Acceleration Due to Gravity (g) - Ninth Grade Science
When you release an object from your hand, it does not hover in mid-air or drift sideways; it accelerates directly downward toward the center of the Earth. While the Universal Law of Gravitation describes the mutual force between any two masses, we frequently want to know the exact acceleration experienced by an object falling near Earth's surface. This parameter is the acceleration due to gravity, universally represented by the symbol $g$. A critical milestone in your scientific development is recognizing that $g$ is fundamentally independent of the falling object's mass. In this chapter, you will derive $g$ directly from Newton's laws, examine how Earth's geographic variations and altitude alter its value, and distinguish the local acceleration $g$ from the universal gravitational constant $G$.
Derivation of $g$ from the Universal Law of Gravitation
Consider an object of mass $m$ resting on or situated near the surface of Earth. The mass of Earth is denoted by $M$, and the radius of Earth is denoted by $R$.
Derivation of Surface Gravity (g)
Object (m)
o
/ \
| R |
\ /
Earth (M)
1. Gravitational Force: F = G * (M * m) / R^2
2. Newton's Second Law: F = m * g
3. Equate Forces: m * g = G * (M * m) / R^2
4. Cancel object mass m: g = G * M / R^2
Step-by-Step Mathematical Formulation
According to Newton's Universal Law of Gravitation, the gravitational force $F$ exerted by Earth on mass $m$ is:
F = G \frac{M m}{R^2}
Simultaneously, according to Newton's Second Law of Motion, this downward gravitational force produces a vertical acceleration $g$, such that:
F = m g
Equating both expressions for force:
m g = G \frac{M m}{R^2}
Dividing both sides of the equation by the object's mass $m$ yields the foundational equation for the acceleration due to gravity:
g = \frac{G M}{R^2}
The Independence of Falling Acceleration from Mass
Notice the most profound consequence of this derivation: the object's mass $m$ cancels out entirely. The acceleration due to gravity $g$ depends solely on the universal gravitational constant ($G$), the mass of the planetary body ($M$), and the radius of the planet ($R$). It does not depend on whether the falling object is a heavy lead cannonball or a delicate bird feather. In the absence of atmospheric drag, all objects released from the same height accelerate toward Earth at the exact same rate.
Calculating the Numerical Value of $g$ on Earth
To compute the standard numerical magnitude of $g$ at sea level on Earth, we substitute known astronomical and physical constants into our derived formula:
Known Astronomical Constants for Earth:
G = 6.674 * 10^-11 N*m^2/kg^2
M = 5.972 * 10^24 kg (Mass of Earth)
R = 6.371 * 10^6 m (Mean Volumetric Radius of Earth)
Substituting these values:
g = \frac{(6.674 \times 10^{-11}) \times (5.972 \times 10^{24})}{(6.371 \times 10^6)^2}
Evaluating the numerator:
(6.674 \times 10^{-11}) \times (5.972 \times 10^{24}) = 3.9857 \times 10^{14}\text{ N}\cdot\text{m}^2/\text{kg}
Evaluating the denominator:
(6.371 \times 10^6)^2 = 4.0589 \times 10^{13}\text{ m}^2
Dividing the terms gives:
g = \frac{3.9857 \times 10^{14}}{4.0589 \times 10^{13}} \approx 9.819\text{ m/s}^2
By international scientific convention, the standard acceleration due to gravity at sea level is taken as $g = 9.8\text{ m/s}^2$ (or $9.81\text{ m/s}^2$).
Factors Affecting the Acceleration Due to Gravity ($g$)
Although $g$ is treated as a constant for introductory textbook calculations, its value varies across Earth's surface and throughout space.
+-----------------------+--------------------------------------------------+
| Influencing Factor | Physical Mechanism & Effect on g |
+-----------------------+--------------------------------------------------+
| Shape of Earth | Oblate spheroid: Radius is larger at Equator. |
| (Poles vs Equator) | g is higher at Poles (~9.83), lower at Equator |
| | (~9.78 m/s^2). |
| Altitude (Height h) | Distance from center increases to (R + h). |
| | g decreases as altitude increases. |
| Depth (Inside Earth) | Only internal sphere mass attracts; g drops |
| | linearly toward zero at Earth's center. |
+-----------------------+--------------------------------------------------+
The Oblate Spheroid Shape of Earth
Earth is not a perfect sphere; because of its rotational spin, it bulges slightly at the equator and flattens at the poles, forming an oblate spheroid. The polar radius ($R_p \approx 6357 ext{ km}$) is approximately 21 kilometers shorter than the equatorial radius ($R_e \approx 6378 ext{ km}$). Since $g \propto 1/R^2$, a smaller radius produces a greater gravitational acceleration:
g_{\text{poles}} \approx 9.832\text{ m/s}^2 \quad > \quad g_{\text{equator}} \approx 9.780\text{ m/s}^2
An object weighs slightly more at the North or South Pole than it does at the equator.
Variation with Altitude ($h$)
When an object ascends to an altitude $h$ above Earth's surface (such as a weather balloon or orbiting spacecraft), its total distance from Earth's center becomes $R + h$. The formula becomes:
g_h = \frac{G M}{(R + h)^2} = g \left( \frac{R}{R + h} \right)^2
As altitude increases, the denominator grows, causing $g$ to decrease. At the altitude of the International Space Station ($h \approx 400 ext{ km}$), $g$ is still approximately $8.7 ext{ m/s}^2$ (nearly ninety percent of its sea-level value), demonstrating that astronauts float not because gravity is absent, but because the station is in continuous free fall.
Contrasting $g$ and $G$
A frequent source of confusion among introductory physics students is the distinction between little $g$ and big $G$.
+------------------------------------+------------------------------------+
| Acceleration Due to Gravity (g) | Gravitational Constant (G) |
+------------------------------------+------------------------------------+
| Local acceleration vector | Universal scalar physical constant |
| SI Unit: m/s^2 or N/kg | SI Unit: N*m^2/kg^2 |
| Value varies by planet, altitude, | Invariant across the entire |
| latitude (9.8 m/s^2 on Earth) | universe (6.674 * 10^-11) |
| Can be zero (at Earth's center) | Never zero under any condition |
+------------------------------------+------------------------------------+
Chapter Exercises
Exercise 1: The planet Mars has a mass of approximately $6.42 imes 10^{23}$ kilograms and an equatorial radius of $3.39 imes 10^6$ meters. Using the universal gravitational constant $G = 6.674 imes 10^{-11} ext{ N}\cdot ext{m}^2/ ext{kg}^2$, calculate the acceleration due to gravity on the surface of Mars. Express this Martian $g$ as a percentage of Earth's standard gravity ($9.8 ext{ m/s}^2$).
Exercise 2: Calculate the acceleration due to gravity experienced by a communications satellite in orbit at an altitude of 36000 kilometers above Earth's surface. Take Earth's radius as 6400 kilometers and sea-level gravity as 9.8 meters per second squared.
Exercise 3: Explain why a gold prospector using an ultra-precise spring balance would prefer to weigh and sell their gold at the equator rather than at the polar regions, whereas a prospector using a two-pan beam balance would find no financial difference whatsoever.
Solutions and Explanations
Solution 1: We use the surface gravity formula $g = G M / R^2$ with Martian values: $M = 6.42 imes 10^{23} ext{ kg}$, $R = 3.39 imes 10^6 ext{ m}$, and $G = 6.674 imes 10^{-11} ext{ N}\cdot ext{m}^2/ ext{kg}^2$. The numerator is $6.674 imes 10^{-11} imes 6.42 imes 10^{23} = 4.2847 imes 10^{13}$. The denominator is $(3.39 imes 10^6)^2 = 1.1492 imes 10^{13}$. Dividing numerator by denominator yields $g_{ ext{Mars}} = 4.2847 imes 10^{13} / 1.1492 imes 10^{13} \approx 3.73$ meters per second squared. As a percentage of Earth's gravity: $(3.73 / 9.8) imes 100 \approx 38.0\%$. An object on Mars accelerates at approximately 38 percent of its rate on Earth.
Solution 2: Earth's radius is $R = 6400 ext{ km}$, and the altitude is $h = 36000 ext{ km}$. The total orbital distance from Earth's center is $r = R + h = 6400 + 36000 = 42400 ext{ km}$. Using the altitude scaling formula: $g_h = g (R / (R + h))^2 = 9.8 imes (6400 / 42400)^2 = 9.8 imes (0.15094)^2 = 9.8 imes 0.02278 \approx 0.223$ meters per second squared. At geostationary orbital altitude, Earth's gravitational acceleration has dropped to approximately 0.22 meters per second squared.
Solution 3: Earth's equatorial radius is larger than its polar radius, making $g$ smaller at the equator ($\approx 9.78 ext{ m/s}^2$) than at the poles ($\approx 9.83 ext{ m/s}^2$). A spring balance measures true weight ($W = mg$), which is the downward gravitational pull stretching a calibrated spring. If the prospector buys gold by weight at the equator where $g$ is low and sells it at the poles where $g$ is high, the spring stretches more at the poles, registering a higher weight reading and earning extra profit. Conversely, selling at the equator yields a lower weight reading. However, a two-pan beam balance operates by comparing the unknown gold mass against standardized brass balance masses. Because local gravity $g$ acts equally on both pans, the gravitational acceleration cancels out ($m_{ ext{gold}} g = m_{ ext{brass}} g \implies m_{ ext{gold}} = m_{ ext{brass}}$). Thus, a beam balance measures invariant mass, showing identical readings at the equator, the poles, or on the Moon.