this, the question is basically asking about the value of the fraction $M_{sun}/M_{moon}$. I wouldn't round the two masses separately and then take the fraction then round the result. I would do the rounding only at the final stage i.e. after the division. So I'd get for the ratio $R$:

$$ R = \frac{1.99 \times 10^{30}}{7.32 \times 10^{22}} = 2.72 \times 10^7 $$

And that gives seven orders of magnitude.

One way to get the orders of magnitude is to take the $\log_{10}$ of the ratio, which in this case gives us:

$$ \log_{10}(R) \approx 7.43 $$

then the part to the left of the decimal point gives the exponent ($7$ in this case) and the fractional part is the log of the number i.e. $0.43 \approx \log_{10}(2.72)$. This neatly explains where that number $3.16$ comes from because it is $\sqrt{10}$ i.e. $\log_{10}(3.16) = 0.5$. So for example:

$$ \log_{10}(3.16 \times 10^7) = 7.5 $$

So below $3.16$ we would round down to $7$ and above $3.16$ we would round up to $8$. If our number is exactly $3.16 \times 10^n$ then you can round up or down as you wish. Given that we are using orders of magnitude as an approximate description it really doesn't matter whether you round up or down.

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  • $\begingroup$ Exactly! Sometimes I am happy when my result is within an order of magnitude of the hypothetical case. $\endgroup$ – Mick Feb 1 at 8:58
  • $\begingroup$ Thank you a lot Mr.Rennie! But before I accept this answer, can you just confirm that numbers below 3.16 are rounded down, instead of keeping the same order of magnitude? Because if i round up the order of mag of $1.99$ x $10^{30}$, it would be $10^{29}$ and if i round up the order of mag of $7.32$ x $10^{22}$, it would be $10^{23}$. Then if i minus them i will get $29 - 23=6$? $\endgroup$ – Fred Weasley Feb 1 at 9:22
  • $\begingroup$ @FredWeasley if you have a number $x \times 10^n$ then if $x < \sqrt{10}$ it should be rounded down to $10^n$ while if $x > \sqrt{10}$ it should be rounded up to $10^{n+1}$. The reason this gives you $6$ in your example is that when you round before doing the division you get two sets of rounding errors. If you do the division first then round you get only one rounding error. $\endgroup$ – John Rennie Feb 1 at 11:45

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