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The Earth rotates eastward at speed s=500m/s at equator. Airplane flies in due west direction along equator at speed v=s=500m/s wrt earth.

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When I do this problem in the frame of refence rotating with the earth, then:

forces acting on the plane are:
gravity mg (downwards)
centrifugal force mω²R = ms²/R (upwards)
airlift F (upwards)

dynamic equation for the airplane is
ma = F - mg + ms²/R
Accleration a is centripetal accleration:
a = -v²/R (minus signifies downward direction)
Final equation for airlift:
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F = m(g - 2s²/R)
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But when I consider this problem in inertial frame of reference, the airplane is at rest, and I get simply
****************
F = mg
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2007-08-27 07:18:39 · 4 answers · asked by Alexander 6 in Science & Mathematics Physics

4 answers

I don't think there is any centrifugal force acting on the plane per se. Gravity provides the centripetal force which results in circular motion.

The dynamic equation should be
ma = -mv²/R = F - mg
where a = acceleration in +ve r direction

For the plane flying on the opposite direction to the earth, the absolute speed in space is zero.
So F - mg = 0
F = mg

2007-08-27 15:48:47 · answer #1 · answered by Dr D 7 · 0 0

Inertially the plane is at rest but neither the Earth nor its atmosphere is, so the aero lift equation applies. And at 1800 km/hr it'll apply a lot! You're right that the centrifugal term vanishes.

2007-08-27 10:57:45 · answer #2 · answered by kirchwey 7 · 0 0

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2016-12-16 06:18:47 · answer #3 · answered by ? 4 · 0 0

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http://www.groupsrv.com/science/about266577-0.html

2007-08-27 08:01:55 · answer #4 · answered by Anonymous · 0 1

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