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 m/s^{2}. It has both magnitude and direction; hence, it’s a vector quantity. Acceleration due to gravity is represented by g. The standard value of g on the surface of the earth at sea level is 9.8 m/s^{2}.
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Acceleration Due to Gravity – Formula, Unit and Values
Acceleration Due to Gravity (g) | |
Symbol | g |
Dimensional Formula | M^{0}L^{1}T^{-2} |
SI Unit | ms^{-2} |
Formula | g = GM/r^{2} |
Values of g in SI | 9.806 ms^{-2} |
Values of g in CGS | 980 cm s^{-2} |
Table of Contents
- What Is Gravity?
- Formula
- g on Earth
- Value of g with Height
- g with Depth
- g Due to the Shape of the Earth
- g Due to Rotation
What Is Gravity?
Gravity is the force with which the earth attracts a body towards its centre. Let us consider two bodies of masses, m_{a} and m_{b}. Under the application of equal forces on two bodies, the force in terms of mass is given by
m_{b }= m_{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}/r^{2},
- F_{B }= GMm_{B}/r^{2},
- m_{B }= [F_{B}/F_{A}] × m_{A}
⇒ More on Gravitation:
The above mass is called a gravitational mass of a body. According to the principle of equivalence, the inertial mass and gravitational mass are identical. We will be using this while deriving acceleration due to the gravity given below.
Suppose a body [test mass (m)] is dropped from a height ‘h’ above the surface of the earth [source mass (M)]; it begins to move downwards with an increase in velocity as it reaches close to the earth’s surface.
We know that the velocity of an object changes only under the action of a force; in this case, the force is provided by gravity.
Under the action of gravitational force, the body begins to accelerate toward the earth’s centre, which is at a distance ‘r’ from the test mass.
Then, ma = GMm/r^{2} (Applying principle of equivalence)
⇒ a = GM/r^{2} . . . . . . . (1)
The above acceleration is due to the gravitational pull of the earth, so we call it acceleration due to gravity; it does not depend upon the test mass. Its value near the surface of the earth is 9.8 ms^{-2}.
Therefore, the acceleration due to gravity (g) is given by = GM/r^{2}.
Formula of Acceleration Due to Gravity
Force acting on a body due to gravity is given by f = mg
Where f is the force acting on the body, g is the acceleration due to gravity, and m is the mass of the body.
According to the universal law of gravitation, f = GmM/(r+h)^{2}
Where,
- f = Force between two bodies
- G = Universal gravitational constant (6.67×10^{-11} Nm^{2}/kg^{2})
- m = Mass of the object
- M = Mass of the earth
- r = Radius of the earth
- h = Height at which the body is from the surface of the earth
As the height (h) is negligibly small compared to the radius of the earth, we re-frame the equation as follows:
f = GmM/r^{2}
Now, equating both expressions,
mg = GmM/r^{2}
⇒ g = GM/r^{2}
Therefore, the formula of acceleration due to gravity is given by g = GM/r^{2}
Note: It depends on the mass and radius of the earth.
This helps us understand the following:
- All bodies experience the same acceleration due to gravity, irrespective of their mass.
- Its value on earth depends upon the mass of the earth and not the mass of the object.
Acceleration Due to Gravity on the Surface of Earth
Earth is assumed to be a uniform solid sphere with a mean density. We know that,
Density = mass/volume
Then, ρ = M/[4/3 πR^{3}]
⇒ M = ρ × [4/3 πR^{3}]
We know that g = GM/R^{2}.
On substituting the values of M, we get,
g = 4/3 [πρRG]
At any distance ‘r’ from the centre of the earth,
g = 4/3 [πρRG]
The value of acceleration due to gravity ‘g’ is affected by
- Altitude above the earth’s surface.
- Depth below the earth’s surface.
- The shape of the earth.
- Rotational motion of the earth.
Variation of g with Height
Acceleration due to gravity at a height (h) from the surface of the earth
Consider a test mass (m) at a height (h) from the surface of the earth. Now, the force acting on the test mass due to gravity is
F = GMm/(R+h)^{2}
Where M is the mass of the earth, and R is the radius of the earth. The acceleration due to gravity at a certain height is ‘h’, then
mg_{h}= GMm/(R+h)^{2}
⇒ g_{h}= GM/[R^{2}(1+ h/R)^{2} ] . . . . . . (2)
The acceleration due to gravity on the surface of the earth is given by
g = GM/R^{2} . . . . . . . . . (3)
On dividing equations (3) and (2), we get
g_{h }= g (1+h/R)^{-2}. . . . . . (4)
This is the acceleration due to gravity at a height above the surface of the earth. Observing the above formula, we can say that the value of g decreases with an increase in the height of an object, and the value of g becomes zero at an 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 g_{h }= g/(1 – 2h/R)
Variation of g with Depth
Consider a test mass (m) taken to a distance (d) below the earth’s surface, the acceleration due to gravity at that point (g_{d}) is obtained by taking the value of g in terms of density.
On the surface of the earth, the value of g is given by
g = 4/3 × πρRG
At a distance (d) below the earth’s surface, the acceleration due to gravity is given by
g_{d }= 4/3 × πρ × (R – d) G
On dividing the above equations, we get,
g_{d }= g (R – d)/R
- When the depth d = 0, the value of g on the surface of the earth g_{d }= g.
- When the depth d = R, the value of g at the centre of the earth g_{d }= 0.
Variation of g Due to the Shape of the Earth
As the earth is an oblate spheroid, its radius near the equator is more than its radius near the poles. Since for a source mass, the acceleration due to gravity is inversely proportional to the square of the radius of the earth, it varies with latitude due to the shape of the earth.
g_{p}/g_{e} = R^{2}_{e}/R^{2}_{p}
Where g_{e} and g_{p} are the accelerations due to gravity at the equator and poles, R_{e }and R_{p} are the radii of the earth near the equator and poles.
From the above equation, it is clear that acceleration due to gravity is more at the poles and less at the equator. So if a person moves from the equator to the poles, their weight decreases as the value of g decreases.
Variation of g Due to Rotation of Earth
Consider a test mass (m) is on a latitude making an angle with the equator. As we have studied, when a body is under rotation, every particle in the body makes circular motions about the axis of rotation. In the present case, the earth is under rotation with a constant angular velocity ω, then the test 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 there exists a centrifugal force on the test mass (mrω^{2}). Gravity is acting on the test mass towards the centre of the earth (mg).
As both these forces are acting from the same point, these are known as co-initial forces, and as they lie along the same plane, they are termed co-planar forces.
We know from the parallelogram law of vectors if two co-planar vectors are forming two sides of a parallelogram, then the resultant of those two vectors will always be along the diagonal of the parallelogram.
Applying the parallelogram law of vectors, we get the magnitude of the apparent value of the gravitational force at the latitude.
(mg′)^{2 }= (mg)^{2 }+ (mrω^{2})^{2 }+ 2(mg) (mrω^{2}) cos(180 – θ) . . . . . . (1)
We know ‘r’ is the radius of the circular path and ‘R’ is the radius of the earth, then r = Rcosθ.
Substituting r = R cosθ. we get,
g′ = g – Rω^{2}cos^{2}θ
Where g′ is the apparent value of acceleration due to gravity at the latitude due to the rotation of the earth, and g is the true value of gravity at the latitude without considering the rotation of the earth.
At poles, θ = 90°⇒ g’ = g.
At the equator, θ = 0° ⇒ g′= g – Rω^{2}.
Important Conclusions on Acceleration Due to Gravity
- For an object placed at a height of 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 (g) falls.
- The value of g is more at the poles and less at the equator.
Gravitation
Gravitational Potential Energy
Frequently Asked Questions on Acceleration Due to Gravity
What does the value 9.8 m/s^{2} for acceleration due to gravity imply?
The value 9.8 m/s^{2} for acceleration due to gravity implies that for a freely falling body, the velocity changes by 9.8 m/s every second.
Does mass have any effect on acceleration due to gravity?
The acceleration due to gravity is independent of the mass of the body.
What is the formula to calculate the force of attraction between two objects?
If m_{1} and m_{2} are the two masses separated by a distance r. According to the universal law of gravitation, the force of attraction between them is
F = G (m_{1} m_{2}/r^{2})
What is free fall?
If the object moves only under the influence of gravity, it is called free fall.
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