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Everybody on Earth experiences a strong gravitational force of attraction from the centre of the Earth. This force is responsible for keeping bodies stuck on the surface of the Earth. As you know, acceleration due to gravity has a value of 9.81 m/s2. But this value of gravitational acceleration is not constant over the surface of the Earth. It is found that the value of ‘g’ (gravitational acceleration) decreases linearly with increasing depth.
What is Acceleration due to Gravity?
Gravitational force is one of the four fundamental forces. This force is attractive in nature and is very weak in magnitude for bodies you encounter daily. The study of gravity is very important in rocket science, navigation, geophysics, climate, etc.
Everybody produces its own field of gravity. For smaller bodies, this field is too weak to be detected by an instrument. But for big celestial bodies like planets and stars, the gravitational force becomes significant, and it can be measured with the help of instruments. The mutual force of attraction is responsible for the planetary motion in an elliptical orbit around the Sun.
According to Newton’s law of gravitation, if two bodies have mass k and k’ and if a distance x separates them, then they will experience a force of attraction as per the following formula:
F=Gkk’/x2
Where,
G = Universal gravitational constant = 6.6710-11 Nm2/Kg2
The above expression gives the gravitational force of attraction between two masses.
In the case of the Earth, when a body is sitting on the surface, it will experience an attractive force from the Earth.
If the mass of a body is k and mass of the Earth is K, and the radius of the Earth is X, then the force on that body is,
F=GKk/X2 …………….(1)
By comparing this equation with Newton’s second law,
F=ka ………………(2)
You will get the value of the gravitational acceleration of a body on the surface as,
g=a=GK/X2…………….(3)
If you substitute the values as Mass of Earth (K)=5.972191024 Kg, Universal gravitational constant = G = 6.6710-11 Nm2/Kg2 and radius of Earth = X = 6378.1 Km, you will get the value of g as 9.8 m/s2.
Variation of the Value of Gravitational Acceleration with Depth
Equation (3) gives you the value of acceleration caused by gravity on the surface of the Earth. This value is around 9.8 m/s2. But this value is not constant. This value varies with change in the altitude on the Earth’s surface and with depth below the surface.
This section will help you derive an expression for the variation of acceleration caused by gravitational force with depth inside the Earth. You can assume the Earth to be made up of concentric spherical shells. In the case of a spherical body (for instance, the Earth), the entire mass of the body is considered to be situated at the centre of the Earth.
Assuming that the density of the Earth () is uniform throughout the volume. This is given as,
=Mass/Volume = M/(4/3)X3
Thus, the mass of the Earth is given as
M=(4/3)X3
Hence, the acceleration on the surface from equation (3) is given as
a=GK/X2
∴ a=G(4/3)X3/X2=(4/3)XG …………….(4)
Now, consider a sphere of radius R-d below the surface at a depth of d. Assume that a body of mass m is sitting on point B (on the surface of a sphere with radius (X-d). The gravitational force of attraction this body will now be,
F’=GK’k/(X-d)2 ………………..(5)
Where,
K’ = Mass of the sphere inside the Earth with radius X-d
Thus, the gravitational acceleration of the body on the surface of the inner sphere will be,
a’=GK’/(X-d)2…………(6)
Where,
K’=(4/3)(X-d)3…………(7)
From equations (6) and (7),
a’=(4/3)G(X-d)3/(X-d)2
∴ a’=G(4/3)(X-d) …………..(8)
Equation (8) is the expression for the acceleration due to gravity inside the Earth at a depth of d.
By dividing equation (8) by equation (4), you get,
a’=a (4/3)G(X-d)]/(4/3)XG
∴ a’=a (X-d)/X
∴ a’=a[1-(d/X)] …………..(9)
Equation (9) describes how acceleration caused by gravity changes with depth (d) below the surface of the Earth. The equation also states that the value of acceleration due to gravity (a) decreases with depth inside the Earth’s surface.
Special Cases
Case 1
When d=0 (at the surface of the Earth),
In this case, d/X=0 and a’=a
Case 2
When d=X (at the centre of the Earth),
In this case, d/X=1 ∴ a’=0
This tells you that when a body is at the centre of the Earth and if the Earth has uniform density, then the body cannot feel any force of attraction due to gravitation from the Earth.
You can conclude the following points:
-
Gravitational acceleration of a body near the surface is maximum.
-
Gravitational acceleration of a body near the centre is minimum.
The following graph shows the variation of a with depth.
This is a linear graph with y-intercept at a=GK/X2 and x-intercept at d=X. This graph has a negative slope.
