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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.
Saad,
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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
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?????
Using the numbers you gave, I get 99.82688422
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.
Yes your mass is right and the answer is reasonable.
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