# First laboratory - Fall 2002

### Due: Friday, September 20, at 11:30pm

This laboratory is concerned with the question:

If we drop an object from a distance 10,000 km from the centre of the Earth, how long does it take for it to fall to the surface of the Earth?

There are four exercises to be done. You will need a straight edge of some kind (paper or cardboard OK) and a calculator to complete them.

## Warning

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## Preliminaries

• the surface of the Earth is R = 6,370 km from the centre;
• the acceleration of gravity at the Earth's surface is 9.8 m/sec2 ;
• the force of gravity is proportional to an object's mass and inversely proportional to the square of the distance from the Earth's centre.

Hence the force on an object at radius r is mg/(r/R)2 . Newton's Second Law F=ma therefore gives us the differential equation

r'' = -g/(r/R)2 = -gR2/r2.

Note that the sign is negative. If you drop an object, it falls. So r decreases with time.

In this laboratory, we shall find an approximate solution to this equation by calculating a sequence of approximations to position and velocity at times 0, dt, 2 dt, ... separated by intervals of length dt . The position, velocity and acceleration at time n dt are called rn, vn and an respectively. We shall assume that the position and velocity at time zero are

r0=10000
v0=0

The position and velocity at subsequent times are computed approximately using the rules

• an = -gR2/rn2 : the differential equation determines the acceleration at radius rn.
• rn+1 = rn + vn dt : this approximates the velocity by the constant vn throughout the time interval from n dt to (n+1) dt .
• vn+1 = vn + an dt : this approximates the acceleration by the constant an throughout the time interval from n dt to (n+1) dt .

These approximations should be better if dt is smaller. The problem this laboratory addresses is how the accuracy of this approximation process depends on the size of the time interval.

## The calculation

In the following applet, the object's state is calculated and displayed according to the approximation procedure described above. While the Run key is pressed, position, velocity, and acceleration are displayed every ten seconds during the object's descent. Pressing the Step button goes through a single step of the process. The Restart button ... restarts the process. The time interval dt can be chosen arbitrarily.

To run the process for a given inteval dt , set that value of dt in the window; restart if necesary; press Run until the value of t is close to where you want it; then Step through until the exact value is hit.

## Graphing

You can plot points on the following graph by clicking the left mouse button at an unselected location, or delete points by clicking the left mouse button at an already selected location. You can also choose which colour to plot - the top button is red for t=1263 , the middle one is green for t=1264 , the bottom blue for t=1265 .

## Exercise 1

Use the two applets above to get an idea of how the accuracy of the approximation depends on the time interval dt . Make a plot on the graph above of approximate value of r(t) versus the time interval dt used to compute the approximation. Do so for a few values of t . For example, if we set dt = 1 we get approximations rapprox(1263) = 6383.84, rapprox(1264) = 6377.13, rapprox(1265) = 6370.41, which are already plotted for you. Precisely,
• Set dt = 0.5. Run and step through the process above to get approximate values for r(1263), r(1264), r(1265).
• For each of these three values of t, record the pairs (dt, r(t)) on the graph and also in the table below.
• Do the same for dt = 0.25 and dt = 0.125.
• Use a straight edge to estimate what the value of rapprox(t) would be if dt were equal to 0 . Plot the estimated points (0, r(1263)), (0, r(1264)), (0, r(1265)), on the graph too.
• Use a calculator to get more accurate estimates, and enter these new data in the following matrix as well.
• Keep in mind that what you would like to do is set dt=0 or something very small, but that this is not practical. You must try to guess what would happen for small values of dt by using the data for large ones.

Answers in the table below must be as accurate as possible. You can do the plotting with the mouse, but to get the data in the first column below you will have to do some calculation.

## Exercise 2

The estimate of, for example, r(1264) is approximately the true value plus a term of the form C dt . In other words, the error in the estimate of r(1264) is linearly proportional to the time interval dt . At least approximately, what is the error coefficient C?

## Exercise 3

Now use the results of your calculations to get a good estimate of the moment of contact with the Earth's surface.

## Exercise 4

Use the same technique, but now working without the plotting applet, to make the best estimate you can of r(1000) .

## Submission

There is no reason, in principle, that you will not be able to submit your answers from anywhere in the Internet, but we cannot guarantee success. If a submission from within the Mathematics Department system is not successful, tell the TA.