Early skeptics of the idea of a rotating Earth said that the fast spin of Earth would throw people at the equator into space. The radius of Earth is about 6400 km. Show why this objection is wrong by determining the following information.

(a) Calculate the speed of a 94-kg person at the equator.

(b) Calculate the force needed to accelerate the person in the circle.
(c) Calculate the weight of the person.
(d) Calculate the normal force of Earth on the person, that is, the person's apparent weight.

To calculate the speed of a person at the equator, we can use the formula for the circumference of a circle:

C = 2πr

Where C is the circumference and r is the radius of the Earth.

(a) The speed of the person at the equator can be found by dividing the distance traveled (circumference) by the time taken for one rotation (24 hours or 86,400 seconds).

Speed = Circumference / Time

Speed = (2π * 6400 km) / (86,400 s)

Speed ≈ 464.8 m/s

(b) The force needed to accelerate the person in a circle is the centripetal force. The formula for centripetal force is:

F = m * v^2 / r

Where F is the force, m is the mass of the person, v is the speed, and r is the radius of the circle (in this case, the radius of the Earth).

F = 94 kg * (464.8 m/s)^2 / 6400 km

Note: For consistent units, convert km to meters.

F ≈ 3321 N

(c) The weight of the person can be calculated using the formula:

Weight = m * g

Where m is the mass of the person and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Weight = 94 kg * 9.8 m/s^2 ≈ 922.2 N

(d) The normal force of Earth on the person (the person's apparent weight) is equal in magnitude but opposite in direction to the weight.

Normal Force = - Weight

Normal Force ≈ -922.2 N

To determine the speed of a person at the equator, we can use the formula for the linear velocity of a rotating object:

v = ω * r,

where v is the linear velocity, ω is the angular velocity, and r is the radius.

(a) To find the speed of a person at the equator, we need to know the angular velocity of the Earth's rotation. The Earth completes one rotation in approximately 24 hours, which is equivalent to 2π radians. Therefore, the angular velocity is:

ω = 2π radians / (24 hours * 3600 seconds/hour)
≈ 7.27 × 10^(-5) radians/second.

The radius of the Earth is given as 6400 km, which is equivalent to 6400*1000 m. Plugging these values into the equation, we can calculate the speed:

v = (7.27 × 10^(-5) radians/second) * (6400 * 1000 m)
≈ 465.6 m/second.

So, the speed of a person at the equator is approximately 465.6 m/second.

(b) To calculate the force needed to accelerate the person in a circle, we can use the formula for centripetal force:

F = m * a_c,

where F is the force, m is the mass of the person, and a_c is the centripetal acceleration.

The centripetal acceleration can be calculated using the formula:

a_c = v^2 / r,

where v is the speed and r is the radius.

Plugging in the given values, we find:

a_c = (465.6 m/second)^2 / (6400 * 1000 m)
≈ 3.39 m/second^2.

Now we can calculate the force:

F = (94 kg) * (3.39 m/second^2)
≈ 318.66 N.

So, the force needed to accelerate the person in the circle is approximately 318.66 Newtons.

(c) The weight of the person can be calculated using the formula:

Weight = mass * gravity,

where mass is the person's mass and gravity is the acceleration due to gravity, approximately 9.8 m/second^2.

Plugging in the given value:

Weight = (94 kg) * (9.8 m/second^2)
≈ 921.2 N.

So, the weight of the person is approximately 921.2 Newtons.

(d) The normal force of Earth on the person, also known as the apparent weight, is equal to the person's weight plus the centripetal force:

Normal force = Weight + F,

where Weight is the person's weight and F is the centripetal force.

Plugging in the calculated values:

Normal force = 921.2 N + 318.66 N
≈ 1239.86 N.

So, the normal force of Earth on the person, or the person's apparent weight, is approximately 1239.86 Newtons.

The order of these makes this very simple. I will be happy to critique your work.