Sub Topics

To know about the concept behind the "Acceleration due to gravity", just drop two objects with same velocity. Normally, two objects will touch the ground at the same time because, both accelerates at same rate. This is denoted as 'g'. When we consider the acceleration due to gravity, must neglect the air resistance. To clarify this just consider an example. When a coin and a feather are dropped from the same height at the same time. The coin will reach the ground first because it is less susceptible to resistance of air. The acceleration due to gravity may refer to gravitational acceleration, gravity of earth, and standard gravity. In general,gravitational acceleration is the acceleration caused by the attraction of two bodies of different masses. Gravity of earth is the acceleration caused by the attraction of the earth to a body. And standard gravity is the standard value of gravitational acceleration at sea level.The gravitational acceleration is the acceleration on a body caused by gravity. Avoiding the friction like air resistance, every objects accelerate in a gravitational field at the same rate. This is true for all bodies regardless of its masses or compositions of the bodies. 
Acceleration due to gravity is defined as the inherent property of earth to attract every object to its center. There is a force known as force of gravity and the acceleration generated by this force is known as the acceleration due to gravity. It is denoted by the letter 'g'.
We know that the acceleration of any object due to gravitational force is acceleration due to gravity. So the equation of 'g' is given by,
The force due to gravity is
F = mg
Where m is the mass of the object
g is the gravity
The equation for Universal law of gravitation is
F = $\frac{GMm}{r^{2}}$
Where G is the gravitational constant
M is the mass of the first body
m is the mass of the second body
r is the distance between two bodies
Equating the two equations we get,
mg = $\frac{GMm}{r^{2}}$
Hence, the equation of acceleration due to gravity is
g = $\frac{GM}{r^{2}}$
The SI unit of acceleration due to gravity is m/s^{2} . And it is also expressed in N/Kg.g = $\frac{GM}{r^{2}}$
Gravitational acceleration is defined the acceleration on a body caused by force of gravity. According to Newton's Universal law of gravity, the gravitational force between two bodies is directly proportional to the product of magnitude of the masses and inversely proportional to the square of the distance between them. Mathematically it is represented as,
F $ \propto \frac{Mm}{r^{2}}$
or,
F = $\frac{GMm}{r^{2}}$
or,
F = $\frac{GMm}{r^{2}}$
Where G is the gravitational acceleration constant and it is given by
G = 6.67 × 10^{11} Nm2/Kg^{2}
M is the mass of the first body
m is the mass of the second body
r is the distance between the two bodies
The following graph represents the relation between the mass, gravity, and distance.
Acceleration or the rate of change of velocity is existing because of a net force acting on a body. A permanent force is present on earth which is pulling downward to every body to its center. If we consider gravity is the only force which is acting on a body, then we can find the body will accelerate at a rate of 9.8 m/s^{2} vertically downward direction to the center of the earth. Every object fall at this same rate if there is no air friction is present.
An easy proof for this is to drop a coin and a feather from the same height at same time, coin will reach the ground first than the feather. Because small force of gravity which opposed by the air resistance on the feather. But the force of gravity of coin is larger than the air friction. So, the coin will reach first.
An easy proof for this is to drop a coin and a feather from the same height at same time, coin will reach the ground first than the feather. Because small force of gravity which opposed by the air resistance on the feather. But the force of gravity of coin is larger than the air friction. So, the coin will reach first.
1. Free fall adapter experiment
This experiment consists of free fall adapter, support rod, meter scale, balls, stand etc.
The aim of this experiment is to find out the acceleration due to gravity of a free falling object assuming that the force acting on this body is only the gravitational force.
We know that the equation of motion of a body from rest and undergo constant acceleration is given by
d = $\frac{1}{2}at^{2}$
where d is the distance the object has traveled from its beginning point, a is the acceleration of the body and t is the time taken.
So acceleration is given by,
2. Pendulum experiment
A pendulum consists of a small mass or a bob suspended on a string. It can oscillate to and fro periodically. Such periodic motion is known as the simple harmonic motion. The time of the to and fro motion is known as the period of the pendulum. Period does not depend up on the mass of the pendulum, the size of the arc through which it oscillates or initial angular displacement. Whereas it depends only on the length of the pendulum and the acceleration due to gravity. According to the equation of the period,
T = $2\Pi \sqrt{\frac{L}{g}}$
Where, T = $\frac{Time\ of\ oscillation}{Number\ of\ oscillation}$
So,
The first procedure of this experiment is measure the length of the pendulum using a meter scale. And note down the time taken for say, 30 oscillation and find out the value of T using the above equation. Substitute these values in the given equation, we will get the value of acceleration due to gravity.
We know that the general expression for acceleration due to gravity isThis experiment consists of free fall adapter, support rod, meter scale, balls, stand etc.
The aim of this experiment is to find out the acceleration due to gravity of a free falling object assuming that the force acting on this body is only the gravitational force.
We know that the equation of motion of a body from rest and undergo constant acceleration is given by
d = $\frac{1}{2}at^{2}$
where d is the distance the object has traveled from its beginning point, a is the acceleration of the body and t is the time taken.
So acceleration is given by,
a = $\frac{2d}{t^{2}}$
The free fall adapter measures the time of fall for a ball. Using the meter scale we can measure the height directly. Then substitute all values in the above equation will give the acceleration due to gravity of a free falling body. 2. Pendulum experiment
A pendulum consists of a small mass or a bob suspended on a string. It can oscillate to and fro periodically. Such periodic motion is known as the simple harmonic motion. The time of the to and fro motion is known as the period of the pendulum. Period does not depend up on the mass of the pendulum, the size of the arc through which it oscillates or initial angular displacement. Whereas it depends only on the length of the pendulum and the acceleration due to gravity. According to the equation of the period,
T = $2\Pi \sqrt{\frac{L}{g}}$
Where, T = $\frac{Time\ of\ oscillation}{Number\ of\ oscillation}$
So,
g = $\frac{4\Pi ^{2}L}{T^{2}}$
g = $\frac{GM}{R^{2}}$
Where G = 6.67 × 10^{11} Nm^{2}/Kg^{2}
1. Acceleration Due to Gravity on Mars
Mass = 6.42 $\times$ 10^{23} Kg , Radius = 3396 Km
g = $\frac{6.67\times 10^{11}\times 6.39\times 10^{23}}{3396000^{2}}$
g = 3.77 m/s^{2}
2. Acceleration Due to Gravity on the Moon
Mass = 7.347 $\times$ 10^{22} Kg , Radius = 1738 Km
g = $\frac{6.67\times 10^{11}\times 7.347\times 10^{22}}{1738000^{2}}$
g = 1.6 m/s^{2}
3. Acceleration Due to Gravity on Earth
Mass = 5.972 $\times$ 10^{24} Kg, Radius = 6400 Km
g = $\frac{6.67\times 10^{11}\times 5.972\times 10^{24}}{6400000^{2}}$
g = 9.82 m/s^{2}
4. Acceleration Due to Gravity on Jupiter
Mass = 1.898 $\times$ 10^{27}Kg, Radius = 7149.2 Km
g = $\frac{6.67\times 10^{11}\times 1.898\times 10^{27}}{71492000^{2}}$
g = 23.6 m/s^{2}
5. Acceleration Due to Gravity on Mercury
Mass = 3.285 $\times$ 10^{23} Kg , Radius = 2440 Km
g = $\frac{6.67\times 10^{11}\times 3.285\times 10^{23}}{2240000^{2}}$
g = 3.76 m/s^{2}
6. Acceleration Due to Gravity on Venus
Mass = 4.87 $\times$ 10^{24} Kg , Radius = 6051 Km
g = $\frac{6.67\times 10^{11}\times 4.87\times 10^{24}}{6051000^{2}}$
g = 9.04 m/s^{2}
7. Acceleration due to Gravity on Pluto
Mass = 1.31 $\times$ 10^{22} Kg, Radius = 1160 Km
g = $\frac{6.67\times 10^{11}\times 1.31\times 10^{22}}{1160000^{2}}$
g = 0.701 m/s^{2}
8. Acceleration Due to Gravity on Neptune
Mass = 1.03 $\times$ 10^{26} Kg, Radius = 2476.4 Km
g = $\frac{6.67\times 10^{11}\times 1.03\times 10^{26}}{24762000^{2}}$
g = 11.23 m/s^{2}
9. Acceleration Due to Gravity on Saturn
Mass = 5.69 $\times$ 10^{26}Kg , Radius = 6026.8 Km
g = $\frac{6.67\times 10^{11}\times 5.69\times 10^{26}}{60268000^{2}}$
g = 10.06 m/s^{2}
Some of the problems related to acceleration due to gravity is given below:
Solved Examples
Question 1: Determine the value of g on earth, the radius of the earth is given by 6400 Km and the density 5.5 g/cc. The value of G = 6.66 × 10^{11} Nm^{2}/Kg^{2}
Solution:
The given parameters are:
R = 6400 × 10^{11} Nm^{2}/Kg^{2} , Density, ρ = 5.5 g/cc
From the equation of gravity, we know that
g = $\frac{GM}{R^{2}}$
And M = Volume × Density
That is, M = V × ρ
M = $\frac{4}{3}$ $\pi$ R^{3}$
M = $\frac{4}{3}$ $G\pi R\rho$
M = $\frac{4\times 6.66\times 10^{11}\times 3.14\times 6400\times 10^{3}\times 5.5\times 10^{3}}{3}$
M = 9.82 m/s^{2}
Solution:
The given parameters are:
R = 6400 × 10^{11} Nm^{2}/Kg^{2} , Density, ρ = 5.5 g/cc
From the equation of gravity, we know that
g = $\frac{GM}{R^{2}}$
And M = Volume × Density
That is, M = V × ρ
M = $\frac{4}{3}$ $\pi$ R^{3}$
M = $\frac{4}{3}$ $G\pi R\rho$
M = $\frac{4\times 6.66\times 10^{11}\times 3.14\times 6400\times 10^{3}\times 5.5\times 10^{3}}{3}$
M = 9.82 m/s^{2}
Question 2: What is the gravitation force on an object which weighs 63 N at a height equal to the half of the radius of earth?
Solution:
Assume that g′ is the acceleration due to gravity. Then
g'= g $\frac{R}{R + h}^{2}$
Here, h = $\frac{R}{2}$
So, g'= g $\frac{R}{R+\frac{R}{2}}^{2}$
g'= g $\frac{R^{2}}{\frac{9}{4}R^{2}}$
g'= $\frac{4}{9}$ g
Solution:
Assume that g′ is the acceleration due to gravity. Then
g'= g $\frac{R}{R + h}^{2}$
Here, h = $\frac{R}{2}$
So, g'= g $\frac{R}{R+\frac{R}{2}}^{2}$
g'= g $\frac{R^{2}}{\frac{9}{4}R^{2}}$
g'= $\frac{4}{9}$ g
Question 3: Calculate the height of an object above the surface of earth if its acceleration due to gravity is reduced by 36% of its value when it is in the earth surface. The radius of earth is given by 6400 Km.
Solution:
If the acceleration due to gravity reduces by 36%, then the value becomes 64%. It means that, g'= $\frac{64g}{100}$. Let h be the height of the object above the surface of earth.
Then,
g'= g $\frac{R}{R+h}^{2}$
$\frac{64g}{100}$ = g $\frac{R}{R+h}^{2}$
$\frac{8}{10}$ = $\frac{R}{R+h}$
8R + 8h = 10R
10R  8R = 8h
R = 4h
h = $\frac{R}{4}$
h = $\frac{6400 \times 10^{3}}{4}$
h = 1.6 $\times 10^{6}m$
Solution:
If the acceleration due to gravity reduces by 36%, then the value becomes 64%. It means that, g'= $\frac{64g}{100}$. Let h be the height of the object above the surface of earth.
Then,
g'= g $\frac{R}{R+h}^{2}$
$\frac{64g}{100}$ = g $\frac{R}{R+h}^{2}$
$\frac{8}{10}$ = $\frac{R}{R+h}$
8R + 8h = 10R
10R  8R = 8h
R = 4h
h = $\frac{R}{4}$
h = $\frac{6400 \times 10^{3}}{4}$
h = 1.6 $\times 10^{6}m$