(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 centripetal acceleration of an object on the Earth's equator due to the rotation of Earth, we need to use the formula for centripetal acceleration:
(a) The centripetal acceleration of an object on the Earth's equator can be determined using the formula:
ac = (v^2) / r
Where:
ac is the centripetal acceleration,
v is the velocity of the object,
r is the radius of the object's circular path.
In this case, the object on the Earth's equator is at a distance equal to the radius of the Earth from its center. The radius of the Earth (r) is approximately 6,371 kilometers (or 6,371,000 meters).
To find the velocity (v) of the object on the equator, we can use the formula:
v = rω
Where:
v is the velocity,
r is the radius of the object's circular path,
ω is the angular velocity of the Earth's rotation.
The angular velocity (ω) can be obtained by dividing the total angle covered by the Earth in one rotation by the time taken for one rotation. The Earth completes a full rotation in approximately 24 hours or 86,400 seconds.
ω = (2π radians) / (time taken for one rotation in seconds)
Now we can calculate the centripetal acceleration:
1. Calculate the angular velocity:
ω = (2π radians) / (86,400 seconds) ≈ 7.27 × 10^(-5) rad/s
2. Calculate the velocity:
v = rω = (6,371,000 m) × (7.27 × 10^(-5) rad/s) ≈ 464.5 m/s
3. Calculate the centripetal acceleration:
ac = (v^2) / r = (464.5 m/s)^2 / (6,371,000 m) ≈ 0.034 m/s^2
Therefore, the centripetal acceleration of an object on the Earth's equator due to the rotation of Earth is approximately 0.034 m/s^2.
(b) To find the Earth's rotation period for objects on the equator to have a centripetal acceleration equal to 9.8 m/s^2, we can rearrange the above formula to solve for the time taken for one rotation:
time taken for one rotation = (2π r) / v
Where:
r is the radius of the Earth,
v is the velocity of the object (which is given by the formula v = rω, as explained in part a).
Using the values mentioned above, we can calculate the rotation period:
1. Calculate the velocity:
v = rω = (6,371,000 m) × (7.27 × 10^(-5) rad/s) ≈ 464.5 m/s
2. Calculate the time taken for one rotation:
time taken for one rotation = (2π × 6,371,000 m) / 464.5 m/s ≈ 86,391 seconds
Therefore, for objects on the Earth's equator to have a centripetal acceleration equal to 9.8 m/s^2, the Earth's rotation period would have to be approximately 86,391 seconds or about 24.05 hours.