Here the particle is performing circular motion due to the rotation of the earth. And for the circular motion it requires centripetal force then which force provides the necessary centripetal force. I think it is mgcos(lamda). Am i right?
Yes, that's correct.Hardik Batra said:I think it is mgcos(lamda). Am i right?
No. As Cutter Ketch stated, mgcosθ is much greater than the needed centripetal force.Hardik Batra said:Can I do this
mgcos(lamda) = View attachment 109856 (centrifugal force) (In PQ direction)
PR = mgcos^2(lamda)
and resultant force = mg - mgcos^2(lamda).
Variation in g with latitude refers to the change in the acceleration due to gravity (g) as one moves from the equator to the poles on Earth's surface. It is caused by various factors such as the Earth's shape, rotation, and distribution of mass.
G varies with latitude due to the oblate shape of the Earth, which is slightly flattened at the poles and bulging at the equator. This uneven distribution of mass causes a variation in the distance between the Earth's center and a point on its surface, leading to a change in the gravitational force at different latitudes.
As one moves from the equator to the poles, g decreases due to the increasing distance from the Earth's center. At the equator, g is approximately 9.78 m/s², while at the poles it is around 9.83 m/s². This difference may seem small, but it can have significant effects on things like the weight of objects and the measurement of time.
Apart from the Earth's shape and rotation, the distribution of landmasses and changes in elevation can also contribute to the variation in g with latitude. For example, areas with high elevations will experience a slightly stronger gravitational pull than those at sea level.
The variation in g with latitude is typically measured using instruments such as gravimeters or pendulums. These devices can detect small changes in g and are often used in geophysical surveys to map out the Earth's gravitational field and understand its variations at different latitudes.