Determine the magnitude of the effective value of at a latitude of 60 on the Earth. Assume the Earth is a rotating sphere.
You seem to have omitted a word in your question.
<<...the effective value of___ at a latitude of 60 (degrees) >>>
To determine the magnitude of the effective value of the acceleration due to gravity (g) at a latitude of 60 degrees on the Earth, we need to take into account the rotation of the Earth and the variation of the gravitational force with latitude.
The formula for calculating the effective value of gravity is given by:
geff = g * (1 - (2 * omega * h / g))
Where:
geff is the effective value of gravity
g is the acceleration due to gravity at the equator (approximately 9.81 m/s^2)
omega is the angular velocity of the Earth (approximately 7.2921159 x 10^-5 rad/s)
h is the height above the Earth's surface (in this case, the radius of the Earth)
Assuming the radius of the Earth (h) is approximately 6,371 kilometers, we can substitute the values into the formula:
geff = 9.81 * (1 - (2 * 7.2921159 x 10^-5 * 6371000 / 9.81))
Simplifying the equation further:
geff = 9.81 - (0.14 * 10^-3 * 6371000)
Calculating the value:
geff ≈ 9.81 - 0.89374 ≈ 9.81 - 0.89 ≈ 8.92 m/s^2
Therefore, the magnitude of the effective value of gravity at a latitude of 60 degrees on Earth is approximately 8.92 m/s^2.