Variation in the Acceleration>Variation in Gravity with Latitude and Rotation of the Earth

The acceleration due to gravity (g) is not perfectly constant across the Earth's surface. One of the main reasons for this variation is the rotation of the Earth. Because the Earth spins, an outward-acting centrifugal effect reduces the effective gravity, and this reduction depends on your location. This effect is maximum at the equator and zero at the poles. Therefore, the value of g changes as one moves from the equator to the poles.

Latitude is an angle made by the radius vector of any point from the centre of the Earth with the equatorial plane. Obviously it ranges from 0° at the equator to 90° at the poles.

The effective acceleration due to gravity (g') at a point on the Earth's surface at latitude θ is given by:

Where:

• g': The effective acceleration due to gravity at latitude θ (m/s²).
• g: The true acceleration due to gravity (without rotational effect) (m/s²).
• R: The radius of the Earth (m).
• ω: The angular velocity of rotation of the Earth (rad / s or s^-1).
• θ: The latitude of the point (in degrees or radians).

• g' increases as we move away from the equator towards any pole due to the rotation of the Earth.
• The maximum reduction in g occurs at the equator (θ = 0°).
• No reduction in g occurs at the poles (θ = 90°).
• The variation is due to a component of the centrifugal acceleration acting opposite to the gravitational force.

The variation of g with latitude (θ) arises because the rotation of the Earth creates an apparent outward (centrifugal) force on an object, which partially cancels the inward (gravitational) force.

Derivation steps:

1. Define Rotation Radius (r):

• Consider a mass m at a point P on the Earth's surface at latitude θ.
• The mass moves in a circle of radius r parallel to the equator.
• From the geometry in the diagram (where R is the radius of the Earth OP):
  \[r = R \cos \theta\]

2. Calculate Centripetal Acceleration (a):

• The centripetal acceleration, directed towards the center of rotation O', is:
  \[a = r\omega^2\]
• Substituting for r:
  \[a = R\omega^2 \cos \theta\]

3. Find the Component Towards the Earth's Center (ar):

• Only the component of this acceleration that acts outward along the radius vector PO (opposite to the true gravity g) affects the effective gravity.
• The angle ∠OPO' is equal to the latitude θ.
• The component of centripetal acceleration along PO (radially outward) is:
  \[a_r = a \cos \theta\]
• Substituting for a:
  \[a_r = (R\omega^2 \cos \theta) \cos \theta\]
  \[a_r = R\omega^2 \cos^2 \theta\]

4. Determine Effective Gravity (g'):

• Part of the true gravitational acceleration (g) is used to provide this outward radial component of centripetal acceleration.
• The effective gravitational force is the true gravitational force minus the required centripetal force component:
  \[mg' = mg - m a_r\]
  \[mg' = mg - m(R\omega^2 \cos^2 \theta)\]
• Dividing by mass $m$ gives the effective acceleration due to gravity:
  \[g' = g - R\omega^2 \cos^2 \theta\]

Aim:
To demonstrate and verify the variation of acceleration due to gravity at different latitudes on Earth's surface and to measure the reduction in effective gravity due to Earth's rotation.

Requirements:

• Precision pendulum or spring balance (gravimeter)
• Protractor or latitude measuring instrument
• Stopwatch (accurate to 0.01 seconds)
• Measuring scale or meter
• Data recording sheets
• Reference table of latitude values for different locations

Procedure:

1. Select Multiple Locations: Choose or reference data from at least 5-6 different latitudes (0°, 15°, 30°, 45°, 60°, 90°)

2. Measure g Using Pendulum Method:

• Set up a simple pendulum of known length L
• Measure the time period T for 20 complete oscillations
• Calculate g using: g = \[\frac{4\pi^2L}{T^2}\]
• Record the value of g at each latitude

3. Alternative Method (Spring Balance):

• Take a known mass object
• Measure its weight at different latitudes using a spring balance
• Calculate g = Weight/Mass for each location

4. Record Latitude Values: Note the exact latitude of each measurement location

5. Calculate Effective Reduction:

• For each latitude θ, calculate: Δg = Rω²cos⁡²θ.
• Predicted effective g: g_predicted′ = g_pole − Rω²cos⁡²θ 

6. Compare and Analyze: Plot measured g values against latitude and compare with theoretical values

Observations:

• The period of the pendulum increases (and thus g decreases) as we move toward the equator
• The weight of an object decreases when measured at the equator compared to the poles
• Spring balance readings are higher at poles and lower at equator for the same object
• The variation follows the cos²θ pattern

Result:

The experimental data confirms that:

• At Equator (0°): g = 9.78 m/s² (minimum)
• At 30° Latitude: g ≈ 9.79 m/s²
• At 45° Latitude: g ≈ 9.81 m/s²
• At Poles (90°): g = 9.83 m/s² (maximum)

The reduction in g at the equator is approximately 0.03386 m/s², which matches the theoretical prediction of Rω². This confirms the formula g′ = g − Rω²cos⁡²θ.