Back in Galileo's day, one of the objections to the heliocentric model of the solar system is that if the earth is spinning, we should all be "thrown off the earth." Actually, you do weigh a bit less on the equator than you would at the poles. Calculate how much. (Hint: Construct force diagrams for a 100 kg person standing on a bathroom scale at the equator and at the pole, and do the Fnet calculations.\

at the equator the normal force will be 3.4N less and at the North Pole = 980N

Explanation:

So, we will start the Calculation by considering the equation below;

Fnet = F(g) - F(N); where Fnet = mv^2/ r.

Hence, F(g) - F(N) = mv^2/ r (1).

Making F(N) the subject of the formula, we have;

F(N) = Gm1m2/r^2 - mv^2/r.

Thus, we have;

Gm1m2/r^2 = 6.68 × 10^-11 × (5.98 × 20^24) × 100/ 6.38 × 10^6 = 980N.

Also, we have;

mv^2/r = m4π^2r/T^2.

100 × 6.38 ×10^6 × 4π^2/(24 × 3600)^2 = 3.4N

Therefore, at the equator the normal force will be 3.4N less and at the North Pole = 980N.

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this should answer you're question