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 x 106 m, determine the centripetal acceleration of a person situated (a) at the equator and (b) at a latitude of 52.0 ° north of the equator.
To determine the centripetal acceleration of a person situated at the equator, we need to find the linear speed of the person first, since the centripetal acceleration depends on the linear speed.
The linear speed of a point on the surface of the Earth can be calculated using the formula:
v = rω
Where:
v is the linear speed
r is the radius of the Earth
ω is the angular velocity
Since the Earth rotates once per day, the angular velocity can be calculated as:
ω = 2π / T
Where:
T is the period of rotation, which is 24 hours or 86400 seconds.
Substituting the values into the equation:
ω = 2π / 86400
Next, we substitute the angular velocity into the linear speed formula:
v = (6.38 x 10^6 m) * (2π / 86400 s)
Simplifying the equation gives us the linear speed, which is the same as the centripetal speed of a person at the equator.
(a) v = (6.38 x 10^6 m) * (2π / 86400 s) ≈ 463.83 m/s
Now, we can find the centripetal acceleration using the formula:
a = v^2 / r
Substituting the values:
(a) a = (463.83 m/s)^2 / (6.38 x 10^6 m)
Simplifying the equation gives us the centripetal acceleration at the equator.
(a) a ≈ 0.0337 m/s^2
To determine the centripetal acceleration of a person at a latitude of 52.0° north of the equator, we need to consider the fact that the radius of the circular path on which the person is located will be smaller.
The radius of the circular path depends on the latitude θ and can be calculated using the equation:
rθ = r * cos(θ)
Where:
rθ is the radius at latitude θ
r is the radius of the Earth, 6.38 x 10^6 m
θ is the latitude of the person, 52.0°
Substituting the values:
rθ = (6.38 x 10^6 m) * cos(52.0°)
(b) rθ ≈ (6.38 x 10^6 m) * 0.6156 ≈ 3.93 x 10^6 m
Now, we can calculate the linear speed at the person's latitude using the same formula as before:
v = rθ * ω
(b) v = (3.93 x 10^6 m) * (2π / 86400 s) ≈ 288.72 m/s
Finally, we can find the centripetal acceleration using the formula:
a = v^2 / rθ
Substituting the values:
(b) a = (288.72 m/s)^2 / (3.93 x 10^6 m)
Simplifying the equation gives us the centripetal acceleration at a latitude of 52.0° north of the equator.
(b) a ≈ 0.0213 m/s^2
To determine the centripetal acceleration of a person situated at the equator, we can use the equation:
a = ω^2 * r
Where:
- a is the centripetal acceleration
- ω is the angular velocity
- r is the distance from the axis of rotation
At the equator, the person is located on the circumference of the Earth, so the distance from the axis of rotation (r) is equal to the radius of the Earth (6.38 x 10^6 m).
The angular velocity (ω) can be calculated using the formula:
ω = 2π / T
Where:
- ω is the angular velocity
- T is the period of rotation
Since the Earth completes one full rotation in 24 hours, the period of rotation (T) is 24 hours or 24 x 3600 seconds.
Now, we can substitute the values in the equations to find the centripetal acceleration at the equator:
ω = 2π / (24 x 3600 s) = 7.27 x 10^-5 rad/s
a = (7.27 x 10^-5 rad/s)^2 * 6.38 x 10^6 m
≈ 0.034 m/s^2
Therefore, the centripetal acceleration of a person at the equator is approximately 0.034 m/s^2.
To determine the centripetal acceleration of a person at a latitude of 52.0° north of the equator, we need to consider the reduced distance from the axis of rotation due to the latitude.
The distance from the axis of rotation can be calculated as:
r = R * cos(θ)
Where:
- r is the distance from the axis of rotation
- R is the radius of the Earth
- θ is the latitude
Substituting the values:
r = (6.38 x 10^6 m) * cos(52.0°)
≈ 4.83 x 10^6 m
Using the same formula for centripetal acceleration:
ω = 2π / (24 x 3600 s) = 7.27 x 10^-5 rad/s
a = (7.27 x 10^-5 rad/s)^2 * (4.83 x 10^6 m)
≈ 0.016 m/s^2
Therefore, the centripetal acceleration of a person at a latitude of 52.0° north of the equator is approximately 0.016 m/s^2.
What do we know?
r=6.38x10^6m
T=1 day=86,400s
Part A:
T=2(pi)r/V ==> V=2(pi)r/T
a=V^2/r (Plug & chug)
Part B:
Now, what do we know?
Only T=1 day=86,400s. We no longer know r.
Latitude we'll say is represented by "l".
So r(l) = r cos l
Once you find that radius, you follow the same series of equations as Part A.