Acceleration due to Gravity

Updated 22 Mar 2026
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  1. 1Variation of g

Acceleration due to gravity, denoted by gg, is the acceleration experienced by an object solely due to the gravitational force exerted by a celestial body, typically Earth. It is a vector quantity, always directed towards the center of mass of the attracting body. According to Newton's Universal Law of Gravitation, the gravitational force between two objects is directly proportional to the produc…

Quick Summary

Acceleration due to gravity, denoted by gg, is the acceleration experienced by an object solely under the influence of a planet's gravitational force. On Earth's surface, its average value is approximately $9.

8\,\text{m/s}^2(oroftenapproximatedas(or often approximated as10\,\text{m/s}^2forsimplercalculations).Itisavectorquantity,alwaysdirectedtowardsthecenteroftheEarth.Thefundamentalformulaforfor simpler calculations). It is a vector quantity, always directed towards the center of the Earth. The fundamental formula forgatthesurfaceofaplanetofmassat the surface of a planet of massMandradiusand radiusRisisg = GM/R^2,where, whereG$ is the universal gravitational constant.

A key takeaway is that gg is independent of the mass of the falling object. However, gg is not constant across the Earth. It decreases with increasing altitude (height above the surface) and with increasing depth (below the surface).

It is maximum at the poles and minimum at the equator, primarily due to the Earth's rotation and its slightly oblate shape. At the Earth's center, gg becomes zero. Understanding these variations and the underlying principles is essential for NEET.

Full explanation

The concept of acceleration due to gravity, denoted by gg, is a cornerstone of classical mechanics and a direct consequence of Newton's Universal Law of Gravitation. It describes the acceleration experienced by an object solely under the influence of gravitational force. For NEET aspirants, a deep understanding of its derivation, standard value, and various factors influencing its magnitude is indispensable.

1. Conceptual Foundation: Linking Gravity and Acceleration

At its heart, acceleration due to gravity is the specific acceleration an object undergoes when the only force acting upon it is gravity. Consider an object of mass mm near the surface of a planet of mass MM and radius RR.

According to Newton's Universal Law of Gravitation, the gravitational force FgF_g exerted by the planet on the object is given by:

Fg=GMmr2F_g = \frac{G M m}{r^2}
where GG is the universal gravitational constant ($6.

67 \times 10^{-11},\text{N m}^2/\text{kg}^2)and) andristhedistancebetweenthecenteroftheplanetandthecenteroftheobject.Iftheobjectisonorverynearthesurface,is the distance between the center of the planet and the center of the object. If the object is on or very near the surface,rcanbeapproximatedastheradiusoftheplanet,can be approximated as the radius of the planet,R$.

Simultaneously, according to Newton's Second Law of Motion, the net force acting on an object is equal to the product of its mass and its acceleration (F=maF = ma). In this case, the acceleration is gg, so the gravitational force can also be written as:

Fg=mgF_g = m g

2. Derivation of Acceleration due to Gravity ($g$)

By equating these two expressions for the gravitational force, we can derive a formula for gg:

mg=GMmR2m g = \frac{G M m}{R^2}
Notice that the mass of the object, mm, cancels out from both sides. This is a profound result: the acceleration due to gravity is independent of the mass of the falling object. This means a feather and a hammer, if dropped in a vacuum, would fall with the same acceleration.

Thus, the acceleration due to gravity at the surface of a planet is:

g=GMR2g = \frac{G M}{R^2}
For Earth, taking ME=5.97×1024,kgM_E = 5.97 \times 10^{24},\text{kg} (mass of Earth) and RE=6.37×106mR_E = 6.37 \times 10^6\,\text{m} (mean radius of Earth), and $G = 6.

67 \times 10^{-11},\text{N m}^2/\text{kg}^2,weget:, we get:$g \approx \frac{(6.67 \times 10^{-11},\text{N m}^2/\text{kg}^2) \times (5.97 \times 10^{24},\text{kg})}{(6.37 \times 10^6\,\text{m})^2} \approx 9.

8\,\text{m/s}^2$$ This is the standard value often used in calculations near the Earth's surface.

3. Factors Affecting Acceleration due to Gravity

While g9.8m/s2g \approx 9.8\,\text{m/s}^2 is a useful approximation, its actual value varies. Understanding these variations is crucial for NEET.

  • **a) Variation with Altitude (Height hh):**

As an object moves above the Earth's surface to a height hh, its distance from the Earth's center becomes RE+hR_E + h. The acceleration due to gravity at this altitude, ghg_h, is:

gh=GME(RE+h)2g_h = \frac{G M_E}{(R_E + h)^2}
We can rewrite this in terms of gg at the surface:
gh=GMERE2(1+h/RE)2=g(1+h/RE)2g_h = \frac{G M_E}{R_E^2 (1 + h/R_E)^2} = \frac{g}{(1 + h/R_E)^2}
For small heights (hREh \ll R_E), we can use the binomial approximation (1+x)n1nx(1+x)^{-n} \approx 1-nx for x1x \ll 1.

Here, x=h/REx = h/R_E and n=2n=2.

ghg(12hRE)g_h \approx g \left(1 - \frac{2h}{R_E}\right)
This shows that gg decreases with increasing altitude. The decrease is approximately linear for small heights.

  • **b) Variation with Depth (Depth dd):**

When an object is taken to a depth dd below the Earth's surface (e.g., inside a mine), its distance from the center is REdR_E - d. However, for calculating gravity inside the Earth, we only consider the mass of the Earth contained within a sphere of radius (REd)(R_E - d).

Assuming the Earth has a uniform density ρ\rho, the mass of the Earth ME=43πRE3ρM_E = \frac{4}{3}\pi R_E^3 \rho. The mass of the inner sphere M=43π(REd)3ρM' = \frac{4}{3}\pi (R_E - d)^3 \rho. The acceleration due to gravity at depth dd, gdg_d, is:

gd=GM(REd)2=G43π(REd)3ρ(REd)2=G43π(REd)ρg_d = \frac{G M'}{(R_E - d)^2} = \frac{G \frac{4}{3}\pi (R_E - d)^3 \rho}{(R_E - d)^2} = G \frac{4}{3}\pi (R_E - d) \rho
We know that g=GMERE2=G43πRE3ρRE2=G43πREρg = \frac{G M_E}{R_E^2} = \frac{G \frac{4}{3}\pi R_E^3 \rho}{R_E^2} = G \frac{4}{3}\pi R_E \rho.

Dividing gdg_d by gg:

gdg=G43π(REd)ρG43πREρ=REdRE=1dRE\frac{g_d}{g} = \frac{G \frac{4}{3}\pi (R_E - d) \rho}{G \frac{4}{3}\pi R_E \rho} = \frac{R_E - d}{R_E} = 1 - \frac{d}{R_E}
So,
gd=g(1dRE)g_d = g \left(1 - \frac{d}{R_E}\right)
This indicates that gg also decreases with increasing depth.

At the center of the Earth (d=REd = R_E), gd=0g_d = 0. This is because at the center, the gravitational forces from all parts of the Earth cancel out.

  • **c) Variation with Shape of Earth (Latitude λ\lambda):**

The Earth is not a perfect sphere; it's an oblate spheroid, meaning it's flattened at the poles and bulges at the equator. The equatorial radius is slightly larger than the polar radius (RE (equator)>RE (pole)R_E \text{ (equator)} > R_E \text{ (pole)}). Since g=GM/R2g = GM/R^2, a larger radius implies a smaller gg. Therefore, gg is slightly less at the equator than at the poles due to the Earth's shape.

  • **d) Variation with Rotation of Earth (Latitude λ\lambda):**

The Earth rotates about its axis. An object on the surface at latitude λ\lambda experiences a centrifugal force (or rather, the gravitational force provides the centripetal force required for circular motion).

This effectively reduces the apparent weight of the object and thus the effective acceleration due to gravity. The effective acceleration due to gravity gg' at latitude λ\lambda is given by:

g=gREω2cos2λg' = g - R_E \omega^2 \cos^2\lambda
where ω\omega is the angular velocity of the Earth's rotation.

At the equator (λ=0\lambda = 0^\circ, coslambda=1coslambda = 1), g=gREω2g' = g - R_E \omega^2. At the poles (λ=90\lambda = 90^\circ, coslambda=0coslambda = 0), g=gg' = g. This shows that gg is minimum at the equator and maximum at the poles due to rotation.

The effect of rotation is more significant than the effect of shape, but both contribute to gequator<gpoleg_{\text{equator}} < g_{\text{pole}}.

4. Real-World Applications and Significance

  • Projectile Motion:The constant (or near-constant for short ranges) acceleration due to gravity is fundamental to understanding the trajectory of projectiles, from a thrown ball to a ballistic missile.
  • Satellite Orbits:While orbital motion involves centripetal force, the gravitational force providing this centripetal force is directly related to gg at that altitude.
  • Weight:An object's weight is defined as W=mgW = mg. Since gg varies, an object's weight also varies slightly depending on its location on Earth.
  • Geodesy and Geophysics:Precise measurements of gg are used to study the Earth's interior, detect mineral deposits, and understand tectonic plate movements.

5. Common Misconceptions

  • $g$ vs $G$:Students often confuse the universal gravitational constant (GG) with acceleration due to gravity (gg). GG is a fundamental constant of nature, always the same everywhere. gg is a local acceleration, specific to a celestial body and varying with location.
  • $g$ is constant:While g9.8m/s2g \approx 9.8\,\text{m/s}^2 is a good approximation for many problems, it's crucial to remember that gg is not truly constant and varies with altitude, depth, and latitude.
  • Mass vs. Weight:Mass is an intrinsic property of an object and remains constant. Weight (mgmg) depends on gg, and thus changes with location.

6. NEET-Specific Angle

For NEET, questions frequently test the understanding of:

  • The basic formula g=GM/R2g = GM/R^2.
  • The approximate value of gg (9.8m/s29.8\,\text{m/s}^2 or 10m/s210\,\text{m/s}^2).
  • The variations of gg with altitude and depth, especially the approximate formulas for small hh and dd.
  • The combined effect of Earth's shape and rotation on gg at poles vs. equator.
  • Conceptual questions distinguishing gg from GG, and mass from weight.
  • Problems involving percentage change in gg due to small changes in hh or dd.
  • Graphical representation of gg variation (e.g., from center to outside the Earth). The value of gg increases linearly from the center (g=0g=0) to the surface (g=GM/R2g=GM/R^2) and then decreases as 1/r21/r^2 outside the surface. Mastery of these nuances is key to scoring well on gravitation questions.

Key Concepts

Derivation of g=GM/R2g = GM/R^2

This fundamental formula connects Newton's Law of Gravitation with Newton's Second Law. The gravitational…

Variation of gg with Altitude

As an object moves to a height hh above the Earth's surface, its distance from the center becomes $(R_E +…

Variation of gg with Depth

When an object is at a depth dd below the Earth's surface, only the mass of the Earth within a sphere of…

Often confused with

Side-by-side differences the NEET paper likes to test.

Acceleration due to Gravity vs Universal Gravitational Constant (G)
AspectAcceleration due to GravityUniversal Gravitational Constant (G)
DefinitionAcceleration due to Gravity ($g$): The acceleration experienced by an object due to the gravitational pull of a celestial body.Universal Gravitational Constant ($G$): A proportionality constant in Newton's Law of Gravitation, representing the strength of the gravitational force.
ValueVaries with location (altitude, depth, latitude, celestial body). On Earth's surface, average $9.8\,\text{m/s}^2$.Constant throughout the universe. Approximately $6.67 \times 10^{-11},\text{N m}^2/\text{kg}^2$.
UnitsMeters per second squared ($\text{m/s}^2$).Newton meter squared per kilogram squared ($\text{N m}^2/\text{kg}^2$).
NatureA vector quantity (has magnitude and direction).A scalar quantity (only has magnitude).
DependenceDepends on the mass and radius of the celestial body, and the object's position relative to it.Independent of the masses or distances of interacting objects; it's a fundamental constant.

While both 'g' and 'G' are fundamental to understanding gravitation, they represent distinct physical quantities. 'G' is a universal constant that dictates the strength of gravity between any two masses, whereas 'g' is a specific acceleration experienced by an object due to the gravitational field of a particular celestial body.

'g' is a local, variable quantity, whereas 'G' is a global, invariant constant. Confusing these two is a common pitfall for students, but their distinct definitions, units, and dependencies make them easily differentiable upon careful study.

Why it is tested: For NEET, understanding the clear distinction between $g$ and $G$ is critical. Questions often test this conceptual clarity, sometimes by asking for units, definitions, or how each quantity varies (or doesn't vary) under different conditions. A strong grasp prevents common errors in calculations and theoretical questions related to gravitation.

Questions students ask

6 answered on this topic.

What is the difference between 'g' and 'G'?

'g' represents the acceleration due to gravity, which is the acceleration experienced by an object due to the gravitational pull of a planet (like Earth). Its value is approximately 9.8m/s29.8\,\text{m/s}^2 on Earth's surface and varies with location.

'G' is the universal gravitational constant, a fundamental constant of nature that quantifies the strength of the gravitational force. Its value is constant everywhere in the universe, approximately $6.

67 \times 10^{-11},\text{N m}^2/\text{kg}^2$. 'g' depends on the mass and radius of the planet, while 'G' is independent of any specific body.

Why does acceleration due to gravity decrease with altitude?

Acceleration due to gravity, gg, is inversely proportional to the square of the distance from the center of the attracting body (g=GM/r2g = GM/r^2). As an object moves to a higher altitude, its distance (rr) from the Earth's center increases. Since rr is in the denominator and squared, even a small increase in altitude leads to a noticeable decrease in gg. Essentially, the gravitational pull weakens as you move further away from the source of gravity.

Why does acceleration due to gravity decrease with depth?

When an object is taken to a depth dd below the Earth's surface, the gravitational force acting on it is only due to the mass of the Earth contained within a sphere of radius (REd)(R_E - d). The mass of the Earth outside this sphere (the hollow shell) does not contribute to the gravitational force on the object inside. As you go deeper, the effective mass pulling you towards the center decreases, and thus gg decreases. At the Earth's center, the effective mass is zero, and gg becomes zero.

Is acceleration due to gravity zero at the center of the Earth?

Yes, the acceleration due to gravity is zero at the exact center of the Earth. If an object were placed at the Earth's center, it would be surrounded by Earth's mass uniformly in all directions. The gravitational pull from every part of the Earth would cancel out, resulting in a net gravitational force of zero. Consequently, the acceleration due to gravity at the center would also be zero.

Why is 'g' different at the poles and the equator?

The Earth's rotation and its oblate spheroid shape (bulging at the equator, flattened at the poles) cause 'g' to vary. Due to the bulge, the radius at the equator is slightly larger than at the poles, leading to a smaller gg (g1/R2g \propto 1/R^2).

More significantly, the Earth's rotation creates a centrifugal effect that reduces the effective gravity at the equator more than at the poles. At the poles, there's no centrifugal effect. Both factors combine to make gg maximum at the poles and minimum at the equator.

Does the acceleration due to gravity depend on the mass of the falling object?

No, the acceleration due to gravity is independent of the mass of the falling object. This is a crucial concept. The derivation mg=GMm/R2mg = GMm/R^2 clearly shows that the mass of the object, mm, cancels out, leaving g=GM/R2g = GM/R^2. This means that in a vacuum, a feather and a bowling ball would fall with the exact same acceleration, reaching the ground simultaneously if dropped from the same height. Air resistance is what makes lighter objects fall slower in everyday experience.

Revise in 30 seconds

  • Definition:Acceleration due to gravity, gg, is the acceleration of an object due to gravitational force.
  • Standard Value:g9.8m/s2g \approx 9.8\,\text{m/s}^2 on Earth's surface.
  • Formula:g=GMR2g = \frac{GM}{R^2} (where MM is planet mass, RR is planet radius).
  • Independence:gg is independent of the mass of the falling object.
  • Variation with Altitude ($h$):gh=g(1+h/RE)2g_h = \frac{g}{(1 + h/R_E)^2}. For hREh \ll R_E, ghg(12hRE)g_h \approx g(1 - \frac{2h}{R_E}).
  • Variation with Depth ($d$):gd=g(1dRE)g_d = g(1 - \frac{d}{R_E}). At center (d=REd=R_E), gd=0g_d = 0.
  • Variation with Latitude ($\lambda$):g=gREω2cos2λg' = g - R_E \omega^2 \cos^2\lambda. Max at poles (λ=90\lambda=90^\circ), Min at equator (λ=0\lambda=0^\circ).
  • Relationship:gpole>gequatorg_{\text{pole}} > g_{\text{equator}}.

GRAVITY: G-M-R-Squared, Altitude 2H, Depth D-R, Rotate Cos-Squared.