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 at each of the following.

(a) the equator
1 m/s
2 m/s2

(b) a latitude of 42.0° north of the equator
3 m/s
4 m/s2

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

Speed (v) = angular velocity (ω) × radius (r)
Centripetal acceleration (a) = ω² × radius (r)

Note: The angular velocity (ω) is the rate at which an object rotates, and is equal to 2π radians divided by the time it takes to complete one full rotation. In the case of the Earth, this time is one day.

(a) At the equator:
The person is situated at the equator, which means they are at a latitude of 0°. Since the axis of rotation passes through the poles, the radius (r) is the same as the Earth's radius, which is given as 6.38 × 10^6 m.

To find the speed (v):
ω = (2π radians) / (time for one rotation)
The time for one rotation is equal to the time for one day, which is 24 hours or 86,400 seconds.
ω = (2π radians) / (86,400 seconds)
v = ω × r

To find the centripetal acceleration (a):
a = ω² × r²

Now we can calculate the values:
v = (2π radians / 86,400 seconds) × (6.38 × 10^6 m) ≈ 464.1 m/s
a = ((2π radians / 86,400 seconds)²) × ((6.38 × 10^6 m)²) ≈ 0.0335 m/s²

Therefore, the speed of a person at the equator is approximately 464.1 m/s, and the centripetal acceleration is approximately 0.0335 m/s².

(b) At a latitude of 42.0° north of the equator:
The person is located at a latitude of 42.0° north, which means they are not at the equator. In this case, the radius (r) needs to be calculated using the latitude and Earth's radius.

To find the new radius (r):
r = Earth's radius × cos(latitude)
r = (6.38 × 10^6 m) × cos(42.0°)

Now we can use the same formulas as above to calculate the speed (v) and centripetal acceleration (a):
v = ω × r
a = ω² × r²

Calculating the values:
r = (6.38 × 10^6 m) × cos(42.0°) ≈ 4.66 × 10^6 m
v = (2π radians / 86,400 seconds) × (4.66 × 10^6 m) ≈ 335.7 m/s
a = ((2π radians / 86,400 seconds)²) × ((4.66 × 10^6 m)²) ≈ 0.0131 m/s²

Therefore, the speed of a person at a latitude of 42.0° north of the equator is approximately 335.7 m/s, and the centripetal acceleration is approximately 0.0131 m/s².