How Far Is the Horizon?Author(s): Philip RobinsonSource: Mathematics in School, Vol. 18, No. 2 (Mar., 1989), pp. 18-19Published by: The Mathematical AssociationStable URL: http://www.jstor.org/stable/30216341 .

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A t."

A by Philip Robinson, St. Stephen's School, Rome

During a discussion about circles and spheres the question arose: How far is the horizon? A bright fifth year group was invited to solve the problem given a sphere of radius R and an observer at height h above the surface. They drew their own diagrams and came up with two approaches.

h

R R

1. By applying Pythagoras' theorem:

(R + h)2 = R2 + X2 leads to x = h2 + 2hR.

We considered h as being small in comparison with R and were able to write x as a good approximation:

x~ 2hR.

x/h

2R

2. The same result was reached by using the tangent- secant theorem. We have directly:

2 = h(h + 2R) and for small h, x 2hR.

18

Using the approximate result, students were invited to draw the graph of x against h, with R = 6400 km and h increasing from 0 to 50 km. Clearly, they obtained the positive branch of a parabola.

x (km) x = constant /h

0 h (km)

One student suggested that, since h was small in com- parison with R, x will be very nearly equal to the are length between the observer and his horizon. [This was accepted as true and can easily be shown using sin a 2 , for small x.] This led to the introduction of an angle of latitude, taking the observer to be at a pole of the sphere.

x h

R R

Immediately we had the result

x = (h + R) cos 8.

This result is interesting in that it gives us the distance of the horizon if our altitude is known and 0 may be measured using a fairly accurate instrument.

As h increases and the observer moves away from the sphere, angle 8 decreases but only becomes zero at infinity. The observer then, will never be able to observe the equator, no matter how far from the sphere he might be.

Mathematics in School, March 1989

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What is the area bounded by the horizon? I led the discussion on to the problem of finding the area bounded by the horizon. The area in question is the area of a cap of a sphere and it allowed me to introduce Archimedes' fascinating result regarding a sphere and its circumscribing cylinder. That is, any two planes at right angles to the axis of the circumscribing cylinder, intercept equal areas on the two surfaces. If we take one of the planes as tangent to the sphere then our problem is solved.

Lateral area = 2nrR.R(1- sin 0) = 2nR2(1-sin 8)

RK

R(1- sin 0)

Rsin 6

The area of the cap is equal to the intercepted lateral area of the cylinder and is therefore given by

A = 2nR2(1 - sin 8).

The area of a hemisphere is 2nR2. We note from the equation that A is only equal to this value when 8 is equal to zero, which is never the case.

From our diagram we have

R sin 8 R

R+h

and therefore

A = 2nR2(1 -sin 0) = 2nR2 1 - = 2R2h R+h h+R'

Thus the area visible is given by

h A h+R

where AH is the area of a hemisphere, 2nR2 Students were asked to find the distance of an observer

from the surface of a sphere if he is to be able to see a certain percentage of the hemisphere.

For example, to see 50%0 of the hemisphere:

h

h+R

leading to h = R, x = /R

and = 300. In conclusion I would simply say that students were able

to a great deal of investigative work using fairly simple ideas. The problems under investigation were easily imagined and the practical side, considering the sphere to be the Earth, drove them on, hungry for real results.

UN I VERS I TY OF MANCiHESTER t:'EPAFRTMENT OF EL:'IUCATICIN

"Practical Investigations in A level Pure and Applied Mathematics"

(A 3 day In-service training, course for all teachers of A level Mathematics.)

The course will cover:

Modelling and Problem Solving.

The use of Graphic Calculators and Computers.

A level Coursework in Pure and Applied Mathematics.

Extended Investigations (e.g. the Alton Towers Project.)

It will include use of class kits of materials which have been tested in schools over the past year by MAP in Manchester.

"A New Approach to GCSE Coursework"

(A 3 day In-service training course for all teachers of 11 - 16 Mathematics.)

The course will look at:

The application of mathematics to problems in the physical world.

Cross Curricular development of work suited to all abilities in the 11 - 16 age group as part of GCSE Course work.

The use of materials and apparatus developed by MAP in Manchester. This includes classroom kits of materials for investigations that have been tried and tested in conjunction with schools from nine LEAs in the North West this year.

Although the course is geared to 11 - 16 age group there will be extensions suitable for A and AS level study.

DATE: 9IqTH - 2STf JUNE, 1989 DATE: 22ND - 24TH JUNE, 1989

Further details and applications forms from: Mrs. A. Kitchen or Mr. J.S. Williams, Mechanics in Action Project, University of Manchester, Humanities Building, Oxford Rd. Manchester, M13 9PL (Tel: 061-275-3409)

Mathematics in School, March 1989 19

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