# Section 3B

...

## Generalities

Sometimes exact values are cumbersome and irrelevant.

Example: World population

Is that indeed accurate? --NO

Does one want an accurate number? --NO

One would better say: around 7 billion, or write $$7.4 \times 10^9$$

## Scientific notation

$$Q \times \text{a power of 10}$$,

$$Q$$ --- a quantity between 1 and 10 ,

a power of 10 is $$10 ^ \text{a whole number}$$

Some powers of ten have their own names:

 $$10^2$$ | hundred $$10^3$$ | thousand $$10^6$$ | million $$10^9$$ | billion $$10^{12}$$ | trillion

## Convenience of scientific notations

Instead of: "Total spending in the new federal budget is \$3,900,000,000,000"

we have: "Total spending in the new federal budget is $$\ 3.9 \times 10^{12}$$", or

"Total spending in the new federal budget is $$\ 3.9$$ trillion".

Instead of: "The diameter of a typical bacteria is 0.000001 meter"

we have: "The diameter of a typical bacteria is $$10^{-6}$$ meter", or "The diameter of a typical bacteria is 1 micrometer".

## Negative powers of ten may be expressed as prefixes.

 $$10^{-1}$$ | deci $$10^{-2}$$ | centi $$10^{-3}$$ | milli $$10^{-6}$$ | micro $$10^{-9}$$ | nano $$10^{-12}$$ | pico

## Price to pay: working with scientific notations (p.136)

Converting to scientific notation

Step 1 Move decimal point to come after the first digit

Step 2 Number of places moved is the power of ten

Examples:

$$3042$$ ...... 3 places to the left ...... $$3.042 \times 10^3$$
$$0.00012$$ ...... 4 places to the right ...... $$1.2 \times 10^{-4}$$

## Price to pay: working with scientific notations (p.136)

Converting from scientific notation

Just move the decimal point in the opposite way

Examples:

$$3.042 \times 10^3$$ ...... 3 places to the right ...... $$3042$$
$$1.2 \times 10^{-4}$$ ...... 4 places to the left ...... $$0.00012$$

## Price to pay: working with scientific notations (p.136)

Multiplication and division in scientific notation

Operate separately with the powers of $$10$$ and the rest.

Examples:

$$(6 \times 10^2) \times (4 \times 10^5)$$ $$= (6 \times 4 ) \times (10^2 \times 10^5 )$$ $$= 24 \times 10^7$$ $$= 2.4 \times 10^8$$

## Price to pay: working with scientific notations (p.136)

Addition and subtraction in scientific notation

Either translate everything into ordinary notations, perform the operations, and translate back, or ... think a bit ...

Example: $$7.4 \times 10^9 + 9.5 \times 10^3$$ $$\approx 7.4 \times 10^9$$

The number in this example has to do with the global population.

One does not know it to this accuracy, and does not really need this precision.

Putting numbers in perspective is our next topic to discuss

## Putting numbers in perspective

The book lists three methods:

• Perspective through estimation

• Perspective through comparison

• Perspective through scaling

## Perspective through estimation

The idea is always to grasp the
order of magnitude,
which is the broad range, within one or two powers of ten,
with as small amount of calculations as possible.

Example (30, p.148) Could a person walk across the United States (New York to California) in a year?

## Could a person walk across the United States (New York to California) in a year?

Collecting data:

$$2,765$$ miles within $$365$$ days ...

means less than $$8$$ miles a day.

## Order of Magnitude estimates

Meaning: very rough estimate, number of digits ...

Example (36, p.148)

The total number of words in the textbook.

About $$700$$ pages, $$50$$ rows on a page, $$15$$ words in a row...

$$700=7 \times 10^2$$ ... $$35 \times 10^3$$ ... $$500 \times 10^3 = 5 \times 10^5$$

## Perspective through comparison

Table 3.1 on p.140 tells us:

Energy released by burning 1 kilogram of coal is $$1.6 \times 10^9$$ joules.

Energy released by fission of 1 kilogram of Uranium-235 is $$5.6 \times 10^{13}$$ joules.

Clearly, fission is more efficient, but how much?

1 kilogram of Uranium-235 gives same amount of energy as

$$\frac{5.6 \times 10^{13}}{1.6 \times 10^9} =3.5 \times 10^4$$, which is 35 tones (thousand kilos) of coal

## Perspective through scaling

Here I refer to Example 7 on p.142

The distance from the Earth to the Sun is 150 million kilometers, the diameter of the Sun is 1.5 million kilometers, while the diamemter of the Earth is 12,760 kilometers. ... So what?

After dividing all these lengths by $$10^{10}$$, they find out the following.

If the Sun was of the size of grapefruit, the Earth would be of size of a ball point in a pen at a distance of $$15$$ meters from the grapefruit.