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Gravitation

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 M1 and M2 are the masses of two objects separated by a distance 'd', force of gravitation can be written as :
f \(\propto\) M1 \(\times\) M2 …………… (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 6.67x10-11Nm2/Kg2
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 M1 = M2 = 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 \(\propto\) \(\frac{1}{R^2}\)).
gpole = 9.83 m/s2
gequator = 9.78 m/s2
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:
gmoon = \(\frac{1}{6}\) gearth
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:

  • When a body falls freely under the influence of force of gravity only i.e. when acceleration produced in a body is equal to acceleration due to gravity.
  • When a body is at the center or at the space i.e. null points, where g = 0.
  • When orbiting around the heavenly body, artificial satellites have weightless condition as the centripetal and centrifugal force becomes balanced.

Parachute Fall

  • While jumping downward, the surface area of parachute causes the air resistance to be more.
  • Slowly, velocity goes on decreasing due to air resistance and finally it may become uniform this makes the acceleration to be almost zero.
  • As a result of slow motion, paratroopers can balance their body and land safely without any hurt.

Questions for practice

  1. State Newton's law of gravitation.
  2. How is acceleration due to gravity related to radius of earth?
  3. State the finding of coin and feather experiment.
  4. How is parachute able to land safely on the earth surface?
  5. What is free fall? What is the value of acceleration during free fall?
  6. Define weightlessness. State any two conditions of weightlessness.
  7. What are gravitational field and its intensity?
  8. What is the value of 'g' at poles, equator, center of earth, in space?

Value and SI units
Gravitational constant = 6.67 \(\times\) 10-11 Nm2/kg2
Acceleration due to gravity on earth's surface = 9.8 m/s2
Acceleration due to gravity at poles = 9.83 m/s2
Acceleration due to gravity at equator = 9.78m/s2
Acceleration due to gravity at center = 0 m/s2
Acceleration due to gravity at space = 0 m/s2
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

  • Weight of object is more at the bottom and less at the top of mountains.
    Value of acceleration due to gravity is inversely proportional to the height [g \(\propto\) \(\frac{1}{(R + h)^2}\)]. So as the altitude increases, acceleration due to gravity decreases and the weight of object also deceases because weight is directly proportional to the value of acceleration due to gravity (W \(\propto\) g)
  • Parachute cannot land safely on the moon's surface.
    As there is no atmosphere on the moon's surface, there is no any effect of air resistance on the parachute. Falling of parachute is exactly like the free fall of the object. So, it cannot land safely.
  • Newton's Law of gravitation is called a universal law.
    It is so because this law can be applied between any two object of any mass and separated by any distance in this universe.

 Try Yourself

  1. Weight of object is more at the poles and less at the equator.
  2. A satellite does not need any fuel to revolve around the planet. (Hint: Centripetal and centrifugal force helps to balance circular motion)
  3. A person is the artificial satellite feels weightless.
  4. Moon has no atmosphere.
  5. When a stone is carried from top to bottom of mountain, does its weight increases or decreases?

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.
(F = \(\frac{GM_1 M_2}{d^2}\))

It depends upon the mass of object and acceleration due to gravity.
(F = mg)

 

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 Nm2/kg2.

Its value is variable around 9.8 m/s2 which differ from place to place.

 

Try Yourself

  • Free all and parachute fall.
  • Weightlessness due to free fall and due to null points.
  • Gravitational field and field intensity.

Some Mathematical conceptual problems:

  1. What will be the effect in force of gravitation when masses of both the objects are doubled and distance between them made constant?
    Solution:
    Here, let M1 and M2 be the masses of two objects respectively and 'd' be the distance between them.
    Then,
    Force of gravitation (F) = \(\frac{GM_1M_2}{d^2}\)
    When masses are doubled, they become 2M1 and 2M2
    then,
    F1 = \(\frac{G \; \times \; 2M_1 \; 2M_2}{d^2}\)
    = 4 \(\times\) \(\frac{GM_1M_2}{d^2}\)
    = 4 \(\times\) F
    Force of gravitation becomes four times the initial force of gravitation.
  2. If masses are halved and distance double, what will be effect in force of gravitation?
    Solution:
    F = \(\frac{GM_1M_2}{d^2}\)
    when masse are halved, they become \(\frac{M_1}{2}\) and \(\frac{M_2}{2}\)
    when distance is doubled, it becomes 2d.
    So, F1 = \(\frac{G \; \times \; \frac{M_1}{2} \; \times \; \frac{M_2}{2}}{(2d)^2}\)
    = \(\frac{\frac{M_1M_2}{4}}{4d^2}\)
    = \(\frac{1}{16}\) \(\frac{GM_1M_2}{d^2}\)
    = \(\frac{1}{16}\) F
    So, Newton's force of gravitation becomes \(\frac{1}{16}\) times the initial force of gravitation.

Try Yourself

  • What will be the effect in force of gravitation when masses are kept constant and distance between them halved?
  • If mass of two objects are doubled and distance also doubled, does there occur any change in force of gravitation verify.

Gravitation and Gravity
Important Expression:
G = Gravitational constant
= 6.67 \(\times\) 10-11 Nm2/kg2

To find force of gravitation between any two bodies
F = \(\frac{GM_1M_2}{d^2}\)
M1 and M2 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

  1. Find the weight of the object at the height of 600 km from earth's surface when its mass is 40 kg.
  2. At what height from the earth's surface the value of acceleration due to gravity becomes 4.9 m/s2 when its value is 9.8 m/s2 on the surface? (Radius of earth = 6400 km)
  3. What is the distance between two spheres each of mass 40 kg when force of attraction between them is 7 \(\times\) 1015 N?
  4. What is the value of acceleration due to gravity at the top of the Everest? (Height of Everest = 8848 m, Radius = 6400 km and Mass of Earth = 6 \(\times\) 1024 kg).

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