Acceleration due to Gravity is the rate at which an object’s speed increases when it falls freely under the influence of gravity. It is equal to 9.8 m/s² on Earth.

Acceleration due to gravity (represented by g) is a vector quantity that is the acceleration gained by an object due to gravitational force. It has a magnitude of 9.8 m/s² on the surface of the earth at sea level and its SI unit is m/s².

Acceleration Due to Gravity: Formula, Unit, and Values

Table of Contents:

• What is Gravity?
• Formula of Acceleration due to Gravity
• Acceleration due to Gravity on the Surface of Earth
• Variation of g with Height
• Variation of g with Depth
• The Effect of Earth’s Shape on the Variation of g
• Variation of g due to the Rotation of the Earth

What is Gravity?

The force between two bodies of masses ma and mb under the application of equal forces is given by: $$F=\frac{G{m}_a{m}_b}{r^2}$$ Where G is the gravitational constant and r is the distance between the two bodies.

The inertial mass of a body is denoted by ${\displaystyle m_b=m_a\left[\frac{a_A}{a_B}\right]}$.

Under the gravitational influence on two bodies,

the bodies will experience a force of attraction between them.

• ${\displaystyle F_A=GM\frac{m_A}{r^2}}$
• ${\displaystyle F_B=GM\frac{m_B}{r^2}}$
• ${\displaystyle m_B=\left[\frac{F_B}{F_A}\right]\times m_A}$

⇒ Learn More About Gravitation:

• Newton’s Law of Gravitation
• Gravitational Potential Energy
• Gravitational Field Intensity

The principle of equivalence states that the gravitational mass and inertial mass of a body are equal. This is important when deriving the acceleration due to gravity, as shown below.

Let us suppose a body test mass (m) is dropped from a height ‘h’ above the surface of the earth source mass (M), it begins to move downwards with an increasing velocity as it approaches the earth’s surface.

We know that velocity of an object can only be altered by the application of a force, in this case, the force is provided by gravity.

Under the action of gravitational force, the body begins to accelerate towards the Earth’s centre, located a distance r away from the test mass.

Then, $$ma=\frac{GMm}{r^2}$$ (Applying principle of equivalence)

$a=G\frac{M}{r^2}$ (1)

We call the above acceleration ‘acceleration due to gravity’ because it is due to the gravitational pull of the Earth. Its value near the surface of the Earth is 9.8 ms-2, and it does not depend on the test mass.

Therefore, the acceleration due to gravity (g) is given by $G\frac{M}{r^2}$

Formula for Acceleration due to Gravity

The force acting on a body due to gravity is given by:

f = mg

Where f is the force acting on the body, g is the acceleration due to gravity, m is the mass of the body.

According to the universal law of gravitation, $f=GmM/(r+h{)}^2$

• F = Force between two bodies
• G = Universal Gravitational Constant ($6.67\times {10}^{-}11N{m}^2/k{g}^2$)
• m = mass of the object,
• M = mass of the Earth,
• r = radius of the Earth.
• h = height from the surface of the earth

As the height (h) is negligibly small compared to the radius of the earth, we re-frame the equation as follows:

$f=\frac{GmM}{r^2}$

Now equating both the expressions:

$\mathrm{mg}=\frac{GmM}{r^2}$

$g=\frac{GM}{r^2}$

Therefore, the formula of acceleration due to gravity is given by:

$g=\frac{GM}{r^2}$

• Note: The gravitational force on the earth depends on its mass and radius.

• This helps us understand:

  • All bodies experience the same acceleration due to gravity, regardless of their mass.
  • The value of the object on Earth is determined by the mass of the Earth and not the mass of the object.

Acceleration Due to Gravity on the Surface of Earth

We know that Earth is not a uniform solid sphere, but rather has a variable density due to its composition.

Density = mass/volume

Then, $\rho =\frac{M}{\frac{4}{3}\pi R^3}$

⇒ ${\displaystyle M=\rho \times \frac{4}{3}\pi R^3}$

We know that, $g=\frac{GM}{r^2}$.

On substituting the values of M, we get:

$g=\frac{4}{3}[\pi \rho RG]$

‘At any distance r from the centre of the Earth’

g = 4/3πρRG

• How ‘g’ (acceleration due to gravity) is Affected
• Altitude above the Earth’s surface
• Depth below the Earth’s surface
• The shape of the earth
• Rotational motion of the Earth.

Variation of g with Height

Variation of Acceleration due to Gravity with Height

Acceleration Due to Gravity at a Height (h) Above the Surface of the Earth

The force acting on the test mass (m) due to gravity at a height (h) from the surface of the earth is:

$F=\frac{GMm}{(R+h{)}^2}$

If the height ‘h’ is above the surface of the earth, then the acceleration due to gravity is given by ${\displaystyle g=G\cdot \frac{M}{R^2}-(G\cdot \frac{M}{{(R+h)}^2})}$ where G is the gravitational constant.

${\displaystyle m{g}_h=\frac{GMm}{{(R+h)}^2}}$

⇒ ${\displaystyle g_h=\frac{GM}{\left[R^2{(1+\frac{h}{R})}^2\right]}}$ (2)

The acceleration due to gravity on the surface of the Earth is 9.8 m/s^2.

${\displaystyle g=\frac{GM}{R^2}}$ (3)

The quotient of dividing equation (3) and (2) is:

${\displaystyle g_h={g(1+\frac{h}{R})}^{-2}}$ (4)

The acceleration due to gravity at a height above the surface of the earth decreases with an increase in height, and becomes zero at an infinite distance from the earth.

⇒ Check: Kepler’s Laws of Planetary Motion

Approximation Formula:

From Equation [4]

When h < < R, the value of g at height h is given by $g_h=\frac{g}{1-\frac{2h}{R}}$

Variation of g with Depth

The acceleration due to gravity at a distance (d) below the Earth’s surface for a test mass (m) can be calculated by taking the value of g in terms of density.

On the surface of the earth, the value of g is given by:

$g=4/3\times \pi \rho RG$

At a distance (d) below the earth’s surface, the acceleration due to gravity is given by g = g₀ (1 - (2d/Rₑ)).

$gd=\frac{4}{3}\pi \rho (R-d)G$

On dividing the above equations, we get:

$g_d=\frac{g(R-d)}{R}$

When d = 0, g_d = g on the surface of the Earth.

When the depth d = R, the value of g at the centre of the earth $g_d=0$.

The Effect of Earth’s Shape on the Variation of g

Since the Earth is an oblate spheroid, its radius near the equator is greater than its radius near the poles. Therefore, the acceleration due to gravity at any given latitude is inversely proportional to the square of the radius of the Earth, and varies with the shape of the Earth.

$g_p/g_e=R_e^2/R_p^2$

Where g_e and g_p are the accelerations due to gravity at the equator and poles, respectively, R_e and R_p are the radii of Earth near the equator and poles, respectively.

From the above equation, it is clear that acceleration due to gravity is greater at the poles than at the equator. Therefore, if a person moves from the equator to the poles, their weight will decrease due to the decrease in the value of g.

Variation of g due to the Rotation of the Earth

Every particle in a body under rotation moves in a circular path with an angular velocity ω, and this is the case for a test mass (m) on a latitude making an angle with the equator. As the Earth rotates with a constant angular velocity ω, the test mass moves in a circular path of radius r with an angular velocity ω.

In a non-inertial frame of reference, the centrifugal force ($mr{\omega }^2$) acts on the test mass, while gravity (mg) pulls it towards the centre of the earth.

As both these forces act from the same point, they are known as co-initial forces. Furthermore, as they lie along the same plane, they are termed as co-planar forces.

According to the Parallelogram Law of Vectors, if two coplanar vectors form two sides of a parallelogram, their resultant will always lie along the diagonal of the parallelogram.

Applying the parallelogram law of vectors, we can calculate the magnitude of the apparent value of the gravitational force at a given latitude.

$(m{g}^{\prime }{)}^2=(mg{)}^2+(mr{\omega }^2{)}^2+2(mg)(mr{\omega }^2)\cos (180-\theta )$ (1)

The radius of the circular path, ‘r’, is equal to the radius of the earth, ‘R’, multiplied by the cosine of the angle, θ.

Substituting $r=R\cos \theta$, we get,

$g^{\prime }=g-R{\omega }^{2}co{s}^2\theta$

Where g’ is the apparent value of acceleration due to gravity at a given latitude due to the rotation of the earth, and g is the true value of gravity at the same latitude without considering the rotation of the earth.

At the poles, when θ = 90°, then g’ = g.

$\theta =0°$ at the equator ⇒ $g\prime =g-R{\omega }^2$

Key Takeaways on Acceleration due to Gravity

• For an object placed at a height h, the acceleration due to gravity is less than that * of an object placed on the surface.
• As the depth increases, the value of acceleration due to gravity (g) decreases.
• The value of g is greater at the poles and lesser at the equator.

Frequently Asked Questions about Acceleration due to Gravity

What does the value 9.8 m/s² for acceleration due to gravity imply?

For a freely falling body, the velocity changes by 9.8 m/s every second, given the value of 9.8 m/s2 for acceleration due to gravity.

Does Acceleration Due to Gravity Change With Mass?

The acceleration due to gravity does not depend on the mass of the body.

What is the formula to calculate the force of attraction between two objects?

The formula to calculate the force of attraction between two objects is $F=G\frac{m_1{m}_2}{r^2}$, where F is the force of attraction, G is the gravitational constant, $m_1$ and $m_2$ are the masses of the two objects, and r is the distance between them.

According to the universal law of gravitation, the force of attraction between two masses, m1 and m2, separated by a distance r, is

$F=G\frac{m_1{m}_2}{r^2}$

What is Free Fall?

Free fall is the motion of a body where it is only acted on by gravity. It is the natural motion of any object that is not subject to any other type of force.

If the object moves only under the influence of gravity, it is called free fall.