Forces and Motion

Estimating the Moon’s orbit time

Teaching Guidance for 14-16 PRACTICAL PHYISCS

The Moon’s orbital period can be estimated by considering the Moon as a satellite of the Earth and using a scale drawing, e.g. the drawing produced in the experiment:

Sketching a satellite orbit and predicting its period

The Moon's distance from the Earth is about 60 Earth-radii. Whereas a scale of 0.5 mm to a kilometre can be used for a near-Earth satellite, it is necessary to use a scale of 0.5 mm to 60 km for the Moon's orbit.

Here’s a possible script for class discussion which might follow immediately after the Sketching a satellite orbit and predicting its period experiment.

Imagine that gravity extends undiminished out to the Moon. Calculate the fall in 2 minutes: 72 km as before.

On a scale drawing with 0.5 mm to 60 km that fall would be only 72/120 mm; too small to work with. Let the Moon travel the same arc on the diagram as for the Earth satellite falling for 2 minutes. (That was somewhere between 40 and 80 cm.)

But now the fall of 36 mm from the tangent to that place on the diagram no longer represents 72 km. It represents a fall of 72 x 60 km. How much time would a falling body need for that under full gravity? Look at s = ½at 2. With s 60 times as big, with the same at 2 must be 60 times as big. Thent must be √60 times as big: that is: 7¾ times as big. [To show that √60 ≈ 7 ¾ you could do this calculation: 

(7¾)2 = (31/4)2 = 96116… within 0.06 % of 96016 = 60

Then the Moon would travel the arc in 7¾ x 2 minutes; and it would travel the whole circle in 7¾ x the orbital period obtained for the near-Earth satellite; that is, between 11 and 12 hours. [Many students will not be worried about 11-12 hours; after all it is half a day and explains why the Moon disappears (sometimes) during the day and only appears at night!]

Even if answers for the near-Earth were far from 90 minutes, the new answer is clearly wrong for the Moon, which takes a month to orbit the Earth. Therefore, if the Moon is constantly pulled from the tangent to its orbit by the Earth's gravity, the force of gravity must be weaker out at the Moon.

[You could then take the Moon's month of 27.3 days for granted and calculate the amount of dilution. But the answer would not look clear and simple to beginners. It is probably better to suggest an inverse-square dilution of gravity and try it in the calculation from the drawing. If it is the pull of gravity that holds the Moon in its orbit, making it fall from the tangent to the orbit again and again; it must be a much weaker gravity. The acceleration must be much less than 10 m/s/s out at the Moon.]

When 17th century astronomers started puzzling about this, several people suggested that gravity may 'thin out’ according to an inverse-square law. According to that, if gravity is so much at a certain distance, it is 1/4 at double distance; 1/9 as much at treble distance; 1/100 as much at 10 times as far away from the attracting body.

An apple near the Earth is pulled so strongly that it falls with acceleration 10 m/s/s .The Earth attracts an apple as if all the Earth were concentrated at the centre 6,400 km below the surface, a whole Earth-radius from the apple. But we know that the Moon is about 60 Earth-radii away from us, 60 times as far from the Earth's centre as an apple. So, if gravity does follow an inverse-square law; it must thin out by a factor (1/60)2 when we change from apple to Moon. If so, free fall under gravity at the Moon would not have an acceleration l0 m/s/s; but it would have acceleration (10/60)2 or 10/3,600 m/s/s.

How would that change affect the calculation of the Moon’s orbit time?

Look ats = ½at 2. With s. If a is 3,600 times smaller, then, for the same st 2 must be 3,600 times bigger: and t must be 60 times longer. As a result of diluting gravity; you might expect the Moon to take 60 times the previous estimate for its whole orbit.

That is 60 x (7¾ x 90) minutes, or 60 x (7¾ x 90) / (60 x 24) days, or very close to 29 days.

It looks as if the Moon may be ‘falling’, to keep its orbit, with inverse-square-diluted gravity.

[It is amazing that the change of scale from the Earth orbit satellite to the Moon gives such a good result. It is easier to do than it is to read the instructions so have a go!]

Newton is credited with doing this calculation and so students who attempt it are following in the footsteps of a great scientist.

appears in the relation T^2 ∝ a^3 f=1/T
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