Measuring the Earth's Radius

Ay 20 – Lab 1:  The Radius of the Earth

By:  Joanna Robaszewski

Purpose of experiment:  To determine the Earth’s radius by using only a stop watch and a clear view of the sun as it sets

Procedure:  We drove to Santa Monica beach on September 30, 2011 and arrived approximately at 6:15 pm.  The sun was supposed to set at 6:40 pm.  Some students lay on the sand while my group stood on the balcony surrounding the lifeguard house.  There were also some students of the pier.  The set-up is shown below: 

We had planned on taking two measurements with the stop watches.  First, the observers on the ground would signal to us standing on the lifeguard balcony when the bottom rim of the sun was tangent with the horizon by waving their arms.  We would then start the stop watches.  We were supposed to stop the stop watches when the bottom edge of the sun hit the horizon from our perspective on the lifeguard balcony.  This first measurement was not completed, however, because to us on the balcony it looked like the bottom of the sun had already touched the horizon before the observers on the ground signaled to us.  We were able to complete the second measurement.  This measurement consisted of starting the stop watches when the observers on the ground signaled that the top edge of the sun had touched the horizon by waving their arms and then stopping the watches when it appeared to us that the top edge had touched the horizon.

Data/Results:  The time measured by the stop watches between when the people on the ground saw the top of the sun set and when those of us on the lifeguard balcony saw the top of the sun set was 4.05 seconds.  We estimated the height of the balcony to be 5 feet.  I am 5’8”.  The total height from the ground was then 10’8” = 3.25 meters.  Looking at the Earth from the side we see:

We know that it takes approximately 24 hours for the Earth to rotate 360°.  How many seconds does it take?

So in the 4.05 seconds we measured, the Earth rotated:

This should be equivalent to θ in the diagram showing Earth from a side view.

We can then find the value of the longer leg of the triangle:

(I apologize that the diagram is slightly blurry, the short leg of the triangle reads "h = 3.25m")

Since x is such a small fraction of the Earth’s circumference, it can be approximated as straight.  The law of sines can then be applied:

Now that x is known, r can be found using the relation:

The Earth fact sheet from the NASA website (nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html) gives the Earth’s radius as 6371 km.  So the accepted value for the Earth’s radius, R, is:

Error Analysis:  According to www.humanbenchmark.com/tests/reactiontime/stats.php the average reaction time to click on a target is 200 ms = .2 s.  We will take this as the reaction time for the people on the ground to wave as they saw the top of the sun set, as the reaction time for us to see the waving and press start, and as the time for us to press stop as we saw the top of the sun set.  The first action and reaction (seeing and waving) would have decreased the time measured, as would the time taken between waving and pressing start.  The time to press stop, however, would have increased the time, so let’s say that the uncertainty on the time measurement was .2 seconds.  In that case the time measured was:

Since we estimated the height of the balcony by comparing it to the height of a person with a known height rather than measuring it with a more precise tool such as a meter stick, the uncertainty in the height is:

To find the uncertainty of our result for the Earth’s radius we need to propagate error:

We know

from:

and we know

from law of sines.  

I apologize for the algebra in advance…

Relating these gives:

Substituting t = 4.05s and simplifying gives:

Remembering that:

Substituting the values calculated and estimated earlier gives:

So our calculated result for the Earth’s radius was:

(Hooray, the uncertainty is smaller than the result!)

To find how many standard deviations our result is from the accepted value:

Our result is within 16 standard deviations of the accepted value.  This comparison is rather poor.  Sources of error when calculating the radius of the Earth may have come from underestimating the reaction times, not having many data points to average out errors, atmospheric and smog interference, and any assumptions about parts of the Earth’s circumference being straight.

Follow-up - Mass of the Earth:  Using our value for the radius of the Earth we want to calculate the mass of the Earth.  We can use the relation:

The density of the Earth can be approximated by the density of the average rock.  Most rocks sink, but some float in water.  So let’s say the density is between 1 and 10 times that of water.  Let’s take an average of the two and approximate the density of the average rock, and the Earth, as 5000 kg/m^3.

In that case:

and the uncertainty is:

So our value for the Earth’s mass is:

The accepted value for Earth’s mass is 5.9736*10^24 kg, as given by the same source used to find the Earth’s radius, NASA’s planetary facts sheet.

Our value of the Earth’s mass is within:

of the accepted value for the Earth’s mass.  This comparison is also poor.  The value found for the Earth’s radius was also far from the accepted value and the error was propagated, so all the sources of error described in the previous section still hold.  Additionally, the density of the Earth was only an approximation and that would add to the error as well.