(a)What is the magnitude of centripetal acceleration due
to the daily rotation of an object at Earth’s equator?
The equatorial radius is 6.38 106 m.
(b)How does this acceleration affect a person’s weight at
the equator?
(a) The centripetal acceleration due to the daily rotation of an object at Earth's equator can be calculated using the formula:
a = rω^2
Where:
a = centripetal acceleration
r = radius of the Earth (6.38 * 10^6 m)
ω = angular velocity
To find the angular velocity, we can use the formula:
ω = 2π / T
Where:
T = period of rotation (24 hours = 86400 seconds)
ω = 2π / 86400 = 7.27 * 10^-5 rad/s
Now, we can calculate the centripetal acceleration:
a = (6.38 * 10^6) * (7.27 * 10^-5)^2
a ≈ 0.0337 m/s^2
So, the magnitude of the centripetal acceleration due to the daily rotation of an object at Earth's equator is 0.0337 m/s^2.
(b) The centripetal acceleration affects a person's weight at the equator by slightly reducing their apparent weight. This is because the centripetal acceleration due to the Earth's rotation counteracts a small part of the gravitational force that is pulling the person towards the Earth's center. As a result, the person would feel slightly lighter at the equator compared to at the poles where there is no such centripetal acceleration.