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I could really use some help with part b, and c here.
1.8. Find the horizontal displacement due to the rotation of the Earth of a body
dropped from a fixed platform at height h at the equator, neglecting the effects
.of air resistance. What is the displacement if h = 5 km?
Notes:
Once again we write down the information we have, the equations we have,
manipulate them as much as necessary and then plug in the numbers.
We learned in the intro classes that there is no coriolis on the equator. This is not
exactly true. There is no horizontal coriolis force, and even that is only true if there
are no vertical motions (all of the terms that have sine phi will go away at the
equator, but the ones that have cosine will not). In meteorology, most motions in the
synoptic scales are horizontal, so the assumption of zero coriolis at the equator is
OK.
Unlike the first problem where we had a constant speed from the throw, this time
the drop will not have a constant speed (since the object will accelerate with gravity,
therefore, w is not a constant but it is the result of an acceleration dw/dt=-g
a. Write down a list of the given and asked information.
b. Write your available equations (in this case the three components of the
coriolis acceleration du/dt, dv/dt, dw/dt and the vertical acceleration due to
gravity as explained in the notes above)
bean rovew
hnex
06 diw qu bs bluode noY
tonla moldorg odi lo bno od ton
L NOW substtiute your
displaceent isshisan
The answerls -7.7m)
C. Cancel out the unnecessary terms in part b – do it right there (there are no
initial horizontal motions) and write down the useful equation.
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Transcribed Image Text:1.8. Find the horizontal displacement due to the rotation of the Earth of a body dropped from a fixed platform at height h at the equator, neglecting the effects .of air resistance. What is the displacement if h = 5 km? Notes: Once again we write down the information we have, the equations we have, manipulate them as much as necessary and then plug in the numbers. We learned in the intro classes that there is no coriolis on the equator. This is not exactly true. There is no horizontal coriolis force, and even that is only true if there are no vertical motions (all of the terms that have sine phi will go away at the equator, but the ones that have cosine will not). In meteorology, most motions in the synoptic scales are horizontal, so the assumption of zero coriolis at the equator is OK. Unlike the first problem where we had a constant speed from the throw, this time the drop will not have a constant speed (since the object will accelerate with gravity, therefore, w is not a constant but it is the result of an acceleration dw/dt=-g a. Write down a list of the given and asked information. b. Write your available equations (in this case the three components of the coriolis acceleration du/dt, dv/dt, dw/dt and the vertical acceleration due to gravity as explained in the notes above) bean rovew hnex 06 diw qu bs bluode noY tonla moldorg odi lo bno od ton L NOW substtiute your displaceent isshisan The answerls -7.7m) C. Cancel out the unnecessary terms in part b – do it right there (there are no initial horizontal motions) and write down the useful equation.
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