Gravitation person weight problem

[melodrameric]

here's the question (from mastering physics):
"The weight of a person at the equator, as determined by a spring balance, is 750 N. By how much does this differ from the true force of gravitational attraction at the same point? Assume that the Earth is spherically symmetric."

Now, i would assume what needs to be done is first finding the mass of the person (750 N / 9.8 m/s^2). then plugging that into the equation:
F = (G*m*M_e)/(R_e^2), using M_e = 5.92*10^24 kg and R_e = 6.38*10^6 m
Then, subtracting that from 750 N. I did this, and came out with 3 different answers depending on my value for acceleration (9.8 or 9.81) and how many figures i used from that figured mass. They were 1.62, 2.09, and 1.32. Mastering Physics told me those were all wrong.
...what?

 

[James R]

Looks like you didn't take the rotation of the Earth into account.

 

[mezarashi]

The question is making you realize the fact that part of our weight is gone to the centripetal acceleration. That statement may be a bit confusing, but if you think about it... if the Earth spun 100 times faster than it did, would you feel lighter? Now if the Earth didn't spin at all, how much heavier would you feel? The question is essentially asking you the latter question.

Hint: Find the centripetal force needed to keep you in circular motion around the equator of the Earth.

 

[melodrameric]

wow, how nice of mastering physics to add in the effect of rotation without the slightest bit of warning.
anyway, i checked my notes, and here's what i have written down:
V_person = (2*pi*R_e)/(T) = (2*pi*(6.38*10^6 m))/(24*3600 s) = 463.97 m/s
ΣF_r=m*a_r
n - m*g_0 = m*(-(V_p^2)/R_e)
n = m*g_0 - m*((V_p^2)/R_e)

so is it that last equation i should be using?
with that, i would come out with 747.4 N, with the difference being 2.58 N.

 

[melodrameric]

well i went ahead and entered that answer and got it right. thanks for your help :)