Today post is all about

**acceleration due to gravity formula**.Because this is very basic concept of Physics , how everything is attracted towards the Earth.

So this is important for you and need to know EVERYTHING acceleration due to gravity formula.

Are you ready to know mystery about acceleration due to gravity experiment.

So In this post you will learn, what is exactly meaning of acceleration due to gravity ?

As well as how, acceleration due to gravity formula varies planet to planet.

So let’s start from very basic questions and concept, So that you can better understand.

Now dive in, what you are going to learn here today.

- What is acceleration due to gravity formula ?
- How to calculate value of acceleration due to gravity any point ?
- What’s acceleration due to gravity on mars ?
- Acceleration due to gravity experiment by Newton
- acceleration due to gravity derivation formula

And many more important concept like, examples of acceleration due to gravity.

If you want to these all questions facts keep reading. You’ll learn here some new concept. Which nobody will tell you.

Well, Do you know ? How its value is 9.8 m/s² ? and acceleration due to gravity has fixed value.

Why and how its value varies at different point location and due to Earth rotation ?

How acceleration due to gravity is a gravitational field, You will learn all concepts about of acceleration due to gravity, so lets start.

But before going to discuss in details. I like to answer some FAQ. Which is given below, So go through this FAQ.

[faq title=”

” number=”1″]The acceleration of a body, in free fall condition under the influence of Earth’s gravity. Which’s rate of increase of velocity per unit time as approaches the ground or Earth. So due to increase in velocity an acceleration is created and its value is 9.8 meter per second square. And this’s called acceleration due to gravity.[/faq_item][/faq]

**Frequently Asked Questions**(**FAQ**)” open1st=”0″ openAll=”0″][faq_item title=”What is the acceleration of gravity?” number=”1″]The acceleration of a body, in free fall condition under the influence of Earth’s gravity. Which’s rate of increase of velocity per unit time as approaches the ground or Earth. So due to increase in velocity an acceleration is created and its value is 9.8 meter per second square. And this’s called acceleration due to gravity.[/faq_item][/faq]

[faq title=”” open1st=”0″ openAll=”0″][faq_item title=”What is the formula for acceleration due to gravity?” number=”2″]Using two most important formula given by Newton F = ma and F = GmM/R*R formula for acceleration due to gravity is calculated g = G*M/R^2. Where g = a = acceleration due to gravity, G is the universal gravitational constant, M is mass, and R is radius of Earth. Note this value is on the surface of the Earth.[/faq_item][/faq]

[faq title=”” open1st=”0″ openAll=”0″][faq_item title=”What does the acceleration due to gravity depend on?” number=”3″]Acceleration due to gravity depends on the mass of the planet. And the distance from the center of planet mass. G is universal constant and Its value is = 6.673 x 10-11 N. [/faq_item][/faq]

## What is acceleration due to gravity Formula ?

Gravity is a force with which Earth attract every other object towards it centre.

Earth has created this gravity around its surrounding. This property of the Earth is due to heavy mass of the Earth.

Because compare to the Earth mass, all other objects masses are small near its surrounding, Hence all other masses falls towards centre of the Earth.

So from the figure above Earth is pulling you with force mg, you also know from the law of gravitation force between two masses.

So from the figure above Earth is pulling you with force mg, you also know from the law of gravitation force between two masses.

Therefore you can refer previous post for

**Gravitational force**the value of force between Earth and man shown in figure will be F= GMₑm/(Rₑ + h)².Because here value of h is very small compare to radius of Earth Rₑ so h is neglected.

Hence F = GMₑm/Rₑ² , Now both force will be equal as per statement.

GMₑm/Rₑ² = mg or

**g =**GMₑ/**Rₑ² ,**this is important formula where G is universal gravitational constant G = 6.67×10⁻¹¹ Nm²/kg², Rₑ is radius of Earth approx Rₑ = 6400 km.Rₑ = 6.4×10⁶ m , Mₑ is mass of Earth approx Mₑ = 6×10²⁴ kg .

Now after putting all the values in above formula g = 6.67×10⁻¹¹x6×10²⁴/(6.4×10⁶)²

g = 9.8 m/s², remember this value of g is near the earth surface only.

Now suppose if h is very high then you can’t neglect value of h, so in this case g = GMₑ/(Rₑ + h)², Now as the value of h increases then value of g will decreases, So you can’t say value of g is constant.

And acceleration due to gravity dimensional formula is [ L/TT].

On increasing the value of h from the surface of the Earth value of g will decrease. You will see how the value of g decreases as moving from the surface of the Earth in this post later.

Now see below picture and find weight.

How you’ll calculate

**acceleration due to gravity on mars**.you can do acceleration due to gravity experiment on different planet, the value will be different.

As from the question, You have to find the weight of same man at Mars, his weight is given at Earth is 70 N.

So very Interesting now scientist are planing to move to Mars booking are going on for Mars journey, But at present very costly in million $, which is not possible for a average men.

Because you should know what will be your weight at Mars ?

You know weight at Earth is mg w = mg, now weight at Mars will be w’ = mg’ here g will be different from Earth so take as (g’).

**w’/w = mg’/mg = g’/g**( You know that mass is not changed anywhere in the universe )

w’/w = g’/g you have calculated g value at Earth surface g = GMₑ/Rₑ² , Similarly for Mars

g’ = GMₘ/Rₘ² put this value of g and g’ in the expression.

w’/w = g’/g = (GMₘ/Rₘ² )/(GMₑ/Rₑ²) = (Mₘ/Mₑ)x(Rₑ/Rₘ)² now from question given that Mars mass is 10 times less than Earth so Mₑ = 10Mₘ and radius of Mars is 1.8 times less than radius of Earth Rₑ = 1.8Rₘ , now put this values in above expression you will get.

w’/w = (Mₘ/Mₑ)x(Rₑ/Rₘ)² = (Mₘ/10Mₘ)x(1.8Rₘ/Rₘ)² = 3.24/10 = 0.324

w’/w = 0.324 = w’ = wx0.324 = 70x0.324 = 22.68 N .

Hence weight at Mars for same person will be w’ = 22.68 N which is approximately 3 times less than weight at Earth .

So you will feel less attracted at Mars, you will feel more free. At moon your weight will be 6 times less than weight at the Earth, So acceleration due to gravity varies.

## Acceleration due to gravity formula Variation in gravity (g) as you go above Earth surface.

Here’s acceleration due to gravity derivation formula, how it’s calculated at different point of the Earth.

So from the above figure mg’ = GMₑm/(Rₑ+h)² or g’ = GMₑ/(Rₑ+h)² see as you will go above the Earth surface, value of h will increase and hence value of g’ will decrease.

and it is also logical that as you will be far distance from Earth surface gravity g field effect will be less on you. So it is clear that above the Earth surface gravity g will decrease.

Now compare the value of g’ with original value of g at Earth surface, So you know the value of g at Earth surface g = GMₑ/Rₑ² now from here you can write GMₑ = gRₑ² now put this value in the new g’ expression .

g’ = GMₑ/(Rₑ+h)² = gRₑ²/(Rₑ+h)² = g/(1+h/Rₑ)² after arranging the expression you can

So in terms of height, you can write new value of g’ like this.

g’ = g/(1+h/Rₑ)² now it is important ,if the height from the earth surface value of h is about 1km up to 100 km value of g’ will be nearly g, So no effect of gravity value is change near the Earth surface within this range.

But when you will go up to 500 km and more then value of g’ will change and its effect will seen. Now i will make some assumption see below

If h<< Rₑ you can write g’ = g(1+h/Rₑ)⁻² now you know from binomial expansion

(1+x)⁻ⁿ = (1-nx) if x<<1 font=”” h=”” of=”” similarly=”” value=””>Rₑ <<1 span=””>

So you can write g’ = g(1-2h/Rₑ) . See below the question.

Now examples of acceleration due to gravity.

**Q Find the height above the Earth surface, where gravity ‘g’ drops to 25% of g at surface .**

Solution for any numerical question, first you should understand well the question then you relate the given parameter So as from the question it is given that new value of gravity g’ 25% of at Earth surface hence g’ = gx25/100 = g/4 (1)

You also know that g’ = g/(1+h/Re)² (2)

hence equate expression (1) and (2) so you can write .

g/(1+h/Re)² = g/4 or (1+h/Re)² = 4 or (1+h/Re) = 2

h/Re = 2-1 = 1 or h = Re So at radius of Earth height value of gravity g’ will be 25% of Earth surface.

### acceleration due to gravity experiment Variation in gravity as you go below the Earth surface.

From figure you are h depth inside the Earth and now you are r distance from the centre of the Earth.

Now from the law of gravitation you can write mg’ = GM’m/r² understand the concept, the mass of the Earth with which now Earth will attract you will be M’ not Mₑ.

only inside circle mass will attract, outer mass will not attract you, Because due to Gauss theorem in a shell gravitational field is zero.

This is also important outer shell mass will not pull you out side because gravitational field is zero inside a shell whether it is solid or hollow in both the cases gravitational field will be zero inside a shell.

Now from expression mg’ = GM’m/r² or g’ = GM’/r²

Suppose Earth is a uniform sphere means equal mass distribution everywhere but it is not like that just we assume.

now the density of Earth is 𝛒 everywhere assumption hence M’ = volume x𝛒. = 4𝛑r³/3x𝛒

M’ = 4𝛑r³/3x𝛒

g’ = GM’/r² = Gx4𝛑r³/3x𝛒/(r²) = g’ = G𝛒4𝛑r/3 here you see g’ is directly proportional to r its mean that as you will come from the centre of the Earth towards the surface of the Earth value of r will increase and hence value of g’ will increase.

At the surface of Earth it will be maximum.

Now you can calculate the density of Earth 𝛒 = mass/volume = Me/4𝛑Rₑ³/3 put this value in above equation g’ = G𝛒4𝛑r/3 = GxMe/4𝛑Rₑ³/3x4𝛑r/3 = GMₑr/Rₑ³ here it is also clear that g’ is directly proportional to r.

You remember for always g = GMₑ/Rₑ² at the surface of Earth this is standard formula use everywhere in this topic, So from here G = gRₑ²/Mₑ , now put this value in g’ expression

g’ = GMₑr/Rₑ³ = (gRₑ²/Mₑ)x(Mₑr/Rₑ³ ) = (g/Rₑ)r or g’ = (g/Rₑ)r

g’ = GMₑr/Rₑ³ = (gRₑ²/Mₑ)x(Mₑr/Rₑ³ ) = (g/Rₑ)r or g’ = (g/Rₑ)r

Hence it clearly show as you will go inside the Earth surface value of gravity g will decrease you have seen here g’ directly proportional to r.

Now you have to write this formula in terms of h so for this see the figure, I can write

Rₑ = r+h or r = Rₑ-h, put this value in above expression

g’ = (g/Rₑ)r = (g/Rₑ)x(Rₑ-h) = g(1-h/Rₑ) or g’ = g(1-h/Rₑ) .

Rₑ = r+h or r = Rₑ-h, put this value in above expression

g’ = (g/Rₑ)r = (g/Rₑ)x(Rₑ-h) = g(1-h/Rₑ) or g’ = g(1-h/Rₑ) .

Now one question are generally asked, what will value of gravity g at the centre of the Earth ?

So you are looking the formula g’ = g(1-h/Rₑ) at the centre of Earth h = Rₑ hence g’ =0

Hence it is very much clear that at the centre of Earth g’ will be zero, and as move from the centre it will increase and at the surface it will be maximum g’ = g = GMₑ/Rₑ², now from surface to above height again value of g’ will decrease, you have already seen above.

So acceleration due to gravity is maximum at Earth Surface.

### Acceleration due to gravity on mars Variation with distance

See below the graph plotted variation of gravity with respect to distance, here you can easily see how value of g is changing with respect to distance.

Variation of gravity g with distance |

Variation in gravity g due to Earth rotation

Due to rotation of Earth value of g decreases but its value is very small due to angular velocity value is very small 𝛚 = 2𝛑/T = 2×3.14/(24x60x60) = 6.28/86400 = 7.28×10⁻⁵ rad/s

Every point on the Earth surface will have different circle of rotation, and due to centrifugal force on body, the effective value of g will decrease.

Important point due to rotation, effective g is not along the centre of the Earth, it will deviate slightly from the centre of the Earth.

After calculation g effective value will be g’ = (g-Rₑ𝛚²cos²ϴ) now at equator put ϴ =0 then

g’ = (g-Rₑ𝛚²) , now at pole ϴ = 90⁰ So g’ = g. refer below picture.

Variation of gravity due to Earth rotation |

The reason of gravity g different at pole and equator is due to rotation of the Earth as you have seen above.

Hence finally you have seen that gravity g varies above the Earth surface, inside the Earth and due to rotation of the Earth, In all the cases gravity g decreases, at the centre of Earth g value is zero.

and at the surface of the Earth maximum g = 9.8 m/s² , not a constant value.

**#Conclusion:**

If you know the basic concept, then it is easy to understand Physics. I hope you have enjoyed learning acceleration due to gravity formula .I will continue post with more hidden concepts, I want Your feedback, comment, likes and shares. If you’ve any doubt ask me, always here for your support.Thanks for sharing, learn and grow.

Dated 23rd Dec 2018.