An early major objection to the idea that

Earth is spinning on its axis was that Earth
would turn so fast at the equator that people
would be thrown into space.
Given : radius earth = 6.37 × 10^6 m ,
mass earth = 5.98 × 10^24 kg ,
radius moon = 1.74 × 10^6 m , and
g = 9.8 m/s^2.
Show the error in this logic by calculating
the speed of a 90.6 kg person at the equator

.

Answer: The speed of a 90.6 kg person at the equator can be calculated using the equation for centripetal acceleration:
a = (v^2)/r
where a is the centripetal acceleration, v is the velocity, and r is the radius of the Earth.

Substituting the given values, we get:
a = (v^2)/(6.37 × 10^6 m)

Solving for v, we get:
v = sqrt(a*r)

Substituting the given values, we get:
v = sqrt((9.8 m/s^2)*(6.37 × 10^6 m))

Therefore, the speed of a 90.6 kg person at the equator is 463.7 m/s, which is much lower than the escape velocity of the Earth (11.2 km/s). This shows that the objection is incorrect, as people would not be thrown into space due to the Earth's rotation.

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

Circumference = 2πr

where "r" is the radius of the Earth.

Given that the radius of the Earth is 6.37 × 10^6 m, we can substitute this value into the formula:

Circumference = 2π(6.37 × 10^6)

Calculating this further, we have:

Circumference = 2 × 3.14159 × 6.37 × 10^6
Circumference = 40.03 × 10^6

The circumference of the Earth at the equator is 40.03 × 10^6 meters.

Next, we can calculate the speed of a 90.6 kg person at the equator using the formula:

Speed = Circumference / Time

Since the Earth completes one full rotation in 24 hours, we can convert this to seconds:

Time = 24 hours × 60 minutes × 60 seconds
Time = 86,400 seconds

Now, we can calculate the speed:

Speed = (40.03 × 10^6 meters) / (86,400 seconds)
Speed = 463.2 meters/second

Therefore, the speed of a 90.6 kg person at the equator would be approximately 463.2 meters/second. As we can see, this speed is not nearly enough to throw a person into space, debunking the earlier objection.

To calculate the speed of a person at the equator, we first need to determine the distance traveled in one complete revolution.

The circumference of a circle is given by the formula:
Circumference = 2πr, where r is the radius.

For Earth, the radius is given as 6.37 × 10^6 m, so the circumference can be calculated as:
Circumference = 2π(6.37 × 10^6) = 2π × 6.37 × 10^6

Next, we need to calculate the time it takes for one complete revolution. The time for one revolution is equal to the period (T) of the Earth's rotation. The period of Earth's rotation is approximately 24 hours (or 86400 seconds).

Now, the speed of the person can be calculated using the formula:
Speed = Distance / Time

Plugging in the values, we have:
Speed = (2π × 6.37 × 10^6) / 86400

To calculate the speed, we multiply 2π by the radius of Earth (6.37 × 10^6) and then divide it by the period of Earth's rotation (86400).

Let's now perform the actual calculation:

Speed = (2π × 6.37 × 10^6) / 86400
= 3959.6 m/s

Therefore, the speed of a 90.6 kg person at the equator is approximately 3959.6 m/s.

By calculating this speed, we can clearly see that the speed at the equator is not even close to being fast enough for an object or person to be thrown into space.