📂Classical Mechanics

gravitational acceleration

Definition

The acceleration of an object moving under the influence of gravity is called gravitational acceleration.

When an object moves solely under the effect of Earth’s gravity, without other external forces such as friction or air resistance, this motion is termed free-fall motion.

Explanation

When gravity or gravitational field is mentioned simply, it usually refers to Earth’s gravity. Earth’s gravitational acceleration is approximately $9.8 \mathrm{m/s^{2}}$, commonly denoted as $g$. This value is independent of the mass of the object.

Motion of an object driven solely by gravity without any external forces is called free-fall motion, which refers to the motion of an object accelerating at gravitational acceleration.

The commonly used term in everyday life, weight, refers to the magnitude of gravitational force acting on an object. In other words, when Earth’s gravitational acceleration is termed $g$, the weight of an object with a mass of $m$ near the Earth’s surface is $mg$.

Gravitational Acceleration of Earth ¹

For convenience, assume the Earth is spherical, and an object with a mass of $m$ is at a distance of $h$ from the Earth’s surface. The magnitude of the gravitational force this object receives from Earth is determined by the law of universal gravitation as follows.

$$F = G \dfrac{M_{E} m}{(R_{E} + h)^{2}} \tag{1}$$

Here, $G$ represents the gravitational constant, $M_{E}$ is the Earth’s mass, and $R_{E}$ symbolizes the Earth’s radius. Let the acceleration of the object due to gravity be denoted by $g$. According to Newton’s second law, the motion of the object can be expressed as follows.

$$F = mg \tag{2}$$

Thus, from $(1)$ and $(2)$, we derive the following relation.

$$mg = G \dfrac{M_{E} m}{(R_{E} + h)^{2}} \implies g = G \dfrac{M_{E}}{(R_{E} + h)^{2}}$$

From this equation, it is apparent that gravitational acceleration is independent of the mass of the object in motion. The distance (height) $h$ from the Earth’s surface is negligible compared to the Earth’s radius $R_{E}$, so we can state it again as below.

$$g = G \dfrac{M_{E}}{R_{E}^{2}}$$

Since these are all constants, substituting them in and calculating yields the following.

$$g = G \dfrac{M_{E}}{R_{E}^{2}} = \dfrac{(6.67 \cdot 10^{-11}\mathrm{m^{3}kg^{-1}/s^{2}}) (5.97\cdot 10^{24} \mathrm{kg})}{(6.37 \cdot 10^{6} \mathrm{m})^{2}} = 9.81344\mathrm{}$$