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Saad
Mar27-04, 12:24 AM
Assume that the Earth is a perfect sphere of diameter 1.274 X 10^7 m. If an object has a weight of 100 N while on a scale at the south pole, how much will it weigh at the equator? Assume that the equatorial spin speed, v = 465 m/s.

enigma
Mar27-04, 01:24 AM
Saad,

Forum rules state that you should at least attempt the problem before you ask for help on a problem.

If you have attempted them, please post what you've tried, so we can find where your misunderstandings or confusions lie.

Us doing the problem for you isn't going to help you learn it.

ShawnD
Mar27-04, 04:54 AM
Gravity pulls stuff towards earth. Velocity of earth tries to throw away from earth. Weight at the poles is due to 100% gravity.

weight at equator = 100N - mv^2/r

Saad
Mar31-04, 04:26 PM
I tried this question the way i was told to..but i dunt think it works..plz help

Variables:
r = 1.274 X 10^7 m
v = 465 m/s
Weight = 100N at South Pole
Mass = F / g = 100N / 9.8 = 10.2kg

Gravity pulls stuff towards earth. Velocity of earth tries to throw away from earth. Weight at the poles is due to 100% gravity.

Weight at equator = 100N - mv^2/r
= 100N – (10.2kg)(465 m/s)^2 / 1.274 X 10^7 m
= 100N – 2205495 / 1.274 X 10 ^7 m
= -2205395 / 1.274 X 10 ^7
= -0.173N
Therefore the weight of the object at the equator is –0.173N?????

ShawnD
Mar31-04, 04:46 PM
Using the numbers you gave, I get 99.82688422

Saad
Mar31-04, 04:52 PM
Ohh my badd..your right...but is the mass i used in the equation rite?
and also..does this answer seem approriate, because it is still approximately 100N.

ShawnD
Mar31-04, 04:54 PM
Yes your mass is right and the answer is reasonable.