Circular Motion of Newton's Laws

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To achieve zero apparent weight for an object at the equator, the rotational period must be calculated using the formula T = 2π√(R/g), where R is the radius of the Earth and g is the acceleration due to gravity. The radius R is approximately 6,371 kilometers. For part (b), the discussion seeks to determine the factor by which the rotational speed must increase to achieve this condition. Participants express confusion about finding R and the specifics of the speed increase factor. The conversation emphasizes the need for numerical answers and clarification on the underlying physics principles.
eunhye732
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The Earth rotates about its axis with a period of 24.0 h. Imagine that the rotational speed can be increased. If an object at the equator is to have zero apparent weight,
(a) what must the new period be?
(b) by what factor would the speed of the object be increased when the planet is rotating at the higher speed? Note that the apparent weight of the object becomes zero when the normal force exerted on it is zero.

For part (a), I know that T= 2(pie)sqrt(R/g)
I was wondering how you find R.
And I don't understand what (b) is asking for.

Thanks!
ps: if this question has already been asked, I'm sorry! :rolleyes:
 
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eunhye732 said:
I was wondering how you find R.

What circle does the object describe in 24 hrs ? What's its radius ?
(ok, correct, the data are NOT in the problem description ; they must be considered general culture :-)
 
well, my teacher told me that it's suposed to be a numberical answer. Thanks anyways
 
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