(a) What is the centripetal acceleration of an object on the Earth's equator due to the rotation of Earth?
(b) What would the Earth's rotation period have to be for objects on the equator to have a centripetal acceleration equal to 9.8 m/s2?
To find the answers to these questions, we need to consider the concepts of centripetal acceleration and the rotation of the Earth.
(a) The centripetal acceleration of an object at the Earth's equator can be calculated using the formula:
a = (v^2) / r
where:
- "a" is the centripetal acceleration,
- "v" is the linear velocity of the object,
- "r" is the radius of the Earth.
To find the linear velocity, we can use the formula:
v = (2πr) / T
where:
- "T" is the rotation period of the Earth.
Substituting the value of linear velocity into the centripetal acceleration formula, we get:
a = ((2πr / T)^2) / r
Simplifying further, we get:
a = (4π^2r) / T^2
The radius of the Earth, "r," is approximately 6371 kilometers or 6371000 meters.
(b) Given that the centripetal acceleration, "a," is equal to 9.8 m/s^2, we can solve for the rotation period, "T," using the same equation:
9.8 = (4π^2 * 6371000) / T^2
Rearranging the equation, we get:
T^2 = (4π^2 * 6371000) / 9.8
Taking the square root of both sides gives us the rotation period, T.
To find the numerical value, you can substitute the values into a calculator.
By following these steps, you can find the centripetal acceleration at the Earth's equator due to its rotation and determine the rotation period required for objects on the equator to have a centripetal acceleration of 9.8 m/s^2.