Math and Measurement Help
The following are worked examples of items LIKE those in Lab 1, Math and Measurement. The
answers are not correct ones for the lab. So don’t turn them in.
But DO use this to check on whether you are doing things correctly. Not all the items are shown
here, only the ones that seem to be the hardest.
To check the measurements with your ruler, print out the pages so that the size will be correct.
Don’t worry if your measurements are a little different (like 1 degree on the angle measurements
or 2 mm on the length measurements.
Section I Measuring the Triangle. Here is an example of what a person might have gotten if
they did the lab correctly bud had a different triangle.
Side a
Side b
Side c
(be sure to include units)
Your Measurements
Length of Side a
149.5 mm
Measurements from another
Group
149.0 mm
Length of Side b
45.2 mm
44.5 mm
Length of Side c
119.0 mm
119.6 mm
Section II Scale Pictures and the Size of the Earth
Measuring distances on a map Maps generally have a scale. It is a line marked the with the
corresponding distance on the map. The scale in the map below is marked to show 1000km.
Measure the line, so that you can write the conversion equation. I measured on the sample map
below and got
1.6 cm= 1000km.
Now measure the distance between Chicago and New Orleans. I got 1.95 cm. Now use the
conversion factor that we found from the scale to convert the measured. For these measurements
you would multiply
#1000km &
1.95cm " %
( = 1218.75km
$ 1.6cm '
= 1.2 )10 3 km to the correct number of significant figures
Check for yourself.
!
Eratosthenes Method to Find the Size of the Earth.
Eratosthenes found the size of the Earth by measuring the length of a gnomon (stick) and its
shadow. The stick and the shadow are at right angles and make two sides of a triangle. The
hypotenuse (slanted side) of the triangle is determined by the first ray of the Sun that gets past
the stick. The angle between the hypotenuse of the triangle and the gnomon is the same as the
Math and Measurement Help
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angle with the vertex at the center of the Earth, one side toward Alexandria and the other toward
Syene.
MAP
SCALE
In the lab, you will draw a triangle with a right angle. One side of the right angle is the gnomon
and the other is the shadow. The dimensions for lab 1 are given in the lab write up. To practice,
you may want to draw a sample triangle and measure the angle between the hypotenuse and the
gnomon.
For the example, make a scale drawing of a gnomon 12.3 m long and a shadow 4 m long. Decide
on a scale and measure the sides and the angle between. Use metric units for the measurement
so that you don’t have to deal with fractions.
A good scale for this would be 2cm in the drawing for each meter in the real world. The triangle
will fit on the paper and will be fairly large, so that the drawing imperfections will be a small
percentage error.
So measure out 12.3m (2 cm/1m) =24.6 cm for the gnomon. Use the end of that line as the
vertex of a right triangle. Draw the other side 4 m (2cm/1m)= 8 cm long. Draw hypotenuse to
o
connect the ends of the two sides. Measure the angle at the end of the gnomon. It will be 18.0 .
The triangle is drawn on this sheet.
Finding the Size of the Earth
o
Suppose that you used triangle above and found that the angle at the top of the gnomon is 18 .
Now you want to find the circumference of the Earth. The angle that you measured is the same
as the angle at the center of the Earth. The ratio
Angle at Center of Earth Distance between Alexandria and Syene
=
360 degrees
Circumference of the Earth
relates this angle and the circumference of the Earth. You know all the values except the
circumference of the Earth. So let’s call the `circumference of the Earth , X. Suppose that the
!
Math and Measurement Help
2
distance
between
the
two
towns
was
900
km.
In
that
case,
the
equation
o
becomes
18.0
900 km
=
o
360
x
Solve for X , multiply both sides by X.
18.0 o 900 km
=
X
360 o
X
!
360 " 900km
X=
18
X = 18000km
X
This would make the circumference of the Earth equal to 18000km.
If you look at the front of the lab write up, it says that the
circumference of a circle =2 π r
!
The values 2 and π are known values. Your calculator has a key for π.
So here 18000 km=2 π r
18000km
2"
r = 2865km
r=
The percentage error formula on the front of the lab is
!
Percentage Error =
(Your Value - Accepted Value)
"100
Accepted Value
The accepted value is 6378 km, so plug into the equation to get
!
!
(2865km - 6378km)
"100
6378km
-3513
=
"100
6378
= -55.1%
Percentage Error =
The percentage error is negative, since your value is LESS than the accepted value. Keep 3
significant figures, since this is the number of figures in the least precise value in the
o
multiplication, the angle measurement, 18.0 . The computation involved several exact values (like
360 degrees and 2) and some values like π, which have at least 8 significant figures. The exact
values can be thought of as having infinitely many digits. Remember, your value and the number
of significant figures will be different.
Section III Finding the Mass of Earth and Moon
This part of the lab assumes that you have already found Earth’s radius in km. The average
density of the Earth is given. The density is the Volume divided by the Mass. So to find the mass,
compute the volume, then multiply by the density. The formula for volume of a sphere is on the
front of the lab. It is
"4%
V = $ '(r 3 where
# 3&
V is the volume of a sphere
r is the radius, in this case the radius of the Earth that YOU found
π, is the constant 3.14159…, but use the π,key on your calculator
!
Math and Measurement Help
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3
Let’s continue, using the value of r from the example above, 2.65x10 km. So
3
3
V=(4/3) x 3.14159… x (2.865x10 km
10
V=(4/3) x 3.14159 x 2.35x10 km3
10
3
V=9.85 x 10 km
Try the computation yourself to be sure that you are doing it correctly. When you cube the radius
of the Earth, you cube the units also.
3
3
Your calculator has a key to help you cube the radius, that is to evaluate (2.865x10 km) . Some
3
y
x
calculators have a key that says x , and others have a key that says x or y . You may need to
press the second key to get this to work. Many Texas Instrument calculators have a caret key, it
looks like ^ . Check out which you have.
3
3
Now calculate 2 . The answer is 8, so you can check the result. If you have the x key, enter 2
y
x
and press the key. If you have one of the other types of key, x or y or ^ , enter 2, hit the key,
then enter 3 (you have to tell the calculator what power you want) and hit enter or answer.
(I apologize if you have a calculator that doesn’t work this way. Keep trying until you get 8 for the
answer.) If you cannot find the key for cubing, leave this part and come for help. You can go on to
the Moon picture in the lab and to the plotting. It will take just a little time to finish up.
Now try the computation for the volume. Do use the E or EE key. Just multiply and divide without
writingout the answers or rounding off in the middle. When you have done the entire computation
write out the answer and round off.
10
3
Once you have the Earth’s volume, 9.85 x 10 km , multiply it by the density to get the mass of
12
3
the Earth. The density is given in the lab write up as 5.5x10 kg/km . Simply multiply
23
Mass=Volume x Density and in this case 5.42 X10 kg.
The percentage error is computed as before,
 Your value - Accepted value 
Percentage Error = 
 × 100
Accepted value


23
24
 5.42 × 10 kg - 5.97 × 10 kg 
 × 100
Percentage Error = 
5.97 × 10 24 kg


= 90. 92%
The error here is very large, because the value for the radius in this example is very small, the
mass turns out very small. When you do this computation in the actual lab, you will have a
smaller percentage error.
The lab asks you to compute the mass of the Moon. The process is just the same as what you
just did, but in the case of the Moon, the values for the radius and the density are given. So the
value for the mass should be very close to the accepted value.
The ratio of the mass of the Earth to the mass of the Moon is obtained by dividing the mass of the
Earth by the mass of the Moon. Be sure to use the exponent key so that the calculator realizes
that the power of 10 is part of the number.
Section IV Measuring Features on a Photo
The lab includes a photo of the Moon. The Moon’s radius is 1738 km, given in the write up.
Measure the diameter of the Moon photo and equate it to the Moon’s diameter, 3476 km ( twice
the radius). This establishes the scale of the photo. Then measure the diameters of the Mare
Imbrium and Mare Crisium and use the scale factor to transform from centimeters on the picure
to kilometers in the real world. Both Mare Imbrium and Mare Crisium are basically circular. Mare
Crisium appears thinner in one direction because it is foreshortened.
Math and Measurement Help
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The image of Mars below is given so that you can follow an example. This is a topographic map.
The colors represent the height, where ]the darker parts are lower, not darker. The radius of Mars
is 3397 km, so the diameter is twice this or 6794 km. The line on the photo measures 11.9 cm
and equals 6794 km in the real world. The oval indicates the Argyre Basin. It isn’t really circular,
but we can find the actual dimensions by measuring the north-south dimension 1.9 cm and the
# 6794km &
( = 1085km . This is the
$ 11.9cm '
east west dimension 3.1 cm. Using the scale, we find 1.9cm " %
total north-south dimension of Argyre Basin. The equivalent “radius” would be half this or 542.4
km (divided before rounding). The east-west extent would be 1770 km. Check for yourself.
!
.
Section V Plotting Refresher
The Laboratory Information Page has information about how to plot. Put your line or curve
through the midst of the points. There is more than one acceptable answer. The idea is to
understand how to make a sensible curve.
For the question about the range of radii, look at the spread of radius values near 1.7 solar
masses. Use the scatter of the points to estimate a highest and lowest radius that could occur
without surprising you (based o the data in the sample).
Math and Measurement Help
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