Calculus II

Lesson 3: Applications of Integration 1: Work

When a force moves an object, we say the force does work. If the force is constant, the work done is given by the equation , where is the distance moved. What happens if the force is not constant?

Suppose, for definiteness, that a force moves an object from to along the -axis. Although depends on , we can divide the interval into small subintervals, and suppose that is almost constant on each subinterval.

Let's see why the expression should be an approximation to the total work done in moving the object from to .

When we divide the interval into equal subintervals, each subinterval has length . If is large, each subinterval will be very short, and so the force on the -th subinterval can be approximated by its value at any point in the subinterval. We will choose the right-hand endpoint of the subinterval, which is the point . The force on the -th subinterval is therefore approximately , and the work done in moving across this subinterval, using the constant-force formula, is approximately . The total work done in moving from to is given by adding up of these terms, one for each subinterval, which gives the formula in the statement of the question.

Let's write a Maple function worksum , which takes a function , an interval and a number , and returns an -subinterval approximation to the work done by the force in moving an object from to (i.e. translate the formula given in Question 1 into Maple syntax).

> worksum := (F,a,b,n)-> sum(F(a + k*(b-a)/n)*(b-a)/n, k=1..n) ;

Example 1: Riemann Sum approach

The force felt by an object of mass at the surface of the Earth is , where m/s^2 is the 'accelerationn due to gravity'. Of course, the force felt by the object lessens as it moves away from the Earth. In fact, the correct force law is given by Newton's Law of Gravitation:

.

Here, is the force felt by the object, is its mass, is the mass of the Earth, is the distance of the object from the centre of the Earth, and is a universal constant. The values of and are known, but we will not need them, because of the following argument.

(a). The kilometre was originally defined as 1/10000 of the distance from the North Pole to the Equator along the meridian which runs through Paris. Hence the circumference of the Earth is almost exactly 40000 kilometres. Find the radius of the Earth and assign it to the variable R.

> R := evalf(40000/(2*Pi));

(b). At the surface of the Earth, . Putting in Newton's Law of Gravitation gives one expression for the gravitational force at the surface of the Earth. Equate this expression to , and hence find the value of the product . (Be careful with units: is expressed in terms of metres/second^2, but other distances are in kilometres.)

Solution.

It is best to start here with pencil and paper. Equating the two expressions for the force at the surface of the Earth gives

. Cancelling and solving for the product gives . We will use the value of computed above, and the value of given in the question, but to make the units consistent we will express in km/s^2:

> g := 0.00982;

> GM := R^2 * g;

(c) At a height of 42377 kilometres above the centre of the Earth, a satellite revolves in a geostationary orbit: it takes exactly 24 hours to revolve once around the Earth, and so it is always directly above the same point on the Earth. Communications satellites, for example, are always placed in geostationary orbits. Use your Maple function from Question 2 to compute approximations to the amount of work that must be done to raise a 250-kilogram satellite from the surface of the Earth to a geostationary orbit. Use approximations with 100 and 1000 subintervals. Your function may return a negative value, although it clearly takes a positive amount of work to raise a satellite into orbit. Explain this.

Solution. We have to raise the satellite from the surface of the Earth to geostationary orbit. Since heights are being measured from the centre of the Earth, we must raise the satellite from a height to a height of 42377 kilometres.

We want the work done against the force of gravity in moving between these heights. Our function worksum will give us (approximations to) the work done by the force. First, of course, we have to tell it what the force is:

> f := r->-GM*250/r^2 ;

> worksum(f,R,42377,100);

> worksum(f,R,42377,1000);

Remember that this is the work done by the force, which is the negative of the work that has to be done in lifting the satellite into orbit. (The gravitational force will do a positive amount of work when the satellite crashes back to Earth.)

We should probably only keep 3 significant figures in our answers, since the value of was only given to this accuracy. Let's say, then, that with 1000 subintervals we estimate the necessary amount of work to be 13200 Newton-kilometres,

or (in more usual units) 13200000 Newton-metres.

We've seen that the expression is an approximation to the work done. On physical grounds, we would expect that this approximation would get better and better as gets larger. Mathematically, on the other hand, we recognise that our approximation is a Riemann sum for the integral , and we know that the Riemann sums will converge to this integral as . It seems reasonable to couclude, therefore, that the integral gives the exact amount of work done by the force.

Example 2: Exact Integration Approach

Use an appropriate integral to compute the exact amount of work required to lift the satellite. Compare with the Riemann sum approximations. (For example, you could ask: How good are the approximations? How many subintervals are necessary for the approximation to be within 1% of the true answer?)

Solution. (The function should still be defined from Question 3.)

> exact_work := int(f, R..42377);

As in Question 1, this is the work done by the force, which is the negative of what we want. Rounding to 3 significant figures, let's say we have to do 13300000 Newton-metres of work to raise the satellite.

Our approximations in Question 1 were pretty good. In particular, a 1% error would mean (in Newton-kilometres) an error of no more than 132 N-km, or an approximation between

> exact_work - 132; exact_work + 132;

Our 1000-interval approximation certainly meets this requirement, but the 100-interval one does not. How many intervals do we need? (Some trial and error was necessary to get the number of intervals in the next commands.)

> worksum(f,R,42377,350);

> worksum(f,R,42377,300);

Somewhere between 300 and 350 intervals would be sufficient---you can experiment further if you want to narrow the number down more accurately.

Question: In this question, it was easy to see how accurate our approximations were, because we could compare them with the exact answer. Of course, this also makes the comparison unnecessary: if we have the exact answer, we don't need to worry about approximations! Riemann sum approximations are most useful when we can't work out the exact answer. In that situation, how do you think we could have confidence that our approximations were sufficiently accurate?