Earth is a spherical body, of radius 6,376 km, which completes a full rotation about its axis in 24.0 h. how long (in hours) should the day on Earth be so that a person at the equator is able to float freely above the ground?

acceleration due to gravity=9.8m/s^2

centripetal acceleration=w^2 *r (r is given, w=2PI/(T*3600) rad/s
set them equal, solve for Period T.

9.8=4PI^2/(T^2*3600^2)

T=2PI/3600 sqrt(1/9.8) check my math, T is in hours.

what is PI?

In order for a person to float freely above the ground at the equator, the centrifugal force due to the Earth's rotation needs to balance out the force of gravity. The centrifugal force is given by the equation:

F = m * ω^2 * r

where F is the centrifugal force, m is the mass of the person, ω is the angular velocity of Earth's rotation, and r is the radius of the Earth.

At the equator, the centrifugal force is equal to the force of gravity, so we can equate the two:

m * ω^2 * r = m * g

where g is the acceleration due to gravity.

We can rearrange the equation to solve for ω:

ω^2 = g / r

ω = √(g / r)

Given that the radius of the Earth is 6,376 km (or 6,376,000 meters) and the acceleration due to gravity is approximately 9.8 m/s^2, we can substitute the values in the equation:

ω = √(9.8 / 6,376,000)

Calculating this, we get:

ω ≈ 0.00000727 rad/s

The angular velocity ω is given by the equation:

ω = 2π / T

where T is the period or length of a day in hours.

Substituting the known values:

0.00000727 rad/s = 2π / T

Rearranging the equation to solve for T:

T = 2π / 0.00000727

Calculating this, we find:

T ≈ 86157.9 hours

So, to be able to float freely at the equator, the length of a day on Earth should be approximately 86157.9 hours.

To determine the length of a day on Earth so that a person at the equator can float freely above the ground, we need to consider the concept of centrifugal force and gravitational force.

Centrifugal force is the apparent force that tends to pull an object away from the center of rotation. In the case of Earth, the rotation creates centrifugal force that slightly counteracts the force of gravity. When the two forces balance, an object can float above the ground without falling down.

The formula for centrifugal force is given by:

Fc = m * ω^2 * r

Where:
Fc is the centrifugal force
m is the mass of the object
ω (omega) is the angular velocity, which is given by 2π divided by the period of rotation (T)
r is the radius of the Earth

To calculate the length of a day (T') needed for a person to float, we can equate the centrifugal force to the gravitational force:

Fc = Fg

m * ω^2 * r = m * g

Where:
g is the acceleration due to gravity which is approximately 9.8 m/s^2

Now, we can substitute the values into the equation:

m * (2π / T')^2 * r = m * g

Since the mass (m) cancels out, we can simplify the equation:

(2π / T')^2 * r = g

Solving for T', we get:

T' = 2π * √(r / g)

Let's substitute the given values:
r = 6,376 km = 6,376,000 meters
g = 9.8 m/s^2

T' = 2π * √(6,376,000 / 9.8)

Now, let's calculate T':
T' ≈ 2π * √(651836.7) ≈ 2π * 807.8 ≈ 5076.66

Therefore, the day on Earth needs to be approximately 5076.66 hours long for a person at the equator to float freely above the ground.