Today post is all about acceleration due to gravity, In this post you will learn, what is acceleration due to gravity ? How its value is 9.8 m/s², 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.
What is acceleration due to gravity ?
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.
From the figure above Earth is pulling you with force mg, you also know from the law of gravitation force between two masses, 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
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.
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.
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.
Weight at Mars
As from the question, You have to find the weight of same man at Mars, his weight is given at Earth is 70 N.
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.
So 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.
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.
Variation in gravity (g) as you go above Earth surface.
Variation in g above Earth Surface
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
So you can write g’ = g(1-2h/Rₑ) . See below the question.
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.
Variation in gravity as you go below the Earth surface.
Variation of gravity g inside the Earth
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
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) . 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.
Variation of acceleration due to gravity g 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.
I hope you have enjoyed learning acceleration due to gravity .I will continue post with more hidden concepts, I want Your feedback, comment, likes and shares, Thanks for sharing, learn and grow.