Refer to Example 10.3 concerning the deflection from the plumb line of a particle falling in Earth's gravitational field. Take g to be defined at ground level and use the zeroth order result for the time-of-fall, T= V2h/g. Perform a calculation in second approximation (i.e., retain terms in w?) and calculate the southerly deflec- tion. There are three components to consider: (a) Coriolis force to second order (C,), (b) variation of centrifugal force with height (C2), and (c) variation of gravi- tational force with height (C3). Show that each of these components gives a result equal to h2 C, w? sin A cos A with C, = 2/3, C2 = 5/6, and Cg = 5/2. The total southerly deflection is therefore (4h?o? sin A cos A)/g.

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10-13. Refer to Example 10.3 concerning the deflection from the plumb line of a particle
falling in Earth's gravitational field. Take g to be defined at ground level and use
the zeroth order result for the time-of-fall, T= V2h/g. Perform a calculation in
second approximation (i.e., retain terms in w?) and calculate the southerly deflec-
tion. There are three components to consider: (a) Coriolis force to second order
(C), (b) variation of centrifugal force with height (C2), and (c) variation of gravi-
tational force with height (Cs). Show that each of these components gives a result
equal to
h2
o sin A cos A
with C 2/3, C2 5/6, and Cs = 5/2. The total southerly deflection is therefore
(4h-w² sin A cos A)/g.
to
Transcribed Image Text:10-13. Refer to Example 10.3 concerning the deflection from the plumb line of a particle falling in Earth's gravitational field. Take g to be defined at ground level and use the zeroth order result for the time-of-fall, T= V2h/g. Perform a calculation in second approximation (i.e., retain terms in w?) and calculate the southerly deflec- tion. There are three components to consider: (a) Coriolis force to second order (C), (b) variation of centrifugal force with height (C2), and (c) variation of gravi- tational force with height (Cs). Show that each of these components gives a result equal to h2 o sin A cos A with C 2/3, C2 5/6, and Cs = 5/2. The total southerly deflection is therefore (4h-w² sin A cos A)/g. to
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