Centripetal Acceleration

Please could someone help:

Calculate the centripetal accelerations as fractions or multiples of g and comment on the results:
i)the acceleration towards the earth's axis of an object resting on the earth's surface at 45degree latitude. the earth's radius is 6378km. [ans]

Thanks

Reply 1

elpaw

at 45 degrees, your radius of rotation will be 6378/root 2 km = 4510 km, and thus the circumfrence will be 28336.7 km.

the earth rotates once every 24 hours, thus the speed of rotation at that latitude is (28336707/24*60*60) = 328 m/s.

centripetal acceleration = v²/r = 328²/4510000 = 0.0239 m/s², which as a fraction of g is 0.00243

comment: at 45 degrees, the acceleration of gravity resolved towards the centre of rotation (not the centre of the earth) = g/root2 = 6.93 m/s². thus we can say that the force of gravity provides enough centripetal acceleration that we do not fly off the earth. which is obvious.

Reply 2

john !!

dotcom

ok at 45 degrees latitude you are halfway between the equator and the pole I guess? So you really want the radius (r) of the circle you are spinning in. Using trigonometry, (call the Eart'hs radius R)

r = R cos 45 = R/(√2) = (R√2)/2 = (6378√2)/2 = 4510 km (4s.f.)
r = 4.510*10^6 m (4s.f.)

Now you can find the acceleration either from ω or v (ω is easier):

ω = Δθ/Δt = 2pi/(24*60*60) = 7.272*10^-5 s^-1 (in radians) (4s.f.)

a = rω² = 4510000 * (4.510*10^6)^2 = 2.39*10^-2 m s^-2

And I'm one place out... ffs
the maths isnt treating me nicely today :mad: