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Ques1) Calculate the change in the value of g at altitude 45 defree. Take radius of earth = 6370 km.
Ques2) Determine the speed with which the earth would have to rotate on it's axis , so that a person on the equator would weigh 3/5 th as much as present. Take R = 6400 km.
Comments (5)
Variation of 'g' with latitude
The value of g changes from place to place due to the elliptical shape of the Earth and the rotation of the Earth. Due to the shape of the Earth,
From equation (4)
Hence, it is inversely proportional to the square of the radius.
It is least at the equator and maximum at the poles, since the equatorial radius (6378.2 km) is more than the polar radius (6356.8 km)
Due to the rotation of the Earth
If 'w' is the angular velocity of the Earth and f is the latitude of the place,
Circular motion
Suppose a body is undergoing circular motion about a circle of radius r, with an angular velocity w rad/s as shown in the figure. From geometry, we can observe that the corresponding distance travelled in 1 second is equal to rw. Since the distance travelled in 1 second is velocity, v = rw. Every body undergoing circular motion with a constant angular velocity is said to be undergoing uniform circular motion. It experiences acceleration towards the centre of the circle of 
Since it is directed towards the centre of the circle, it is called centripetal acceleration.
Therefore, centripetal acceleration is associated with uniform circular motion and directed towards the centre of the circle 
Let us now consider earth as a sphere of radius R, undergoing uniform circular motion about its polar axis, connecting the north and south poles. Equator is the horizontal circle passing through the centre of this axis, P1.
What is latitude?
Every point on the sphere lies on the same latitude, which lie on the base of the cone whose axis coincides with the polar axis and whose generators make an angle f with the horizontal or equatorial plane. The angle f is called the latitude of the place. Latitude of equator = 0o and latitude of north pole = 90o and latitude of south pole = -90o.
We can observe that if the Earth is considered to be a sphere of radius R, the radius of the smaller circle in the latitude f=Rcosf.
A particle on the latitude f which is undergoing uniform circular motion with angular velocity 'w', experiences centripetal acceleration 
directed towards the centre of the small circle OI. This acceleration 'a' can be resolved into two components, tangential and vertical.
Gravitational acceleration 'g' acts on the body.
Since all the forces acting on the body at the latitude f result in uniform circular motion, the net of all forces should be equal to centripetal force.
At the equator,
At the pole,
Hence, the gravitation acceleration is maximum at the poles and minimum at the equator.
We should observe that the net of all the forces acting on the body results in uniform circular motion, which means that uniform circular motion is the result of all the forces acting on the body.
At the equator f = 0
At the poles, f = 90o, cos f = 0
It is less at the equator and maximum at the poles.
1.g =G ME / (R+h)2
g = 6.67 x10-11 x 6 x 1024 / 4 x 67300002
g = 2.2 m/sec2
Although there is another formula to calculate g without putting ME but I am not able to recollect.
2. mg' = mg - mw2R
g -(3/5) g = w2R => w = (2g/5R)1/2 = 7.83 x 10-4 rad/sec
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