How To Find Acceleration In G's

What is Acceleration due to Gravity?

Acceleration due to gravity is the acceleration gained by an object due to gravitational force. Its SI unit is thou/south². It has both magnitude and direction, hence, information technology'due south a vector quantity. Acceleration due to gravity is represented by m. The standard value of yard on the surface of the earth at body of water level is 9.eight m/sⁱⁱ.

JEE Main 2021 Live Physics Paper Solutions 24-Feb Shift-1 Memory-Based

JEE Main 2021 LIVE Physics Paper Solutions 24-Feb Shift-1

Dispatch due to Gravity – Formula, Unit of measurement and Values

Acceleration Due to Gravity (g)
Symbol	one thousand
Dimensional Formula	Yard⁰L¹T^-two
SI Unit	ms^-2
Formula	g = GM/r²
Values of g in SI	9.806 ms^-2
Values of thousand in CGS	980 cm due south^-2

Table of Content:

What is Gravity?
Formula
one thousand on Earth
Value of g with Height
g with Depth
g due to Shape of Earth
g due to Rotation

What is Gravity?

Gravity is the force with which theearth attracts a body towards its centre. Let us consider two bodies of masses m_a and thousand_b. Under the awarding of equal forces on two bodies, the force in terms of mass is given past:

m_b= thousand_a[a_A/a_B] this is called an inertial mass of a body.

Under the gravitational influence on two bodies,

F_A= GMm_A/rⁱⁱ,
F_B= GMm_B/r²,
m_B= [F_B/F_A] × grand_A

⇒ More on Gravitation:

Newton'due south Law of Gravitation
Gravitational Potential Free energy
Gravitational Field Intensity

The higher up mass is called a gravitational mass of a torso. According to the principle of equivalence, the inertial mass and gravitational mass are identical. Nosotros volition be using this while deriving acceleration due to the gravity given below.

Let u.s.a. suppose a body [test mass (k)] is dropped from a height 'h' above the surface of the earth [source mass (M)], it begins to motility downward with an increase in velocity as information technology reaches close to the globe surface.

We know that velocity of an object changes just under the action of a force, in this case, the forcefulness is provided by the gravity.

Under the activeness of gravitational force, the body begins to accelerate toward the earth's middle which is at a distance 'r' from the test mass.

Then, ma = GMm/r² (Applying principle of equivalence)

⇒ a = GM/rⁱⁱ . . . . . . . (1)

The above acceleration is due to the gravitational pull of earth so we call information technology dispatch due to gravity, it does non depend upon the test mass. Its value well-nigh the surface of the world is 9.viii ms^-2.

Therefore, the dispatch due to gravity (g) is given by = GM/r².

Formula of Acceleration due to Gravity

Force acting on a body due to gravity is given past, f = mg

Where f is the force acting on the trunk, 1000 is the acceleration due to gravity, k is mass of the body.

According to the universal law of gravitation, f = GmM/(r+h)^two

Where,

f = force betwixt 2 bodies,
G = universal gravitational constant (vi.67×10^-xi Nm²/kg²)
m = mass of the object,
M = mass of the earth,
r = radius of the earth.
h = top at which the body is from the surface of the world.

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

f = GmM/r²

At present equating both the expressions,

mg = GmM/r²

⇒ m = GM/r²

Therefore, the formula of acceleration due to gravity is given by, g = GM/r^two

Notation: It depends on the mass and radius of the earth.

This helps us empathise the following:

All bodies experience the aforementioned acceleration due to gravity, irrespective of its mass.
Its value on earth depends upon the mass of the globe and not the mass of the object.

Dispatch due to Gravity on the Surface of Earth

Earth every bit causeless to exist a uniform solid sphere with a mean density. Nosotros know that,

Density = mass/volume

So, ρ = M/[4/3 πR^three]

⇒ 1000 = ρ × [four/3 πRⁱⁱⁱ]

We know that, grand = GM/R².

On substituting the values of M we go,

g = iv/3 [πρRG]

At whatever distance 'r' from the centre of the earth

g = four/three [πρRG]

The value of acceleration due to gravity 'chiliad' is afflicted past

Altitude in a higher place the globe's surface.
Depth below the earth'southward surface.
The shape of the earth.
Rotational motion of the earth.

Variation of yard with Height

Variation of Acceleration due to Gravity with Height

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

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

F = GMm/(R+h)ⁱⁱ

Where Thou is the mass of world and R is the radius of the earth. The acceleration due to gravity at a certain tiptop is 'h' and so,

mg_h= GMm/(R+h)^two

⇒ g_h= GM/[R²(1+ h/R)^two ] . . . . . . (ii)

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

g = GM/R² . . . . . . . . . (3)

On dividing equation (3) and (2) we become,

one thousand_h= m (i+h/R)^-2. . . . . . (4)

This is the dispatch due to gravity at a height above the surface of the globe. Observing the higher up formula nosotros can say that the value of thousand decreases with increase in height of an object and the value of thou becomes zero at 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 thou_h= grand/(ane – 2h/R)

Variation of m with Depth

Variation of Acceleration due to Gravity with Depth

Consider a test mass (m) taken to a distance (d) beneath the earth'southward surface, the acceleration due to gravity that bespeak (g_d) is obtained by taking the value of g in terms of density.

On the surface of the globe, the value of one thousand is given by;

g = iv/three × πρRG

At a distance (d) below the earth's surface, the acceleration due to gravity is given past;

g_d= four/3 × πρ × (R – d) M

On dividing the higher up equations we go,

g_d= thou (R – d)/R

When the depth d = 0, the value of thou on the surface of the earth chiliad_d= g.
When the depth d = R, the value of g at the centre of the world k_d= 0.

Variation of chiliad due to Shape of Earth

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

g_p/yard_e = R^two _e/R^two _p

Where thousand_{due east} and g_p are the accelerations due to gravity at equator and poles, R_eand R_p are the radii of globe nearly equator and poles, respectively.

From the above equation, information technology is clear that acceleration due to gravity is more at poles and less at the equator. Then if a person moves from the equator to poles his weight decreases as the value of chiliad decreases.

Variation of g due to Rotation of Earth

Consider a examination mass (k) is on a latitude making an angle with the equator. Equally we accept studied, when a trunk is nether rotation every particle in the body makes circular motions nigh the centrality of rotation. In the present case, the earth is under rotation with a constant angular velocity ω, and so the examination mass moves in a circular path of radius 'r' with an angular velocity ω.

This is the case of a non-inertial frame of reference so at that place exists a centrifugal force on the test mass (mrω²). Gravity is acting on the examination mass towards the middle of the earth (mg).

As both these forces are acting from the same betoken these are known as co-initial forces and equally they lie along the same aeroplane they are termed equally co-planar forces.

We know from parallelogram law of vectors, if two coplanar vectors are forming ii sides of a parallelogram then the resultant of those 2 vectors volition always along the diagonal of the parallelogram.

Applying parallelogram police force of vectors we get the magnitude of the apparent value of the gravitational force at the latitude

(mg′)^two= (mg)²+ (mrωⁱⁱ)^two+ ii(mg) (mrω²) cos(180 – θ) . . . . . . (1)

Nosotros know 'r' is the radius of the circular path and 'R' is the radius of the earth, so r = Rcosθ.

Substituting r = R cosθ nosotros get,

thou′ = g – Rω²cos²θ

Where g′ is the apparent value of dispatch due to gravity at the latitude due to the rotation of the globe and one thousand is the true value of gravity at the latitude without considering the rotation of the earth.

At poles, θ = xc°⇒ 1000' = grand.

At the equator, θ = 0° ⇒ g′= g – Rω².

Important Conclusions on Acceleration due to Gravity :

For an object placed at a height 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 (grand) falls.
The value of g is more at poles and less at the equator.