# Science   Astronomy of Scale   There are two things you have to be able to do in thinking about the size of the…

Science

Astronomy of Scale

There are two things you have to be able to do in thinking about the size of the universe in scale; one, think and work on scientific notation, and two, work with ratios.

Scientific notation.

This is just the way we right really big or small numbers.  It gets to cumbersome to write them out the normal way.  You simply put down the important digits, then x 10power.  The power is how many places you had to move the decimal. + if to the left, – if to the right.

Ex. 135,000,000 is 1.35 x 108      260,000  is 2.6 x 105    5257  is 5.257 x 103

0.00034 is 3.4 x 10-4              0.0000000067 is 6.7 x 10-9       0.040 is 4.0 x 10-2

Multiplying

When multiplying two nubers in scientific notation, the rule is to multiply the coefficients (the number in front) and add the exponents (the powers of ten).

Ex. 1.2 x 103 x 4.5 x 106 = (1.2 x 4.5) x 10(3 + 6) = 5.4 x 109

3.5 x 105 x 7.2 x 10-3 = (3.5 x 7.2) x 10(5 + -3) = 25.2 x 102

Dividing

Similar to multiplying, the rule for dividing is to divide the coefficients, but this time you subtract the bottom power from the top power.

Ex. 1.2 x 103 / 4.5 x 106 = (1.2 / 4.5) x 10(3 – 6) = 0.27 x 10-3

3.5 x 105 / 7.2 x 10-3 = (3.5 / 7.2) x 10(5 – -3) = 0.49 x 108

Express the following numbers in scientific notation.

1)  356,000 = _______________         2)  0.000084 = _________________

3) 4.5 x 102 x 6.7 x 105 = ______________        4) 5.6 x 106 x 9.2 x 10-2 = ___________________

5) 7.1 x 103 / 2.5 x 108 = ______________         6) 9.4 x 108 / 6.3 x 10-5 = ___________________

Now, Let’s look at the scale part.  The universe is huge and really incomprehensible for us humans.  We do better if we can put those big distances in terms we can understand, compare it to things we see every day.  If you know that if one thing can be represented by an egg, then the other thing is the size of a basketball, then you get a better understanding of how they compare.  This is a scale model; using different, but understandable sizes to represent the real things.  But everything still has to be relatively the same relationship.

Suppose you want to make a scale model of your town on your counter.  You know (because you measured them) the width of your house is 50 feet and the width of your town is 5000 ft.  This means the town is 100 times bigger (5000 ft/50 ft) than your house.  To keep the proper proportion in your model, the model town needs to be 100 times bigger than your model house.  If you choose a monopoly hous to use for your house, it is 1 cm wide.  How wide should the town be on this scale?

Set up the ratios, one for the real sizes, one for the model sizes.  5000 ft/50 ft = dist/1 cm.  Notice how the house measurements are on the bottom of both ratios and the town distances are on the top.  You don’t know the distance yet for the model town, that’s what we need to find out, although you might already have guessed what it should be.

Solving the equation for the unknown distance means you have to be able to do basic algebra.

first, cross multiply the two ratios.

5000 ft/50 ft = dist/1 cm

5000 ft (1 cm) = 50 ft (dist)

second, divide both sides by the number multiplying the unknown dist, in case 50 ft)

5000 ft (1 cm) = 50 ft (dist)

50 ft                50 ft

100 (1 cm) = dist

Now finish the multiplication

100 (1 cm) = dist

100 cm = dist

So, if the house is a monopoly house (1 cm) then the model town has to be 100 cm wide.  This keeps the proper proportions.

You try these.

If the earth was being represented by a golf ball (about 2.5 cm diameter), what would be best to represent the moon?  (the real diameter of the earth is 1.3 x 104 km, the moon’s diameter is 3.5 x 103 km)  Hint: first figure out the model size in cm, then think of something real that is that size.

If the earth was being represented by a golf ball (about 2.5 cm diameter), what would be best to represent the sun?  (the real diameter of the earth is 1.3 x 104 km, the sun’s diameter is 1.4 x 106 km)

Ok, this one is a little different, but try to think about it like a model.

It takes 8.3 min for light from the sun to travel the 1.5 x 108 km to the earth.  If uranus is 2.9 x 109 km from the sun, how long will it take the sun’s light to get there?

Mathematics homework help

## Why US?

##### 100% Confidentiality

Information about customers is confidential and never disclosed to third parties.

##### Timely Delivery

No missed deadlines – 97% of assignments are completed in time.

##### Original Writing

We complete all papers from scratch. You can get a plagiarism report.

##### Money Back

If you are convinced that our writer has not followed your requirements, feel free to ask for a refund.