Interactive physics simulator
Acceleration Due to Gravity
Understand the gravitational acceleration constant (g) and its variations. Explore how altitude and latitude shape planetary gravity, track a probe falling through Earth's core using core-density modeling, and drop objects through viscous fluids to witness drag forces and terminal velocity.
Acceleration Due to Gravity Lab
Modify parameters like planetary altitude, shaft depth models, or fluid densities to observe gravity field vectors and velocity curves.
Live Telemetry
- Altitude (h)
- 6000 km
- Latitude (φ)
- 45°
- Orbital Speed (v)
- 5.68 km/s
- Base Gravity (g₀)
- 2.60 m/s²
- Centrifugal Accel (a_c)
- 0.03 m/s²
- Effective Gravity (g_eff)
- 2.57 m/s²
- Weight Ratio
- 26.2% of surface
What is Acceleration Due to Gravity?
The acceleration due to gravity (denoted by g) is the constant downward acceleration experienced by an object falling freely under the sole influence of a body's gravitational field. According to Newton's second law and his law of universal gravitation, this local acceleration is given by:
Where G is the universal gravitational constant, M is the mass of the planetary body, and r is the distance from the center of mass of the planetary body to the falling object. On the surface of the Earth, this gives an average value of:
However, the value of g is not uniform. It varies dynamically based on altitude (falling off as an inverse square), latitude (due to Earth's rotation creating centrifugal offsets and its ellipsoidal bulge), and interior depth (governed by core density layers).
Altitude & Latitude Offsets
Gravity weakens with height and varies from equator to poles:
- Altitude Decrease: Since g(h) = GM / (R + h)2, gravity decreases rapidly. At 400 km (ISS orbit), gravity drops to about 8.7 m/s² (89% of surface).
- Centrifugal Speed: Earth's spin pushes mass outward by ac = ω2r cos2(φ), reducing effective gravity at low latitudes.
- Oblateness Bulge: The Earth is 21 km wider at the equator than the poles, making the polar surface closer to the core and gravity stronger.
Interior Earth Core Gravity
Gravity inside a planet depends on internal density distribution:
- Shell Theorem: As you descend, all mass in the spherical layers above you exerts zero net force. Only mass inside your current radius pulls you.
- Constant Density: If density was uniform, gravity would decrease linearly with depth: g(r) = gsurf · (r / R).
- PREM Model: Earth's iron-nickel core is highly dense. Gravity actually increases from 9.81 m/s² to a peak of 10.7 m/s² at the core-mantle boundary before dropping to 0.
Fluid Drag & Terminal Speed
Falling in a medium creates resistance forces opposing gravity:
- Viscous Drag: In fluids, drag force is modeled as Fd = 1/2 ρCdAv2. It increases quadratically with speed.
- Terminal Velocity: When drag force equals gravity (Fd = Fg), net acceleration is zero (a = 0). The velocity becomes constant.
- Medium Comparison: Terminal speed is high in thin air (~50 m/s for a human), low in water (~3 m/s), and extremely slow in viscous castor oil.
The Equivalence Principle
A core foundation of Einstein's General Relativity:
- Inertial vs Gravitational Mass: Inertial mass resists force (F = ma); gravitational mass responds to gravity (F = mg).
- Exact Equivalence: Experiments show these masses are identical. As a result, m cancels in ma = mg, making acceleration independent of mass.
- Hammer & Feather: In a vacuum, a hammer and a feather accelerate at the exact same rate and strike the ground together.
Solved Examples
Calculate the acceleration due to gravity g(h) at the altitude of a geostationary orbit (h = 35,786 km above Earth's surface). (Earth radius R = 6,371 km, Earth mass M = 5.972 × 10<sup>24</sup> kg, G = 6.674 × 10<sup>-11</sup> N·m²/kg²)
- Identify initial values: M = 5.972 × 1024 kg, G = 6.674 × 10-11 N·m²/kg², R = 6,371,000 m, h = 35,786,000 m.
- Calculate orbital radius: r = R + h = 6,371,000 + 35,786,000 = 42,157,000 m = 4.2157 × 107 m.
- Apply the altitude gravity formula: g(h) = G * M / (R + h)². ⇒ g(h) = G * M / r².
- Multiply G and M: G * M = (6.674 × 10-11) * (5.972 × 1024) ≈ 3.9857 × 1014 m³/s².
- Calculate the square of the orbital radius: r² = (4.2157 × 107)² ≈ 1.7772 × 1015 m².
- Divide to find gravity: g(h) = 3.9857 × 1014 / (1.7772 × 1015) ≈ 0.224 m/s².
Answer: g(h) ≈ 0.224 m/s² (about 2.3% of surface gravity)
Determine the effective acceleration due to gravity g<sub>eff</sub> at Earth's equator (φ = 0°), accounting for Earth's axial rotation. (Surface gravity without rotation g<sub>0</sub> = 9.81 m/s², equatorial radius R = 6.378 × 10<sup>6</sup> m, rotation period T = 86,164 seconds)
- Identify variables: g0 = 9.81 m/s², R = 6.378 × 106 m, T = 86,164 s.
- Calculate angular velocity of Earth's rotation: ω = 2π / T = 2 × 3.14159 / 86,164 ≈ 7.292 × 10-5 rad/s.
- Apply the centrifugal acceleration formula at the equator (φ = 0°, cos(φ) = 1): ac = ω² * R.
- Substitute values: ac = (7.292 × 10-5)² * (6.378 × 106) ≈ (5.317 × 10-9) * (6.378 × 106) ≈ 0.0339 m/s².
- Subtract centrifugal force from base gravity: geff = g0 - ac = 9.81 - 0.0339 = 9.776 m/s².
Answer: geff ≈ 9.776 m/s² (reduced by 0.35% compared to non-rotating base)
Find the terminal velocity of a spherical steel ball bearing falling through water. (Mass m = 0.5 kg, radius r = 2.5 cm, drag coefficient C<sub>d</sub> = 0.47, water density ρ = 1,000 kg/m³, g = 9.81 m/s²)
- Identify the given values: m = 0.5 kg, r = 0.025 m, Cd = 0.47, ρ = 1,000 kg/m³, g = 9.81 m/s².
- Calculate the cross-sectional area of the sphere: A = π * r² = 3.14159 * (0.025)² ≈ 1.963 × 10-3 m².
- State the terminal velocity condition where drag force equals gravity: Fd = Fg ⇒ 0.5 * ρ * Cd * A * v² = m * g.
- Solve for terminal velocity: vt = √(2 * m * g / (ρ * Cd * A)).
- Substitute values: Numerator = 2 * 0.5 * 9.81 = 9.81 N. Denominator = 1,000 * 0.47 * (1.963 × 10-3) ≈ 0.9226 kg/m.
- Divide: 9.81 / 0.9226 ≈ 10.633 m²/s².
- Take the square root: vt = √(10.633) ≈ 3.26 m/s.
Answer: vt ≈ 3.26 m/s
Common Misconceptions
- "Astronauts float because there is zero gravity in space": False. Earth's gravity at ISS height is 90% of surface gravity. Astronauts float because they are falling around the Earth in permanent free fall.
- "Heavier objects always fall faster": False. In a vacuum, all objects fall at the same speed. On Earth, difference in fall rates is caused entirely by air drag, not gravity.
- "Gravity is constant inside the Earth": False. Gravity fluctuates, first rising to 10.7 m/s² due to core density transitions, then dropping to zero at the exact center.
Practice Questions
1. Explain the physical difference in how gravity (g) behaves inside the Earth under a constant density assumption versus the Preliminary Reference Earth Model (PREM).
Under a constant density assumption, the Shell Theorem states that mass outside the current radius r exerts zero net force, and the mass inside behaves as a point mass proportional to volume (M_inside ∝ r³). Thus, gravity increases linearly from 0 at the center to 9.81 m/s² at the surface: g(r) ∝ r. In contrast, the PREM model accounts for Earth's layered structure: the core is extremely dense (up to 13,000 kg/m³) while the mantle is less dense. As a probe descends, it gets closer to the massive iron core, so gravity actually increases at first, peaking at approximately 10.7 m/s² at the core-mantle boundary (depth of ~2,900 km) before dropping steeply to zero at the center.
2. Explain why acceleration due to gravity varies with latitude on Earth and write down the qualitative contributions.
There are two physical contributions that cause gravity (g) to vary with latitude (from ~9.78 m/s² at the equator to ~9.83 m/s² at the poles):
1. Centrifugal Force: Earth's rotation creates an outward acceleration perpendicular to the axis of rotation: a_c = ω² * R * cos(φ). At the equator (φ = 0°), this outward force is maximized (~0.034 m/s²) and opposes gravity. At the poles (φ = 90°), centrifugal force is zero.
2. Oblateness of Earth: Earth's rotation makes it bulge at the equator. The equatorial radius is about 21 km larger than the polar radius. Because the poles are closer to the Earth's center of mass, the base gravitational attraction is stronger there.
3. Derive the equation of motion for an object falling in a fluid with quadratic drag and show how it approaches terminal velocity.
The net force on a falling object of mass m in a fluid is the difference between gravity and drag: F_net = F_g - F_d.
Writing this in terms of acceleration: m * a = m * g - 0.5 * ρ * C_d * A * v².
Dividing by mass gives the differential equation: a = dv/dt = g - (0.5 * ρ * C_d * A / m) * v².
As velocity v increases, the drag term increases, which reduces the acceleration. When acceleration drops to zero (a = 0), the object is in dynamic equilibrium and no longer speeds up. Setting a = 0 gives: g = (0.5 * ρ * C_d * A / m) * v_t² ⇒ v_t = √(2mg / ρC_dA), which is the formula for terminal velocity.
4. Prove using Newton's laws of motion that all objects accelerate at the exact same rate under gravity in a vacuum, regardless of mass.
According to Newton's Second Law, the acceleration of an object is proportional to the net force acting on it and inversely proportional to its inertial mass: a = F_net / m_inertial.
According to Newton's Law of Universal Gravitation, the gravitational force acting on an object is proportional to its gravitational mass: F_g = G * M * m_gravitational / R².
Substituting F_g for F_net in a vacuum: a = (G * M * m_gravitational / R²) / m_inertial.
By the Equivalence Principle, inertial mass is equal to gravitational mass (m_inertial = m_gravitational). They cancel out completely, yielding: a = G * M / R² = g. Because the object's mass is eliminated, all objects fall at the same rate.
FAQ
Frequently Asked Questions
What is the acceleration due to gravity (g)?
Acceleration due to gravity is the constant acceleration experienced by any object falling freely in a vacuum under the sole influence of a body's gravitational pull.
What is the standard value of g on Earth?
The standard value of g on the surface of the Earth is defined as exactly 9.80665 m/s² (commonly rounded to 9.8 or 9.81 m/s²).
Why does g vary at different places on Earth?
Earth is not a perfect sphere; it is an oblate spheroid (bulging at the equator). The equator is farther from the center of mass than the poles, making gravity weaker. Additionally, Earth's rotation generates centrifugal force, reducing effective gravity at the equator.
How does altitude affect the value of g?
Acceleration due to gravity decreases with altitude. According to the formula g = GM / (R + h)², as altitude (h) increases, the distance from Earth's center increases, which exponentially weakens the gravitational pull.
How does latitude affect the value of g?
Latitude increases g from the equator (approx 9.78 m/s²) to the poles (approx 9.83 m/s²). At higher latitudes, the distance to Earth's center is shorter, and the centrifugal speed from Earth's rotation drops to zero.
What is the value of g at the center of the Earth?
The acceleration due to gravity is exactly 0 m/s² at the center of the Earth. According to the Shell Theorem, the gravitational pulls from all the mass of the surrounding spherical shell cancel each other out in all directions.
How does g change as you go deeper into the Earth?
In a uniform density sphere, g decreases linearly with depth. However, because Earth's core is extremely dense, g actually increases from 9.81 m/s² at the surface to a peak of about 10.7 m/s² at the core-mantle boundary before dropping to 0 at the center.
Does mass affect the acceleration due to gravity in a vacuum?
No. According to the equivalence principle, in a vacuum all objects accelerate downward at the exact same rate under gravity (g), regardless of their mass, shape, or composition.
How does acceleration due to gravity (g) differ from the gravitational constant (G)?
G is a fundamental universal constant (6.674 x 10^-11 N·m²/kg²) that determines the strength of the gravitational force everywhere in the universe. g is a local acceleration field (9.81 m/s² on Earth) that depends on the mass and radius of the host body.
What is the acceleration due to gravity on other planets?
g depends on the mass and radius of the planet. For example, g is 1.62 m/s² on the Moon (16% of Earth), 3.7 m/s² on Mars (38% of Earth), and 24.79 m/s² on Jupiter (253% of Earth).
How does drag affect the acceleration of falling objects in a medium?
In a medium like air or water, drag pushes upward opposing gravity. The net acceleration is a = g - F_drag / m. As velocity increases, drag increases until it equals gravity, halting acceleration and reaching terminal velocity.