Acceleration Gravity Latitude

Overview

Engineering reference data for Acceleration Gravity Latitude in dynamics.

Key Formulas

Newton's Second Law

$F = ma$

Force = mass × acceleration.

Kinetic Energy

$E_k = \frac{1}{2}mv^2$

Energy of motion.

Momentum

$p = mv$

Mass × velocity.

Work

$W = Fd\cos\theta$

Force × displacement × cos(angle).

Variables

Symbol	Description	Unit
$F$	Force	N
$m$	Mass	kg
$a$	Acceleration	m/s²
$v$	Velocity	m/s
$E_k$	Kinetic energy	J

Practical Examples

Time for Falling Object to Hit Ground

The time required for an object to fall a specific distance can be calculated using the formula: $t = \sqrt{\frac{2s}{a_g}}$

Where:

• $t$ is the time in seconds (s)
• $s$ is the distance fallen in meters (m) or feet (ft)
• $a_g$ is the local acceleration of gravity (m/s² or ft/s²)

Example Calculation: For an object falling 1 meter at the poles ($a_g$ = 9.832 m/s²) versus the equator ($a_g$ = 9.780 m/s²):

• At the pole: $t = \sqrt{\frac{2 \cdot 1 \, \text{m}}{9.832 \, \text{m/s}^2}} \approx 0.4510$ seconds
• At the equator: $t = \sqrt{\frac{2 \cdot 1 \, \text{m}}{9.780 \, \text{m/s}^2}} \approx 0.4522$ seconds

Weight Variation with Latitude

The weight (gravitational force) of an object changes with the local acceleration of gravity. Weight is calculated as: $F_g = m \cdot a_g$

Where:

• $F_g$ is the gravitational force (weight) in Newtons (N)
• $m$ is the mass in kilograms (kg)
• $a_g$ is the local acceleration of gravity (m/s²)

Example Calculation: For a person with a mass of 100 kg:

• In Canada (latitude ~60°): $a_g \approx 9.818$ m/s², Weight = $100 \cdot 9.818 = 982$ N
• In Venezuela (latitude ~5°): $a_g \approx 9.782$ m/s², Weight = $100 \cdot 9.782 = 978$ N

This demonstrates that the same person would weigh approximately 4 N more in Canada than at the equator due to the variation in gravitational acceleration.

References

• Original Source

Physical Principles

The variation in gravitational acceleration with latitude is primarily due to two factors:

1. Earth's Rotation: The centrifugal force resulting from Earth's rotation is strongest at the equator and decreases with latitude, effectively reducing the measured gravitational acceleration at the equator.
2. Earth's Oblate Shape: Earth is an oblate spheroid, meaning the radius is larger at the equator than at the poles. Since gravitational force decreases with the square of the distance from the center, gravity is weaker at the equator.

Practical Examples (Additional)

Example: Falling Time at the Pole vs. Equator

The time $t$ for an object to fall a distance $s$ from rest is given by: $t = \sqrt{\frac{2s}{a_g}}$

For an object falling from a height of $s = 1$ m:

• At the Pole ($a_g \approx 9.832 \, \text{m/s}^2$): $t_{\text{pole}} = \sqrt{\frac{2 \times 1 \, \text{m}}{9.832 \, \text{m/s}^2}} \approx 0.4510 \, \text{s}$
• At the Equator ($a_g \approx 9.780 \, \text{m/s}^2$): $t_{\text{equator}} = \sqrt{\frac{2 \times 1 \, \text{m}}{9.780 \, \text{m/s}^2}} \approx 0.4522 \, \text{s}$

Example: Weight Variation with Location

The weight (gravitational force) $F_g$ of an object with mass $m$ is $F_g = m \cdot a_g$.

For a person with mass $m = 100 \, \text{kg}$:

• In Canada (latitude ~60°, $a_g \approx 9.818 \, \text{m/s}^2$): $F_g = 100 \, \text{kg} \times 9.818 \, \text{m/s}^2 = 982 \, \text{N}$
• In Venezuela (latitude ~5°, $a_g \approx 9.782 \, \text{m/s}^2$): $F_g = 100 \, \text{kg} \times 9.782 \, \text{m/s}^2 = 978 \, \text{N}$

This difference of ~4 N demonstrates how location affects measured weight.

Physical Principles

The variation in gravitational acceleration with latitude is primarily due to two factors:

1. Earth's Rotation: The centrifugal force resulting from Earth's rotation is strongest at the equator and decreases with latitude, effectively reducing the measured gravitational acceleration at the equator.
2. Earth's Oblate Shape: Earth is an oblate spheroid, meaning the radius is larger at the equator than at the poles. Since gravitational force decreases with the square of the distance from the center, gravity is weaker at the equator.

Practical Examples (Additional)

Example: Falling Time at the Pole vs. Equator

The time $t$ for an object to fall a distance $s$ from rest is given by: $t = \sqrt{\frac{2s}{a_g}}$

For an object falling from a height of $s = 1$ m:

• At the Pole ($a_g \approx 9.832 \, \text{m/s}^2$): $t_{\text{pole}} = \sqrt{\frac{2 \times 1 \, \text{m}}{9.832 \, \text{m/s}^2}} \approx 0.4510 \, \text{s}$
• At the Equator ($a_g \approx 9.780 \, \text{m/s}^2$): $t_{\text{equator}} = \sqrt{\frac{2 \times 1 \, \text{m}}{9.780 \, \text{m/s}^2}} \approx 0.4522 \, \text{s}$

Example: Weight Variation with Location

The weight (gravitational force) $F_g$ of an object with mass $m$ is $F_g = m \cdot a_g$.

For a person with mass $m = 100 \, \text{kg}$:

• In Canada (latitude ~60°, $a_g \approx 9.818 \, \text{m/s}^2$): $F_g = 100 \, \text{kg} \times 9.818 \, \text{m/s}^2 = 982 \, \text{N}$
• In Venezuela (latitude ~5°, $a_g \approx 9.782 \, \text{m/s}^2$): $F_g = 100 \, \text{kg} \times 9.782 \, \text{m/s}^2 = 978 \, \text{N}$

This difference of ~4 N demonstrates how location affects measured weight.