What is the value of acceleration due to gravity(g) on the earth's surface?

The value of acceleration due to gravity at the surface of the earth is approximately 9.8m/s^2

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Acceleration Due To Gravity

Q: What is acceleration due to gravity?

Acceleration due to gravity is the rate at which objects accelerate when they fall towards Earth under the influence of gravity.

It is the rate at which objects accelerate when they fall towards Earth under the influence of gravity. It is represented by 'g' and is approximately 9.8 m/s² near the Earth's surface. This means that every second, the velocity of a falling object increases by 9.8 meters per second. All objects experience the same acceleration due to gravity, regardless of their mass, when air resistance is not considered.

1.0Gravitation and Gravity

Gravitation: Every body in the universe attracts every other body with a force called the force of gravitation. It is the force of attraction between any two bodies in the universe.

Example-The attraction between sun and the Earth, The attraction between table and the chair.

Gravity-It is a special case of gravitation, if one of the attracting bodies is the earth then the gravitation is called the gravity. It is the force of attraction between the earth and any object lying on or near its surface.

Example-The body thrown up falls back on the earth surface of the earth due to earth’s force of gravity.

2.0Acceleration Due To Gravity(g)

When a body falls freely towards the surface of earth, its velocity continuously increases. The acceleration developed in its motion is called acceleration due to gravity
The acceleration felt by a freely falling body due to Earth's gravitational pull is called acceleration due to gravity.
Represented by ‘g’.
It is a vector quantity, pointed towards the Earth's center.
It does not depend on mass, size and shape of the body.
It depends on altitude, depth, rotation of the earth and shape of the earth.
Near the earth surface $g = 9.8 m / s^{2}$ or $g = 32 f t / s^{2}$
If a body is placed in a gravitational field with intensity I, the gravitational force exerted on it is $F = m I$

Earth exerts a force on every object, pulling it towards its center.

F=mg

On comparing we get

$g = ∣ I ∣$

Here, g is the acceleration due to gravity, I is the intensity of Earth's gravitational field.

Assuming Earth to Be a Solid Sphere

Dimensions of earth assuming it be a solid sphere

$g_{S} = 9.81 m / s^{2}$

In terms of Density( $ρ$ )

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

$M a ss = V o l u m e \times De n s i t y$

$M = ρ \times \frac{4}{3} π R^{3}$

$g = \frac{G ( ρ \times \frac{4}{3} π R ^{3} )}{R ^{2}} = \frac{4}{3} π GRρ$

3.0Relation Between Acceleration Due To Gravity(g) and Gravitational Constant (G)

Relation between acceleration between gravity (g) and gravitational Constant (G)

As Per the Universal Law of Gravitation,

$F = G \frac{M m}{R ^{2}}$ …….(1)

The force of gravity F produces an acceleration g in the body of mass m

$F = m g$ ……….(2)

From equations (1) and (2)

$m g = G \frac{M m}{R ^{2}} \Rightarrow g = \frac{GM}{R ^{2}}$

4.0Variation of Acceleration Due to Gravity With Height

variation of acceleration of gravity with height

$\frac{g _{h}}{g} = \frac{\frac{G m _{e}}{( R _{e} + h ) ^{2}}}{\frac{G m _{e}}{R _{e}^{2}}} = \frac{R _{e}^{2}}{( R _{e} + h ) ^{2}}$

$\frac{g _{h}}{g} = \frac{R _{e}^{2}}{( R _{e} + h ) ^{2}} = \frac{R _{e}^{2}}{R _{e}^{2} ( 1 + \frac{h}{R _{e}} ) ^{2}} = \frac{1}{( 1 + \frac{h}{R _{e}} ) ^{2}}$ $= \frac{g _{h}}{g} = (1 + \frac{h}{R _{e}})^{- 2}$

$By Binomial Expansion$

$\frac{g _{h}}{g} = (1 + \frac{h}{R _{e}})^{- 2} \Rightarrow (1 - \frac{2 h}{R _{e}})$

$If (h << R_{e})$

$g_{h} = g (1 - \frac{2 h}{R _{e}})$

Note:

This formula is valid if h is up to 50% of rath’s radius (320 km from earth’s Surface)
If h is greater than 5% earth radius we use $g_{h} = \frac{G m _{e}}{( R _{e} + h ) ^{2}}$

Example-1.At which height from surface of earth acceleration due to gravity becomes $\frac{1}{4}$ times of its value at earth surface.

Solution: g $\to$ at earth surface $g^{'} = \frac{g}{4}$

$g_{h} = \frac{g}{( 1 + \frac{h}{R _{e}} ) ^{2}}$

For more than 5%

$\frac{g}{4} = \frac{g}{( 1 + \frac{h}{R _{e}} ) ^{2}} \Rightarrow \frac{1}{2} = \frac{1}{( 1 + \frac{h}{R _{e}} )}$ $= 2 = 1 + \frac{h}{R _{e}} \Rightarrow \frac{h}{R _{e}} = 1$

$h = R_{e} \Rightarrow h = 6400 km$

5.0Variation of Acceleration Due to Gravity With Depth

At any general point inside the earth

$g = \frac{GM}{R ^{3}} r$ R $\to$ Radius of Earth

here, r is distance from centre of Earth,

$g_{d} = \frac{GM}{R ^{3}} (R - d)$

$g_{d} = \frac{GM}{R ^{2}} (1 - \frac{d}{R})$ $\frac{GM}{R ^{2}} = g_{S} \to$ Gravity on the surface of earth

$g_{d} = g_{S} (1 - \frac{d}{R})$

For any depth below Earth Surface.

Fractional Decrement $\frac{( Δ g ) _{d}}{g} = \frac{d}{R}$ (Valid for all depths)

Note:

Valid for small or large depth
Fractional change in value of $g \to \frac{Δ g}{g} = \frac{d}{R}$
At any depth, $g_{d} < g_{s}$
Another form that can be used, $g_{d} = \frac{GM}{R ^{3}} r \Rightarrow g_{d} = g_{s} \frac{r}{R}$

Graph

For r<R	For r ≥ R
$g_{d} = \frac{GM}{R ^{3}} r \to g = g_{s} \frac{r}{R}$	$g = \frac{GM}{R ^{2}} \to g = g_{s} \frac{R ^{2}}{r ^{2}}$

Note: Value of g is maximum at Earth's surface

Example-2.A body weighs 72N on the surface of the Earth. If it is taken to a depth of R/2 from the Earth's surface ( R = radius of Earth), what would its weight be?

Solution:

$g_{d} = g_{s} (1 - \frac{d}{R}) = g_{s} (1 - \frac{R}{2 \times R})$ $= g_{d} = g_{s} (1 - \frac{1}{2}) = \frac{1}{2} g_{s}$

$m g_{d} = \frac{1}{2} m g_{s} \Rightarrow m g_{d} = \frac{1}{2} \times 72 = 36 N$

6.0Relation Between Height and Depth for Same change in Acceleration Due to Gravity(g)

$g_{h} = g (1 - \frac{2 h}{R})$ …….(1)

$g_{d} = g (1 - \frac{d}{R})$

For same change in g

$g_{h} = g_{d}$

$1 - \frac{2 h}{R} = 1 - \frac{d}{R} \Rightarrow \frac{2 h}{R} = \frac{d}{R} \Rightarrow d = 2 h$

But it holds valid for when $h << r$

7.0Alteration of Acceleration Due to Gravity Due to Rotation of Earth

$N = M g - M ω^{2} R_{e} C o s^{2} λ$

Acceleration due to gravity's variation due to the earths rotation

$M g_{e ff} = M g - M ω^{2} R_{e} C o s^{2} λ$

$g_{e ff} = g - ω^{2} R_{e} C o s^{2} λ$

Where, g : acceleration due to gravity, ignoring Earth's rotation.

$g_{e ff}$ : Acceleration due to gravity, accounting for Earth's rotation.

$λ$ :Angle of latitude

For Poles $λ$ =90°

For Equator $λ$ =0°

$N = M g_{e ff} = M g - M ω^{2} R_{e} C o s^{2} λ$ , As one moves towards pole, apparent weight increases

8.0Alteration of Acceleration Due to Gravity Due to Shape of Earth

$Δ g = g_{e} - g_{p} \approx 0.02 m / s^{2}$ , Earth is not a perfect sphere

$g_{s u r f a ce} = \frac{GM}{R ^{2}}$

$g_{p} > g_{e q}$ The surface at the poles is closer to the Earth's center than at the equator.

Earth is an oblate shape, flattened at the poles and bulging at the equator.
The equatorial radius of earth is greater than the polar radius.

Polar radius and equatorial radius of the earth

The value of gravity (g) is lowest at the equator and highest at the poles, causing a body's weight to increase when moving from the equator to the poles. The variation in g between the equator and poles is about 0.5%.

9.0Condition of Weightlessness on Earth's Surface

If Apparent weight of the body is zero than angular speed of earth can be calculated as,

$m g^{'} = m g - m R_{e} ω^{2} C o s^{2} λ \Rightarrow ω = \frac{1}{cos λ} = \frac{g}{R _{e}}$

For Equator $λ = 0°,$

$ω = \frac{1}{cos λ} = \frac{g}{R _{e}} = \frac{1}{800} r a d / s = 0.00125 r a d / s = 1.25 \times 1 0^{- 3} r a d / s$

10.0Sample Questions on Acceleration Due to Gravity

Q1. Where would you find more sugar in 1 kg: at the North Pole or the Equator?

Solution:

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

$g_{p} > g_{e q} [R_{e q} > R_{P}]$

$W = w e i g h t = m g$

At pole

$W_{P o l e} = m g_{P}$

At Equator

$W_{Eq u a t or} = m g_{e q}$

$W_{P o l e} > W_{Eq u a t or}$

So at the poles weight is more and at equator weight is less so at the equator we will get more amount of sugar in 1 kg.

Q2. What is the value of acceleration due to gravity at a certain height? 2 times the Earth's radius $(2 R_{e})$ above the Earth's surface.

Solution:

$\frac{g _{h}}{g} = \frac{1}{( 1 + \frac{h}{R _{e}} ) ^{2}}$ $h = 2 R_{e} (g i v e n)$

$g_{h} = g \frac{1}{( 1 + \frac{h}{R _{e}} ) ^{2}} = \frac{g}{( 1 + 2 ) ^{2}} = \frac{g}{9}$

$g_{h} = \frac{g}{9}$

Q3. At what depth from the earth surface acceleration due to gravity decreases by 75% of its value at the surface of earth?

Solution:

$g_{d} = g (1 - \frac{d}{R})$

Here 75% decreases means 25% become

$\frac{25}{100} g = g (1 - \frac{d}{R}) = \frac{1}{4} = 1 - \frac{d}{R} \Rightarrow \frac{d}{R} = \frac{3}{4} \Rightarrow d = \frac{3 R}{4}$

Q4. At what depth from earth surface acceleration due to gravity is decreased by 2% ?

Solution:

$\frac{Δ g}{g} = \frac{d}{R}$

$\frac{2}{100} = \frac{d}{6400} \Rightarrow d = 128 km$

Q5. What should be the angular velocity of the Earth for a person's weight at the equator to become half of its current value? (Express the answer in terms of g and R)

Solution:

Weight of the equator $W ’ = \frac{W}{2}$

$M g^{'} = M g - M R_{e} ω^{2} C o s^{2} λ$

At equator $λ$ =0

$M g^{'} = M g - M R_{e} ω^{2} C o s^{2} λ$

$M g^{'} = M g - m R_{e} ω^{2} (1)^{2} = \frac{W}{2} = W - M ω^{2} R_{e}$

$M ω^{2} R_{e} = \frac{W}{2} = \frac{M g}{2}$

$ω^{2} = \frac{g}{2 R _{e}} \Rightarrow ω = \frac{g}{2 R _{e}}$

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Acceleration Due To Gravity

It is the rate at which objects accelerate when they fall towards Earth under the influence of gravity. It is represented by 'g' and is approximately 9.8 m/s² near the Earth's surface. This means that every second, the velocity of a falling object increases by 9.8 meters per second. All objects experience the same acceleration due to gravity, regardless of their mass, when air resistance is not considered.

1.0Gravitation and Gravity

Gravitation: Every body in the universe attracts every other body with a force called the force of gravitation. It is the force of attraction between any two bodies in the universe.

Example-The attraction between sun and the Earth, The attraction between table and the chair.

Gravity-It is a special case of gravitation, if one of the attracting bodies is the earth then the gravitation is called the gravity. It is the force of attraction between the earth and any object lying on or near its surface.

Example-The body thrown up falls back on the earth surface of the earth due to earth’s force of gravity.

2.0Acceleration Due To Gravity(g)

When a body falls freely towards the surface of earth, its velocity continuously increases. The acceleration developed in its motion is called acceleration due to gravity
The acceleration felt by a freely falling body due to Earth's gravitational pull is called acceleration due to gravity.
Represented by ‘g’.
It is a vector quantity, pointed towards the Earth's center.
It does not depend on mass, size and shape of the body.
It depends on altitude, depth, rotation of the earth and shape of the earth.
Near the earth surface $g = 9.8 m / s^{2}$ or $g = 32 f t / s^{2}$
If a body is placed in a gravitational field with intensity I, the gravitational force exerted on it is $F = m I$

Earth exerts a force on every object, pulling it towards its center.

F=mg

On comparing we get

$g = ∣ I ∣$

Here, g is the acceleration due to gravity, I is the intensity of Earth's gravitational field.

Assuming Earth to Be a Solid Sphere

$g_{S} = 9.81 m / s^{2}$

In terms of Density( $ρ$ )

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

$M a ss = V o l u m e \times De n s i t y$

$M = ρ \times \frac{4}{3} π R^{3}$

$g = \frac{G ( ρ \times \frac{4}{3} π R ^{3} )}{R ^{2}} = \frac{4}{3} π GRρ$

3.0Relation Between Acceleration Due To Gravity(g) and Gravitational Constant (G)

As Per the Universal Law of Gravitation,

$F = G \frac{M m}{R ^{2}}$ …….(1)

The force of gravity F produces an acceleration g in the body of mass m

$F = m g$ ……….(2)

From equations (1) and (2)

$m g = G \frac{M m}{R ^{2}} \Rightarrow g = \frac{GM}{R ^{2}}$

4.0Variation of Acceleration Due to Gravity With Height

$\frac{g _{h}}{g} = \frac{\frac{G m _{e}}{( R _{e} + h ) ^{2}}}{\frac{G m _{e}}{R _{e}^{2}}} = \frac{R _{e}^{2}}{( R _{e} + h ) ^{2}}$

$\frac{g _{h}}{g} = \frac{R _{e}^{2}}{( R _{e} + h ) ^{2}} = \frac{R _{e}^{2}}{R _{e}^{2} ( 1 + \frac{h}{R _{e}} ) ^{2}} = \frac{1}{( 1 + \frac{h}{R _{e}} ) ^{2}}$ $= \frac{g _{h}}{g} = (1 + \frac{h}{R _{e}})^{- 2}$

$By Binomial Expansion$

$\frac{g _{h}}{g} = (1 + \frac{h}{R _{e}})^{- 2} \Rightarrow (1 - \frac{2 h}{R _{e}})$

$If (h << R_{e})$

$g_{h} = g (1 - \frac{2 h}{R _{e}})$

Note:

This formula is valid if h is up to 50% of rath’s radius (320 km from earth’s Surface)
If h is greater than 5% earth radius we use $g_{h} = \frac{G m _{e}}{( R _{e} + h ) ^{2}}$

Example-1.At which height from surface of earth acceleration due to gravity becomes $\frac{1}{4}$ times of its value at earth surface.

Solution: g $\to$ at earth surface $g^{'} = \frac{g}{4}$

$g_{h} = \frac{g}{( 1 + \frac{h}{R _{e}} ) ^{2}}$

For more than 5%

$\frac{g}{4} = \frac{g}{( 1 + \frac{h}{R _{e}} ) ^{2}} \Rightarrow \frac{1}{2} = \frac{1}{( 1 + \frac{h}{R _{e}} )}$ $= 2 = 1 + \frac{h}{R _{e}} \Rightarrow \frac{h}{R _{e}} = 1$

$h = R_{e} \Rightarrow h = 6400 km$

5.0Variation of Acceleration Due to Gravity With Depth

At any general point inside the earth

$g = \frac{GM}{R ^{3}} r$ R $\to$ Radius of Earth

here, r is distance from centre of Earth,

$g_{d} = \frac{GM}{R ^{3}} (R - d)$

$g_{d} = \frac{GM}{R ^{2}} (1 - \frac{d}{R})$ $\frac{GM}{R ^{2}} = g_{S} \to$ Gravity on the surface of earth

$g_{d} = g_{S} (1 - \frac{d}{R})$

For any depth below Earth Surface.

Fractional Decrement $\frac{( Δ g ) _{d}}{g} = \frac{d}{R}$ (Valid for all depths)

Note:

Valid for small or large depth
Fractional change in value of $g \to \frac{Δ g}{g} = \frac{d}{R}$
At any depth, $g_{d} < g_{s}$
Another form that can be used, $g_{d} = \frac{GM}{R ^{3}} r \Rightarrow g_{d} = g_{s} \frac{r}{R}$

Graph

For r<R	For r ≥ R
$g_{d} = \frac{GM}{R ^{3}} r \to g = g_{s} \frac{r}{R}$	$g = \frac{GM}{R ^{2}} \to g = g_{s} \frac{R ^{2}}{r ^{2}}$

Note: Value of g is maximum at Earth's surface

Example-2.A body weighs 72N on the surface of the Earth. If it is taken to a depth of R/2 from the Earth's surface ( R = radius of Earth), what would its weight be?

Solution:

$g_{d} = g_{s} (1 - \frac{d}{R}) = g_{s} (1 - \frac{R}{2 \times R})$ $= g_{d} = g_{s} (1 - \frac{1}{2}) = \frac{1}{2} g_{s}$

$m g_{d} = \frac{1}{2} m g_{s} \Rightarrow m g_{d} = \frac{1}{2} \times 72 = 36 N$

6.0Relation Between Height and Depth for Same change in Acceleration Due to Gravity(g)

$g_{h} = g (1 - \frac{2 h}{R})$ …….(1)

$g_{d} = g (1 - \frac{d}{R})$

For same change in g

$g_{h} = g_{d}$

$1 - \frac{2 h}{R} = 1 - \frac{d}{R} \Rightarrow \frac{2 h}{R} = \frac{d}{R} \Rightarrow d = 2 h$

But it holds valid for when $h << r$

7.0Alteration of Acceleration Due to Gravity Due to Rotation of Earth

$N = M g - M ω^{2} R_{e} C o s^{2} λ$

$M g_{e ff} = M g - M ω^{2} R_{e} C o s^{2} λ$

$g_{e ff} = g - ω^{2} R_{e} C o s^{2} λ$

Where, g : acceleration due to gravity, ignoring Earth's rotation.

$g_{e ff}$ : Acceleration due to gravity, accounting for Earth's rotation.

$λ$ :Angle of latitude

For Poles $λ$ =90°

For Equator $λ$ =0°

$N = M g_{e ff} = M g - M ω^{2} R_{e} C o s^{2} λ$ , As one moves towards pole, apparent weight increases

8.0Alteration of Acceleration Due to Gravity Due to Shape of Earth

$Δ g = g_{e} - g_{p} \approx 0.02 m / s^{2}$ , Earth is not a perfect sphere

$g_{s u r f a ce} = \frac{GM}{R ^{2}}$

$g_{p} > g_{e q}$ The surface at the poles is closer to the Earth's center than at the equator.

Earth is an oblate shape, flattened at the poles and bulging at the equator.
The equatorial radius of earth is greater than the polar radius.

The value of gravity (g) is lowest at the equator and highest at the poles, causing a body's weight to increase when moving from the equator to the poles. The variation in g between the equator and poles is about 0.5%.

9.0Condition of Weightlessness on Earth's Surface

If Apparent weight of the body is zero than angular speed of earth can be calculated as,

$m g^{'} = m g - m R_{e} ω^{2} C o s^{2} λ \Rightarrow ω = \frac{1}{cos λ} = \frac{g}{R _{e}}$

For Equator $λ = 0°,$

$ω = \frac{1}{cos λ} = \frac{g}{R _{e}} = \frac{1}{800} r a d / s = 0.00125 r a d / s = 1.25 \times 1 0^{- 3} r a d / s$

10.0Sample Questions on Acceleration Due to Gravity

Q1. Where would you find more sugar in 1 kg: at the North Pole or the Equator?

Solution:

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

$g_{p} > g_{e q} [R_{e q} > R_{P}]$

$W = w e i g h t = m g$

At pole

$W_{P o l e} = m g_{P}$

At Equator

$W_{Eq u a t or} = m g_{e q}$

$W_{P o l e} > W_{Eq u a t or}$

So at the poles weight is more and at equator weight is less so at the equator we will get more amount of sugar in 1 kg.

Q2. What is the value of acceleration due to gravity at a certain height? 2 times the Earth's radius $(2 R_{e})$ above the Earth's surface.

Solution:

$\frac{g _{h}}{g} = \frac{1}{( 1 + \frac{h}{R _{e}} ) ^{2}}$ $h = 2 R_{e} (g i v e n)$

$g_{h} = g \frac{1}{( 1 + \frac{h}{R _{e}} ) ^{2}} = \frac{g}{( 1 + 2 ) ^{2}} = \frac{g}{9}$

$g_{h} = \frac{g}{9}$

Q3. At what depth from the earth surface acceleration due to gravity decreases by 75% of its value at the surface of earth?

Solution:

$g_{d} = g (1 - \frac{d}{R})$

Here 75% decreases means 25% become

$\frac{25}{100} g = g (1 - \frac{d}{R}) = \frac{1}{4} = 1 - \frac{d}{R} \Rightarrow \frac{d}{R} = \frac{3}{4} \Rightarrow d = \frac{3 R}{4}$

Q4. At what depth from earth surface acceleration due to gravity is decreased by 2% ?

Solution:

$\frac{Δ g}{g} = \frac{d}{R}$

$\frac{2}{100} = \frac{d}{6400} \Rightarrow d = 128 km$

Q5. What should be the angular velocity of the Earth for a person's weight at the equator to become half of its current value? (Express the answer in terms of g and R)

Solution:

Weight of the equator $W ’ = \frac{W}{2}$

$M g^{'} = M g - M R_{e} ω^{2} C o s^{2} λ$

At equator $λ$ =0

$M g^{'} = M g - M R_{e} ω^{2} C o s^{2} λ$

$M g^{'} = M g - m R_{e} ω^{2} (1)^{2} = \frac{W}{2} = W - M ω^{2} R_{e}$

$M ω^{2} R_{e} = \frac{W}{2} = \frac{M g}{2}$

$ω^{2} = \frac{g}{2 R _{e}} \Rightarrow ω = \frac{g}{2 R _{e}}$

Frequently Asked Questions

What is acceleration due to gravity?

What is the value of acceleration due to gravity(g) on the earth's surface?

Join ALLEN!

Acceleration Due To Gravity

1.0Gravitation and Gravity

2.0Acceleration Due To Gravity(g)

3.0Relation Between Acceleration Due To Gravity(g) and Gravitational Constant (G)

4.0Variation of Acceleration Due to Gravity With Height

5.0Variation of Acceleration Due to Gravity With Depth

6.0Relation Between Height and Depth for Same change in Acceleration Due to Gravity(g)

7.0Alteration of Acceleration Due to Gravity Due to Rotation of Earth

8.0Alteration of Acceleration Due to Gravity Due to Shape of Earth

9.0Condition of Weightlessness on Earth's Surface

10.0Sample Questions on Acceleration Due to Gravity

Table of Contents

Frequently Asked Questions

What is acceleration due to gravity?

What is the value of acceleration due to gravity(g) on the earth's surface?

Join ALLEN!

Acceleration Due To Gravity

1.0Gravitation and Gravity

2.0Acceleration Due To Gravity(g)

3.0Relation Between Acceleration Due To Gravity(g) and Gravitational Constant (G)

4.0Variation of Acceleration Due to Gravity With Height

5.0Variation of Acceleration Due to Gravity With Depth

6.0Relation Between Height and Depth for Same change in Acceleration Due to Gravity(g)

7.0Alteration of Acceleration Due to Gravity Due to Rotation of Earth

8.0Alteration of Acceleration Due to Gravity Due to Shape of Earth

9.0Condition of Weightlessness on Earth's Surface

10.0Sample Questions on Acceleration Due to Gravity

Table of Contents