Variation of g with latitude or rotation of earth

Derivation for Variation of g with latitude or rotation of earth

1. Variation of g with latitude or rotation of the earth Dr.R.Hepzi Pramila Devamani , Assistant Professor of Physics, V.V.Vanniaperumal College for Women, Virudhunagar

2. Variation of g with latitude or rotation of the earth Let us assume that the earth is a uniform sphere of radius R revolving about its polar diameter NS (Fig.3.9). Consider a particle of mass m on the surface of the earth at a latitude λ . If the earth were at rest, a particle of mass m placed at P will experience a force mg along the radius PO towards O.

3. Variation of g with latitude or rotation of the earth Let ω be the angular velocity of the earth. As the earth revolves, the particle at P will execute circular motion with B as centre and BP as radius. A centrifugal force will develop and the centrifugal force acting on P along BP, away from B = m.BP . ω 2 = m( R.cos λ ) ω 2 (BP = R cos λ ) = mR ω 2 cos λ

4. Variation of g with latitude or rotation of the earth Force mg acts along PO. Resolve mg into two rectangular components (i) mg sin λ along PA and (ii) mg cos λ along PB. Out of the resolved component along PB, a portion mR ω 2 cos λ is used in overcoming centrifugal force. Let the net force be represented by PC. Then, PC = mg cos λ - mR ω 2 cos λ and PA = mg sin λ .

5. Variation of g with latitude or rotation of the earth The resultant force (mg’) experienced by P is along PQ, such that (PQ) 2 = (PC) 2 + (PA) 2 or PQ = [(PC) 2 + (PA) 2 ] 1/2 mg’ = [(mg cos λ - mR ω 2 cos λ ) 2 + (mg sin λ ) 2 ] 1/2 mg’ = mg [ 1+((R 2 ω 4 )/g 2 )cos 2 λ – ((2R ω 2 )/g) cos 2 λ ] m g’ = mg [1- ((2R ω 2 )/g ) cos 2 λ ] 1/2 (neglecting ((R 2 ω 4 )/g 2 )cos 2 λ mg’ = mg [1- ((R ω 2 cos 2 λ )/ g )] ( R ω 2 /g is small, its higher powers can be neglected) g’ = g [1-((R ω 2 cos 2 λ )/g)]

6. Alternate Method The earth is a sphere of radius R rotating with angular velocity ω about an axis passing through its poles ( N ,S) and centre O. Consider a body of mass m at P whose latitude is λ (Fig.3.10) The body situated at P moves in a circular path of radius r = R cos λ with angular speed ω , the necessary centripetal force Fc is by a component of gravitational attraction force between the body and the earth.

7. Alternate Method The centripetal force Fc = m(CP) ω 2 = mr ω 2 Fc = mR ( cos λ ) ω 2 The effective force along centre of earth O = (force of gravity F) – component of centripetal force Fc along PO = mg – Fc cos λ = mg – [ m R ( cos λ ) ω 2 ] cos λ = mg- mR ω 2 cos 2 λ Let g’ be the effective value of g due to earth’s rotation. Then mg’ = mg - mR ω 2 cos 2 λ g’ = g - R ω 2 cos 2 λ The equation shows that the value of g decreases due to earth’s rotation.

8. Variation of g with altitude Let P be a point on the surface of earth and Q another point at an altitude h.(Fig.3.11). Mass of the earth is M and radius of the earth is R. Let g be the acceleration due to gravity on the surface of the earth. The force experienced by a body of mass m at P = mg = GMm /R 2 (1) The force experienced by a body of mass m at Q = mg = GMm /( R+h ) 2 (2)

9. Variation of g with altitude Here g’ is the acceleration due to gravity at an altitude h. Dividing (ii) by (i). g ’/g = R 2 / ( R+h ) 2 = R 2 / R 2 [1 + (h/R)] 2 = (1-2h/R) (neglecting higher powers of h/R) Or g’ = g (1-2h/R) This shows that the acceleration due to gravity decrease with increase in altitude.

10. Variation of g with depth Let g and g’ be the values of acceleration due to gravityat P and Q respectively Fig (3.12). At P, the whole mass of the earth attracts the body. Mg = GMm /R 2 (1). Here m = mass of the body M = mass of the earth and R = radius of the earth

11. Variation of g with depth At Q, the body is attracted by the mass of the earth of radius (R-h) m g’ = GM’m /(R-h) 2 (2) Here M = 4/3 π R 3 ρ and M’ = 4/3 π (R-h) 3 ρ Here ρ is the mean density of earth. Diving (2) by (1) g’ = g M’/M (R 2 /(R-h) 2 (4/3 π (R-h) 3 ρ )/(4/3 π R 3 ρ ) x R 2 /(R-h) 2 g ’ = (R-h)/R = 1-h/R g’ = g (1-h/R) Therefore the accelration due to gravity decreases with increase of depth