# Acceleration Due to Gravity

**Acceleration Due to Gravity:**

**Gravity:**

Gravity is the force by which the Earth pulls an object towards its center. Consider two mass bodies, ma, and mb. If equal forces are applied to two objects, the force corresponding to the mass is given by the formula

m_{b }= m_{a [$\frac{{a}_{A}}{{a}_{B}}$}_{]}

This is called the inertial mass of the body.

Due to the gravity of two bodies

The upper mass is called the gravitational mass of the body. According to the equivalence principle, inertial mass and gravitational mass are identical. We use this to derive the acceleration due to gravity shown below.

Suppose a body [test mass (m)] is dropped from a height 'h' above the ground [source mass (M)]; As it approaches the ground, it begins to move downward with increasing speed.

We know that the speed of an object is only changed by a force; in this case, the force is due to gravity.

Under the influence of gravity, the body begins to accelerate towards the center of the Earth, which is at a distance "r" from the test mass. Then ma = GMm/r2 (applying the equivalence principle)

The above acceleration is due to the pull of the earth, so we call it the acceleration due to gravity; it does not depend on the mass of the test. Its value near the ground is 9.8 ms-².

Therefore, the acceleration due to gravity (g) is given by = GM/r². The formula for acceleration due to gravity.

The force acting on the body under the influence of gravity is given by the formula f = mg

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

According to the law of universal gravitation

f = GmM/(r h)²

where,

f = force between two bodies

G = universal gravitational constant (6.67 × 10^{-11} Nm²/kg²)

m = mass of the object

M = Earth's mass

r = radius of the Earth

h = height at which the object is above the ground

Since the height (h) is negligible compared to the radius of the earth, let's reformulate the equation as follows:

f = GmM/r²

Now that both expressions are equated,

mg = GmM/r²

⇒ g = GM/r²

Therefore, the formula for the acceleration due to gravity is g = GM/r²

Note: This depends on the mass and radius of the earth.

This helps us understand the following.

All bodies experience the same acceleration due to gravity, regardless of their mass.

Its value on earth depends on the mass of the earth, not the mass of the object.

**Acceleration Due to Gravity on the Surface of Earth**

Earth is assumed to be a uniform solid sphere with a mean density. We know that,

Density = mass/volume

Then, ρ = M/[4/3 πR^{3}]

⇒ M = ρ × [4/3 πR^{3}]

We know that g = GM/R^{2}.

On substituting the values of M, we get,

g = 4/3 [πρRG]

At any distance ‘r’ from the centre of the earth,

g = 4/3 [πρRG]

The value of acceleration due to gravity ‘g’ is affected by

« 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:**

Acceleration due to gravity at a height (h) from the surface of the earth

Consider a test mass (m) at a height (h) from the surface of the earth. Now, the force acting on the test mass due to gravity is

F = GMm/(R+h)^{2}

Where M is the mass of the earth, and R is the radius of the earth. The acceleration due to gravity at a certain height is ‘h’, then

mgh= GMm/(R+h)^{2}

⇒ gh= GM/[R^{2}(1+ h/R)^{2} ] . . . . . . (2)

The acceleration due to gravity on the surface of the earth is given by

g = GM/R^{2 }. . . . . . . . . (3)

On dividing equations (3) and (2), we get

gh = g (1+h/R)^{-2}. . . . . . (4)

This is the acceleration due to gravity at a height above the surface of the earth. Observing the above formula, we can say that the value of g decreases with an increase in the height of an object, and the value of g becomes zero at an infinite distance from the earth.

**Approximation Formula:**

From equation (4)

when h << R,

the value of g at height ‘h’ is given by gh = g/(1 – 2h/R)

**Variation of g with Depth:**

Variation of Acceleration due to Gravity with Depth

Consider a test mass (m) taken to a distance (d) below the earth’s surface, the acceleration due to gravity at that point (gd) is obtained 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 × πρRG

At a distance (d) below the earth’s surface, the acceleration due to gravity is given by

gd = 4/3 × πρ × (R – d) G

On dividing the above equations, we get,

gd = g (R – d)/R

When the depth d = 0, the value of g on the surface of the earth gd = g.

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

**Variation of g Due to the Shape of the Earth**

As the earth is an oblate spheroid, its radius near the equator is more than its radius near the poles. Since for a source mass, the acceleration due to gravity is inversely proportional to the square of the radius of the earth, it varies with latitude due to the shape of the earth.

From the above equation, it is clear that acceleration due to gravity is more at the poles and less at the equator. So if a person moves from the equator to the poles, their weight decreases as the value of g decreases.

**Variation of g Due to Rotation of Earth**

Consider a test mass (m) is on a latitude making an angle with the equator. As we have studied, when a body is under rotation, every particle in the body makes circular motions about the axis of rotation. In the present case, the earth is under rotation with a constant angular velocity ω, then the test mass moves in a circular path of radius ‘r’ with an angular velocity ω.

^{2}). Gravity is acting on the test mass towards the centre of the earth (mg).

We know from the parallelogram law of vectors if two co-planar vectors are forming two sides of a parallelogram, then the resultant of those two vectors will always be along the diagonal of the parallelogram.

Applying the parallelogram law of vectors, we get the magnitude of the apparent value of the gravitational force at the latitude.

(mg′)^{2} = (mg)^{2} + (mrω^{2})^{2} + 2(mg) (mrω^{2}) cos(180 – θ) . . . . . . (1)

We know ‘r’ is the radius of the circular path and ‘R’ is the radius of the earth, then r = Rcosθ.

Substituting r = R cosθ. we get,

g′ = g – Rω^{2}cos2θ

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

At poles, θ = 90°⇒ g’ = g.

At the equator, θ = 0° ⇒ g′= g – Rω^{2}.

**Important Conclusions on Acceleration Due to Gravity**** **

· For an object placed at a height of h, the acceleration due to gravity is less as compared to that placed on the surface.

· As depth increases, the value of acceleration due to gravity (g) falls.

· The value of g is more at the poles and less at the equator.