Question

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.\

Answer 1

The weight of a person decreases slightly at the equator compared to the poles due to the centrifugal force from the Earth's rotation and the oblate shape of the Earth. To calculate the apparent weight difference, we must account for both the centrifugal force and the difference in gravitational strength caused by the Earth's increased radius at the equator.

Step-by-step explanation:

The question involves calculating the difference in weight (apparent weight) for a 100 kg person standing on a bathroom scale at the equator compared to the poles, accounting for the effect of Earth's rotation. At the equator, the rotational motion of the Earth generates a centrifugal force that slightly reduces the gravitational force experienced by the person, leading to a lower apparent weight on the scale. Additionally, the shape of the Earth causes the gravitational force to be slightly weaker at the equator than at the poles due to the increased radius.

To calculate the apparent weight reduction at the equator, we start with the force diagram. The true weight (W = mg) points toward Earth's center, and the scale force (Fs) points away. At the poles, Fs equals the true weight because there's no rotational effect. At the equator, the net force must be adjusted for the outward centrifugal force. The centrifugal force Fc can be determined by multiplying the mass by the centripetal acceleration (ac) due to Earth's rotation. The apparent weight at the equator is then Fs = mg - Fc. The apparent weight differs at the two locations because of the combination of the centrifugal reduction and the weaker gravity due to Earth's oblate shape.

Answer 2

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

Step-by-step 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.