Gravitational Acceleration at Different Latitudes

AI Thread Summary
The discussion focuses on how Earth's rotation affects gravitational acceleration at different latitudes, specifically comparing weight at the Equator and the North Pole. A man weighing 709.7 N at the Equator would weigh slightly less at the North Pole due to centrifugal force from Earth's rotation. Calculations indicate that the gravitational acceleration would be reduced by approximately 0.034 m/s² at the North Pole. The conversation includes corrections on terminology and units used in the calculations. Overall, the effects of Earth's shape and rotation on weight are emphasized.
duggielanger
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Homework Statement


The rotation of the Earth affects the apparent gravitational acceleration at
different latitudes. At a location on the Equator, a man’s weight is registered
as 709.7 N on a set of very accurate scales. Assuming a perfectly spherical
Earth, determine what the same set of scales would register for the same
man’s weight at the North Pole



Homework Equations


Not sure I have used f=ma and w=mg but don't think I need these as the Earth is perfectly spherical
So it might be a=v^2/r


The Attempt at a Solution


As far as I can figure if Earth is perfectly spherical then there will be no change in weight due to the force of gravity. But maybe because of the rotation and Centrifugal force this might have a slight effect.
 
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duggielanger said:

Homework Statement


The rotation of the Earth affects the apparent gravitational acceleration at
different latitudes. At a location on the Equator, a man’s weight is registered
as 709.7 N on a set of very accurate scales. Assuming a perfectly spherical
Earth, determine what the same set of scales would register for the same
man’s weight at the North Pole



Homework Equations


Not sure I have used f=ma and w=mg but don't think I need these as the Earth is perfectly spherical
So it might be a=v^2/r


The Attempt at a Solution


As far as I can figure if Earth is perfectly spherical then there will be no change in weight due to the force of gravity. But maybe because of the rotation and Centrifugal force this might have a slight effect.
Correct.
So, how much effect ?
 


So if I find the angular velocity first ω=2∏rad/T and then v=rω^2
Plugged in values
2*3.14/(24h*3600s/h)=7.3x10^-5 rads
v=6.378x10^6m/(7.3x10^-5)^2rads=0.034m/s^2
So the force of gravity will be 0.034m/s^2 less
and then I can find the weight .
Dose this seem right
 


duggielanger said:
So if I find the angular velocity first ω=2∏rad/T and then v=rω^2
Plugged in values
2*3.14/(24h*3600s/h)=7.3x10^-5 rads/s2
v=6.378x10^6m/(7.3x10^-5)^2rads/s2=0.034m/s^2
So the [STRIKE]force[/STRIKE] acceleration of gravity will be 0.034m/s^2 less
and then I can find the weight .
Does this seem right
Yes, except for some units and saying force rather than acceleration .

(Corrected above)
 


Ok thanks A lot SammyS your help is appreciated
 


And ... I forgot to say,

Welcome to PF !
 

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