Gravitation
It is defined as the force of attraction between any two bodies in this universe.
Gravity
Force of attraction by which a heavenly body pulls any object towards its center is called gravity.
Newton's Law of Gravitation
Force of gravitation attraction between two objects is directly proportional to the product of their masses and inversely proportional to the square of distance between their centre.
If M_{1} and M_{2} are the masses of two objects separated by a distance 'd', force of gravitation can be written as :
f \(\propto\) M_{1} \(\times\) M_{2} …………… (i)
f \(\propto\) \(\frac{1}{d^2}\) ……………. (ii)
Combining (i) and (ii),
f \(\propto\) \(\frac{M_1 \; \times \; M_2}{d^2}\)
or f = \(\frac{GM_1 \; M_2}{d^2}\)
where 'G' is Universal gravitational constant. It's value is 6.67x10^{-11}Nm^{2}/Kg^{2}
This law is called the universal low because it can be applied between any two objects in the universe.
Universal Gravitational Constant
It can be defined as the force of gravitation between two objects of unit masses (1 kg) each separated by a unit distance (1m).
F = \(\frac{GM_1 \; M_2}{d^2}\)
when M_{1} = M_{2} = 1kg and d = 1m ,
F = \(\frac{G \; \times \; 1 \; \times \; 1}{1^2}\)
\(\therefore\) F = G
Acceleration Due to Gravity (g)
Acceleration produced in a freely falling body due to the effect of force of gravity.
Relation of 'g' with radius of earth or heavenly body (R):
g = \(\frac{GM}{R^2}\)
\(\begin{bmatrix} Radius \; increases \\ 'g' \; decrease \end{bmatrix}\)
\(\therefore\) g \(\propto\) \(\frac{1}{R^2}\) ['G' and 'M' being constant terms]
Relation with height and depth
g \(\propto\) \(\frac{1}{(R + h)^2}\)
Height increase, 'g' decreases
Acceleration due to gravity (g) and weight of body (W)
\(\because\) W = mg [W \(\propto\) g] (When mass is constant)
More is the value of acceleration due to gravity, more will be the value of weight.
g\(\uparrow\) W\(\uparrow\)
Value of 'g' at poles and equator
Earth is not perfectly round, little depressed at poles. Equatorial radius is more than polar radius. Value of acceleration due to gravity is more at poles and less at equator because acceleration due to gravity is inversely proportional to "Radius" of earth (g \(\propto\) \(\frac{1}{R^2}\)).
g_{pole} = 9.83 m/s^{2}
g_{equator} = 9.78 m/s^{2}
So, weight of body is more at pole and less at equator as weight is directly proportional to acceleration due to gravity.
In condition:
R\(\downarrow\) \(\Rightarrow\) g\(\uparrow\) \(\Rightarrow\) W\(\uparrow\)
R\(\uparrow\) \(\Rightarrow\) g\(\downarrow\) \(\Rightarrow\) W\(\downarrow\)
Comparison of Value of 'g' on the moon's surface with that of earth's surface:
g_{moon} = \(\frac{1}{6}\) g_{earth}
Value of acceleration due to gravity on the moon's surface is \(\frac{1}{6}\) times to that of the earth's surface. Weight of object on the moon's surface is \(\frac{1}{6}\)^{th} of the weight on earth's surface because of force of gravity on the moon's surface \(\frac{1}{6}\) times of to that of earth's surface.
Mass and Weight
Mass is the total matter contained in a body. It is constant quantity.
Weight is the force of gravity by which an object is pulled towards the center. This applied force of gravity is different at different regions. So, weight of body differs from one place to another place.
Free Fall
When a body is falling only under the influence of force of gravity without air or other resistance it is called as free fall.
"When the body is in the state of free fall, the acceleration produced in the body is equal to the acceleration due to gravity". However there is no perfect free fall on the earth's surface because air resistance always exists.
Examples:
Falling of object in the vacuum.
Falling of object in the moon's surface.
Weightlessness
Some conditions:
Parachute Fall
Questions for practice
Value and SI units
Gravitational constant = 6.67 \(\times\) 10^{-11} Nm^{2}/kg^{2}
Acceleration due to gravity on earth's surface = 9.8 m/s^{2}
Acceleration due to gravity at poles = 9.83 m/s^{2}
Acceleration due to gravity at equator = 9.78m/s^{2}
Acceleration due to gravity at center = 0 m/s^{2}
Acceleration due to gravity at space = 0 m/s^{2}
Si unit of Gravitational Field Intensity is N/kg.
Coin and Feather Experiment
A coin and feather were put into a closed long glass tube. When the tube was inverted, coin was found to be falling quicker than feather.
Then, the tube was evacuated using a vacuum pump and the tube was inverted again. It could be observed that both the coin and the feather fall simultaneously at the same time.
Conclusion: Acceleration due to gravity of freely falling body is the same in the absence of air resistance irrespective of their masses.
Reasoning Questions
Try Yourself
Question Regarding Differences
S.N |
Gravitation |
Gravity |
1. |
Force of attraction between any two bodies in the universe. |
Force exeted by a heavenly body to pull an object towards its center. |
2. |
It depends upon the masses of two objects and distance between them. |
It depends upon the mass of object and acceleration due to gravity. |
S.N |
Free Fall |
Parachute |
1. |
It is not affected by air resistance. |
Highly affected by air resistance. |
2. |
Velocity goes on increasing while falling. |
Velocity becomes uniform. |
3. |
Acceleration of falling object is equal to acceleration due to gravity. |
Acceleration finally becomes zero. |
S.N |
Mass |
Weight |
1. |
Matter contained in a body. |
Force of gravity that pulls an object towards its center. |
2. |
It is a constant term. |
Its values differ from place to place. |
3. |
SI unit is kg. |
SI unit is Newton's (N). |
S.N |
'G' |
'g' |
1. |
The gravitational constant is defined as the force of attraction between two bodies each of unit masses kept at a unit distance. |
The acceleration produced on a freely falling body under the effect of gravity is called acceleration due to gravity. |
2. |
It denotes universal gravitational constant. |
It denotes acceleration due to gravity. |
3. |
It is a constant quantity and its value is 6.67 \(\times\) 10^{-11} Nm^{2}/kg^{2}. |
Its value is variable around 9.8 m/s^{2} which differ from place to place. |
Try Yourself
Some Mathematical conceptual problems:
Try Yourself
Gravitation and Gravity
Important Expression:
G = Gravitational constant
= 6.67 \(\times\) 10^{-11} Nm^{2}/kg^{2}
To find force of gravitation between any two bodies
F = \(\frac{GM_1M_2}{d^2}\)
M_{1} and M_{2} are the masses of two bodies and 'd' is the distance between them.
Finding force of gravity exerted by heavenly body or weight of object.
F = W = \(\frac{GMm}{R^2}\)
where,
M = Mass of heavenly body
m = Mass of object
R = Radius of heavenly body
Object placed at certain height,
F = W = \(\frac{GMm}{(R+H)^2}\),
'h' = height from the earth's surface
Acceleration Due to Gravity
On the earth's surface; g = \(\frac{GM}{R^2}\)
Object placed at height 'h'; g = \(\frac{GM}{(R+h)^2}\)
Weight of object (W) = mg
Value of field intensity is always equal to the value of acceleration due to gravity at a place
I = g = \(\frac{GM}{R^2}\) or \(\frac{GM}{(R+h)^2}\)
Comparing acceleration due to gravity on the surface and at the height.
\(\frac{g_h}{g}\) = \(\frac{R^2}{(R+h)^2}\)
weight at different regions can be compared as:
\(\frac{W_h}{W}\) = \(\frac{g_h}{g}\) = \(\frac{R^2}{(R+h)^2}\) or, \(g_h\) = \(\frac{R^2}{(R+h)^2}\)\(\times\)g
Practice Yourself