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Physics for Geologists, Second edition

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Notes<br />

1 The centripetal <strong>for</strong>ce required to keep a body in circular orbit is mv2/r, as we<br />

found in the dimensional analysis on page 11. The velocity of the body is 2nr/T<br />

where T is the period, the time taken <strong>for</strong> one orbit. So,<br />

mV2 m4n2r<br />

F,=ma=--- -<br />

T2<br />

4n2r<br />

and a = -<br />

Y T2 '<br />

Kepler's Third Law states that the square of the period is proportional to the cube<br />

of the mean distance, so<br />

Similarly, given the inverse square law, you can infer Kepler's third law of planetary<br />

motion, that r3 cc T2.<br />

2 Work is a weight multiplied by the distance moved in the direction of a <strong>for</strong>ce. So<br />

if you drop a mass m from a height h above the ground, its velocity on impact is<br />

v = gt.<br />

Draw a graph now of v against t, and you have a triangle (you don't need to put<br />

numbers to the graph) the area of which is the distance fallen. Thus the distance<br />

fallen is<br />

The potential energy be<strong>for</strong>e dropping is converted to kinetic energy on impact<br />

(ignoring friction) so<br />

Ep = <strong>for</strong>ce x height = mg x igt2 = imgZt2 = LmU2 2 = E k .<br />

3 If you take the Earth's radius as the unit of distance, then, as stated in the text,<br />

the attraction of the Moon is roughly proportional to 1/602 at the Earth's centre,<br />

1159~ at the sublunar point, and 1/612 at the antipode. If x is the number of Earth's<br />

radii between the centre of the Earth and the centre of the Moon, the <strong>for</strong>ce at the<br />

sublunar point is proportional to l/(x- at the centre of the Earth, to 1/x2; and<br />

at the antipode, to l/(x+ I ) ~ Two . approximations are used. (x - 1)2 = x2 -2x+ 1<br />

Copyright 2002 by Richard E. Chapman

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