How to find the centripetal acceleration of an object that is at a latitude of 52 degree to the equator given the radius of earth=6400km and angular vel of Earth about its axis=7.3 x 10^-5 rad/s ?
find the radius of motion at latitude 52deg. At zero latitude, rad=re, at latitude theta, rad=re*cosineTheta.
So you know ang velocity, radius
Centripetal acceleration=w^2*r
remember the radius of motion is the distance from the Point on the surface to the AXIS of rotaion, not the distance to the center of the Earth.
To find the centripetal acceleration of an object at a given latitude on Earth, you can use the formula:
a = (ω^2) * r * cos(φ)
Where:
a = centripetal acceleration
ω = angular velocity of Earth about its axis
r = radius of Earth at the given latitude
φ = latitude angle
In this case, you are given:
ω = 7.3 x 10^-5 rad/s
r = 6400 km
φ = 52 degrees
To use this formula, you need to convert the given values into consistent units. Let's convert the radius from kilometers to meters and the angle from degrees to radians:
r = 6400 km = 6400 * 1000 m = 6,400,000 m
φ = 52 degrees = (52 * π) / 180 radians ≈ 0.9077 radians
Now, substitute the values into the formula:
a = (7.3 x 10^-5 rad/s)^2 * 6,400,000 m * cos(0.9077 radians)
Calculate the value inside the parentheses first:
(7.3 x 10^-5 rad/s)^2 = 5.329 x 10^-9 rad^2/s^2
Now, calculate the cosine of the latitude angle:
cos(0.9077 radians) ≈ 0.6180
Substitute these values into the formula:
a = (5.329 x 10^-9 rad^2/s^2) * 6,400,000 m * 0.6180
Finally, calculate the centripetal acceleration:
a ≈ 2.048 m/s^2
Therefore, the centripetal acceleration of an object at a latitude of 52 degrees to the equator, given the radius of Earth as 6400 km and the angular velocity of Earth about its axis as 7.3 x 10^-5 rad/s, is approximately 2.048 m/s^2.