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. ω2
 = m(R.cos λ) ω2 (BP = R cos λ)
 = mR ω2cos 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 ω2 cos λ is used in overcoming
centrifugal force.
 Let the net force be represented by PC. Then,
 PC = mg cos λ - mR ω2 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 ω2 cos λ)2 + (mg sin λ)2]1/2
 mg’ = mg [ 1+((R2 ω4)/g2)cos2 λ – ((2Rω2)/g)
cos2 λ]
 mg’ = mg [1-((2Rω2)/g) cos2 λ]1/2 (neglecting ((R2
ω4)/g2)cos2 λ
 mg’ = mg [1-((Rω2 cos2 λ)/g)] (Rω2/g is small, its higher
powers can be neglected)
 g’ = g [1-((Rω2 cos2 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) ω2 =mrω2
 Fc = mR(cos λ) ω2
 The effective force along centre of earth O
 = (force of gravity F) – component of centripetal force
Fc along PO = mg – Fc cos λ
 = mg – [mR(cos λ) ω2] cos λ = mg-mRω2cos2λ
 Let g’ be the effective value of g due to earth’s rotation.
Then mg’ = mg - mRω2cos2λ
 g’ = g - Rω2cos2λ
 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/R2 (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 = R2 / (R+h)2 = R2 /
R2[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/R2 (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)
 mg’ = GM’m/(R-h)2 (2)
 Here M = 4/3 π R3 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 (R2/(R-h)2
 (4/3 π (R-h)3 ρ)/(4/3 π R3 ρ) x R2/(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