Question

What would be the length of a day(that is, the time required forone

rotation of the earth on its axis ) if the rate of rotation ofthe
earth were such that g=0 at the equator?

Answer 1

T = 5,06 10³ s

Step-by-step explanation:

For this exercise, let's use Newton's second law, where force is the force of gravitational attraction

F = m a

The acceleration is centripetal

a = v² / r

Let's replace

G m
`M_(e)` /
`R_(e)`² = m v² /
`R_(e)`

G
`M_(e)` /
`R_(e)` = v²

The velocity module is constant, so we can use the equation of uniform motion

v = d / t

Where the distance is the length of the circle and in this case the time is called the period

d = 2π
`R_(e)`

We replace

(2π
`R_(e)` / T)² = G
`M_(e)` /
`R_(e)`

T² = 4π²
`R_(e)`³ / G
`M_(e)`

T = √ 4π²
`R_(e)`³ / G
`M_(e)`

Let's calculate

T = √ (4π² (6.37 10⁶)³ / 6.67 10⁻¹¹ 5.98 10²⁴)

T = √ (255.83 10⁵) = √ (25.583 10⁶)

T = 5,05796 10³ s