# The rotation of the Earth affects the apparent gravitational acceleration at differ

#### duggielanger

1. The problem statement, all variables and given/known data
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

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

3. 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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#### SammyS

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Re: The rotation of the Earth affects the apparent gravitational acceleration at dif

1. The problem statement, all variables and given/known data
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

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

3. 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 ?

#### duggielanger

Re: The rotation of the Earth affects the apparent gravitational acceleration at dif

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

#### SammyS

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Science Advisor
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Re: The rotation of the Earth affects the apparent gravitational acceleration at dif

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)

#### duggielanger

Re: The rotation of the Earth affects the apparent gravitational acceleration at dif

Ok thanks A lot SammyS your help is appreciated

#### SammyS

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Science Advisor
Homework Helper
Gold Member
Re: The rotation of the Earth affects the apparent gravitational acceleration at dif

And ... I forgot to say,

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