How Does Earth's Rotation Affect Weight at the Equator Compared to the Poles?

[Saad]

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]

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]

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]

Plzz helpp

I tried this question the way i was told to..but i dunt think it works..please 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]

Using the numbers you gave, I get 99.82688422

 

[Saad]

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]

Yes your mass is right and the answer is reasonable.