(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.
(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?
1 answer