Home Page | About Me | Home Entertainment | Home Entertainment Blog | Politics | Australian Libertarian Society Blog | Disclosures

A letter to the Editor explaining the effect of the Moon's gravitational field on falling objects, published in The Australian on 20 Aug 1991, p.10

In answer to A. Yearley (Letters, 14/8), there are two factors influencing how a falling (or tossed) object would act here and on the moon.

They are air resistance (of which the moon has none) and surface gravity (of which the moon's is but one sixth of the Earth's). As the effect of the Earth's air resistance on a coin that is not tossed too high is minimal, it can be ignored for practical purposes.

The gravitational differences would mean that a tossed coin would go six times higher and take six times longer to land on the moon than it would on Earth, given the same initial velocity. To take a typical example, a coin tossed to a height of 60cm on Earth would take about 7/10ths of a second to be caught again. Tossing the same coin at the same speed on the moon would fling it up better than 3½m and it would take over four seconds to come back.

NOTE: Of all the controversial opinions I've expressed on letters' pages, this is the only one -- a matter of simple scientific fact -- that received a published rebuttal!

© 1991 - Stephen Dawson