The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the plane of the equator. Assuming the earth is a sphere with a radius of 6.38 106 m, determine the speed and centripetal acceleration of a person situated

(a) at the equator and
m/s
m/s2
(b) at a latitude of 40.0° north of the equator.
m/s
m/s2

(a) First of all, there are

T = 86,400 seconds
in a day. The distance travelled by a person at the equator in a day is

X = 2 pi R = 4.01*10^7 m
For the speed, use
speed V = X/T
acceleration a = V^2/R

(b) T is the same, but the radiis of the circle travelled at latitude 40 N as the earth turns once is
R' = R cos 40
use the same formulas as on (a), but with
X' = 2 pi R'
V = X'/T, and
V' = X'/T
and a = V'2/R'

To determine the speed and centripetal acceleration of a person situated at different locations on the Earth, we will need to use the following formulas:

Speed (v) = Distance / Time

Centripetal Acceleration (a) = (v² / r)

Given:
Radius of the Earth (r) = 6.38 × 10^6 m

(a) At the Equator:
The person at the equator is at a latitude of 0°, which means they are situated at the widest point of the Earth. Therefore, the distance they would travel in one full rotation is equal to the circumference of a circle with the radius of the Earth.

Circumference = 2πr

Substituting the values:
Circumference = 2π(6.38 × 10^6 m) = 40.08 × 10^6 m

Since the Earth rotates once per day, the time it takes for one rotation is 24 hours or 86,400 seconds.

Speed (v) = Distance / Time
Speed (v) = (40.08 × 10^6 m) / (86,400 s) ≈ 465.1 m/s

Centripetal Acceleration (a) = (v² / r)
Centripetal Acceleration (a) = (465.1 m/s)² / (6.38 × 10^6 m) ≈ 0.0341 m/s²

Therefore, at the equator, the speed is approximately 465.1 m/s, and the centripetal acceleration is approximately 0.0341 m/s².

(b) At a Latitude of 40.0° North of the Equator:
At this latitude, the person is not situated at the widest point of the Earth, so the distance they would travel in one full rotation would be shorter than the circumference of the Earth.

To calculate the distance, we need to find the circumference of the circle formed by the latitude of 40.0° north. This can be done by using the formula:

Circumference = 2πr × (cos(latitude))

Substituting the values:
Circumference = 2π(6.38 × 10^6 m) × (cos(40.0°)) ≈ 3.85 × 10^6 m

Using the same time for one full rotation as before (86,400 seconds), we can calculate the speed:

Speed (v) = Distance / Time
Speed (v) = (3.85 × 10^6 m) / (86,400 s) ≈ 44.6 m/s

Centripetal Acceleration (a) = (v² / r)
Centripetal Acceleration (a) = (44.6 m/s)² / (6.38 × 10^6 m) ≈ 0.03 m/s²

Therefore, at a latitude of 40.0° north of the equator, the speed is approximately 44.6 m/s, and the centripetal acceleration is approximately 0.03 m/s².

To determine the speed and centripetal acceleration of a person situated at different locations on Earth, we can use the formula for centripetal acceleration:

ac = v^2 / r

where ac is the centripetal acceleration, v is the velocity or speed, and r is the radius of rotation.

(a) At the Equator:
At the equator, the person is situated on a circle with a radius equal to the Earth's radius, which is 6.38 * 10^6 m.

1. Speed (v):
To calculate the speed, we need to determine the distance traveled by an object on the equator in one day, which is equal to the circumference of a circle with a radius of 6.38 * 10^6 m.

C = 2πr
C = 2π * 6.38 * 10^6 m

Since the Earth completes one rotation in 24 hours, the speed can be calculated using the formula:

v = distance / time

v = C / 24 hours
v = C / (24 hours * 3600 seconds/hour)

Now we can plug the values into the formula to find the speed:

v = (2π * 6.38 * 10^6 m) / (24 hours * 3600 seconds/hour)

Simplifying the calculation will give us the speed at the equator in meters per second (m/s).

2. Centripetal acceleration (ac):
Using the formula ac = v^2 / r, we can plug in the known values of v and r to find the centripetal acceleration at the equator.

ac = (v^2) / r

(b) At a Latitude of 40.0° North of the Equator:
To determine the speed and centripetal acceleration at a latitude of 40.0° north of the equator, we need to find the radius of rotation at that latitude.

1. Radius (r):
The radius at a given latitude can be calculated as the product of the equatorial radius and the cosine of the latitude angle:

r = (6.38 * 10^6 m) * cos(40.0°)

2. Speed (v):
Using the same formula as before, we can calculate the speed using the revised radius.

v = (2π * r) / (24 hours * 3600 seconds/hour)

3. Centripetal acceleration (ac):
Again, using ac = v^2 / r, we can find the centripetal acceleration.

ac = (v^2) / r

Now, if you substitute the appropriate values into the equations, you will get the speed and centripetal acceleration at the equator and 40.0° north of the equator.