Centripetal acceleration at equator

[Masafi]

The centripetal acceleration of a person standing at the equator is about 0.03 m s–2.

(Radius of the Earth = 6.4 × 106 m.)


The size of the force R provides a measure of the apparent strength of the gravitational field. Show that the apparent field strength g at the equator differs from that at the poles by about 0.3%.

The answer is:

(0.034 ÷ 9.81) × 100% = 0.35%


I don't understand this. How do we know that the acc at the pole is 9.81, and that 0.034 is the difference in the 2 accerelations?

 

[mgb_phys]

The acceleration due to gravity 'g' you can measure.
You can also work it out from GM/r^2 ( M is the mass of the Earth and r is the radius)

ps. 9.81m/s^2 - is an average for the whole planet,
g is around 9.79 m/s^2 at the equator and 9.83 m/s^2 at the poles - it also varies with local rock types

 

[kerimek]

The acceleration due to gravity, denoted by "g", is a constant value of 9.81 m/s^2 at the Earth's surface. This is the value used in the formula to calculate the apparent field strength at the equator (0.034 m/s^2). The 0.034 m/s^2 is the difference between the centripetal acceleration at the equator (0.03 m/s^2) and the acceleration due to gravity (9.81 m/s^2).

To calculate the percentage difference, we take the difference between the two values (9.81 - 0.03 = 9.78 m/s^2) and divide it by the original value (9.81 m/s^2), then multiply by 100 to get a percentage. This calculation results in a 0.35% difference between the apparent field strength at the equator and the poles. This shows that there is a slight variation in the gravitational field strength at different points on the Earth's surface due to the Earth's rotation.