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

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Homework Help Overview

The discussion centers around the effects of Earth's rotation on weight, specifically comparing the weight of an object at the equator to its weight at the poles. The original poster presents a scenario involving an object with a known weight at the south pole and seeks to determine its weight at the equator, incorporating the effects of rotational speed.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between gravitational force and the centrifugal effect due to Earth's rotation. The original poster attempts calculations using the formula for weight adjustment based on rotational speed and radius. Questions arise regarding the validity of the mass used in calculations and the appropriateness of the resulting weight.

Discussion Status

Some participants provide guidance on the calculations and confirm the mass used is correct. There is a recognition of the reasonableness of the final answer, though the discussion reflects a lack of explicit consensus on the interpretation of the results.

Contextual Notes

Participants note the importance of attempting the problem before seeking help, in line with forum rules. The original poster expresses uncertainty about their approach and results, indicating a need for clarification on the concepts involved.

Saad
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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.
 
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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.
 
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
 
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?
 
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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