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 103-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.
a.) To calculate the speed of a person at the equator, we need to use the formula for the linear speed of a point on a rotating object. The linear speed (v) can be calculated using the formula:
v = r * ω
where:
v = linear speed
r = radius of the Earth
ω = angular speed
The radius of the Earth is given as 6400 km, which in meters is 6400 * 1000 = 6,400,000 meters.
The angular speed (ω) can be calculated using the formula:
ω = 2π / T
where:
ω = angular speed
π = pi (approximately 3.14)
T = period of one rotation (24 hours for Earth)
Plugging in the values:
ω = 2π / 24 hours = 2π / (24 * 60 * 60 seconds) = 2π / 86,400 seconds
Now, substituting the values of r and ω into the formula for linear speed:
v = 6,400,000 meters * (2π / 86,400 seconds)
v ≈ 463.3 meters/second
Therefore, the speed of a person at the equator is approximately 463.3 meters/second.
b.) To calculate the force needed to accelerate the person in the circle, we can use Newton's second law of motion:
F = m * a
where:
F = force
m = mass of the person
a = acceleration (centripetal acceleration)
The centripetal acceleration can be calculated using the formula:
a = v^2 / r
Plugging in the values:
a = (463.3 meters/second)^2 / 6,400,000 meters
a ≈ 0.033 meters/second^2
Now, substituting the value of a and the mass of the person (103 kg) into the formula for force:
F = 103 kg * 0.033 meters/second^2
F ≈ 3.399 N
Therefore, the force needed to accelerate the person in the circle is approximately 3.399 Newtons.
c.) The weight of the person can be calculated using the formula:
weight = mass * gravitational acceleration
The gravitational acceleration on Earth is approximately 9.8 meters/second^2.
Substituting the mass of the person (103 kg) into the formula:
weight = 103 kg * 9.8 meters/second^2
weight ≈ 1,008.4 N
Therefore, the weight of the person is approximately 1,008.4 Newtons.
d.) The normal force of Earth on the person, which is the person's apparent weight, can be calculated using the formula:
apparent weight = weight - centripetal force
Substituting the values we calculated earlier:
apparent weight = 1,008.4 N - 3.399 N
apparent weight ≈ 1,005.001 N
Therefore, the normal force of Earth on the person, or the person's apparent weight, is approximately 1,005.001 Newtons.
By calculating the various factors involved, we can see that the force required to keep a person at the equator is relatively small compared to their weight and the downward gravitational force from Earth. Thus, the objection that the fast spin of Earth would throw people at the equator into space is clearly wrong.