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The order of magnitude is the power of 1010 a number is raised to when it's in scientific notation.

For example, the order of magnitude for 19,40019{,}400 would be 44. This is because when we convert it into scientific notation, 1.94×1041.94\times10^4, the 1010 is being raised to the 4th4^{\text{th}} power.

We can also compare orders for two different numbers in scientific notation. Consider for example

A=6.7×108A = 6.7 \times 10^8 B=6.7×109B = 6.7 \times 10^9 C=6.7×1010C = 6.7 \times 10^{10}

One can quickly see that BB is a bigger number than AA, and CC is bigger than BB. However, we can say more. We can say CC is two orders of magnitude larger than AA. Also, CC is ten times the smaller number BB; BB is ten times AA.

Now it's your turn to try a problem yourself.

What is the order of magnitude for 6,032,095?6{,}032{,}095?

Choose one answer

Video · + 1 min read

(Video) Orders of Magnitude

by Truckee

Truckee has a good video about orders of magnitude.

Summary\large{\text{Summary}}

We use orders of magnitude for estimating. If you're looking for the volume of a lake or the area of a field, you can use estimates. Sometimes the numbers will be written in scientific notation. Other times you'll practice writing a number in scientific notation yourself. Regardless, these estimates are useful for small numbers and large numbers alike. The power of 1010 could be positive or negative.

If you're asked to find the order of magnitude for 2.3×1042.3\times10^4 you have two options. The first is rounding the scientific notation to the ones place. 2.3×104≈2×1042.3\times10^4\approx 2\times10^4 The other, is to just write out the power of ten. 2.3×104≈1042.3\times10^4\approx10^4

The numbers 2×1042 \times 10^4 and 10410^4 are certainly different. The first number is 22 times larger than the second. However, the powers of 1010 are the same. They are numbers of the same order. In addition, there are a total of four factors of ten.

She then looks at an example.

Estimate the volume of water in Gilmore Lake in the Desolation Wilderness. The lake is 2 km2 \text{ km} across and has an average depth of 50m50 \text{m}.

We can think of the lake as a cylinder with a height of 50 m50 \text{ m} and a radius of 1 km1\text{ km}. The volume of a cylinder is: V=πr2hV=\pi r^2h We can write 1 km1 \text{ km} as 1×1031\times10^3 and 50 m50 \text{ m} as 5×1015\times10^1. Plugging these into the equation, we get: V=πr2hV=π(103)2(5×101)V=π×106×5×101V=5π×106+1V=5π×107\begin{align} V&=\pi r^2h\\ V&=\pi(10^3)^2(5\times10^1)\\ V&=\pi\times10^6\times5\times10^1\\ V&=5\pi\times10^{6+1}\\ V&=5\pi\times10^7 \end{align} Be careful though. We see that 5π5\pi is approximately 15.715.7. This is not between 11 and 1010. In fact, 5π5\pi is greater than 1010.

V≈1.57×108V\approx1.57\times10^8

We can estimate this more as 2×1082\times10^8 or just 10810^8. These are called examples of approximate comparisons.

Type a concept, like “slope” or “photosynthesis”.

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