# Class 11 Gravitation Notes

The mere idea of an apple falling down on the earth from a tree led to the discovery of a very interesting topic in the field of Physics called Gravitation. Every object seems to fall on the earth, whether a small water particle or a human. Gravitation is much more than just an everyday phenomenon, it encompasses various theories and principles in its working, which are a part of the Class 11 Physics syllabus. Do you want to know the reasons why everything falls back on the earth? Here is an insightful blog on class 11 Gravitation notes!

## What is Gravitational Force?

In this universe, every body attracts another body towards itself. This force is called the force of gravitation. Discovered by Sir Issac Newton, Gravity is nothing but a force of attraction between any two elements that exist in a universe. Elements in the universe attract each other with a specific amount of force. Gravity is a force which has an infinite range. Gravitation is a study of the interaction between two masses:

• Source Mass: The heavier one
• Test Mass: The lighter one

The force of gravitation depends on the distance between the test mass from the source mass. It always acts according to the invisible line joining the centres of the masses.

## Law of Gravitation

Newton’s law of gravitation says, ” Every substance in the universe attracts every other substance with force.”
Its magnitude can be defined as:

• Directly proportional to the product of their masses, i.e. F ∝ (M1M2)
• Inversely proportional to the square of the distance between their centre, i.e. (F ∝ 1/r^2)

Let’s combine the equations of the above two points.
F ∝ (M1M2) x (1/r^2)
So, which gives us the formula,
F ∝ M1M2/r2
F = G × [M1M2]/r2
Or,
f(r) = GM1M2/r2
Here, f(r) is a variable denoting a Conservative Force. The f(r) here is the inverse law force because it varies inversely as a square of ‘r’. In the above equation, ‘G’ is the Gravitational Constant.

The dimension formula of the gravitation constant is [M-1L3T-2], and the units and respective values of the gravitational constant are –

Test your knowledge with our Physics quiz

## Vector Form

The gravitational forces acting between two particles from the action-reaction pair is the vector form of Newton’s law of gravitation.

F12 = [-G m1m2]/ |r12| 2 r12
F12 = [-Gm1m2/ |r1 – r2| 3] (r1 – r2) . . . . . . . . . Equation 1

Where r1,2 is a unit vector pointing from m2 to m1.
The negative sign in equation 1 indicates the direction of force F12 is opposite to that of r12.
Now, the force on m2 due to m1, i.e. F12 is given by,

F12 = -Gm1m2/ |r21 |2 r12 . . . . . . . . . . Equation 2
From equations 1 and 2 we get,

F12 = – F21

As F12 and F21 are directed towards the centres of the two particles, the gravitational force is conservative.

Must Read: How to Study Physics for NEET

## Formula of Gravitation

According to Newton’s Law of Gravitation, every element in the universe always attracts some or the other element. This product of the masses is directly proportional to the force whereas the force can be regarded as inversely proportional to the square of the distance between the objects.

The mathematical formula of the Law of Gravitation is:

F = Gm1m2/ r2
In the above formula:
F: The Gravitational force between two objects measured in Newton (N).
G: It is the Universal Gravitational Constant which has a standard value of 6.674 × 10-11 Nm2kg-2.

m1: The mass of one massive body measured in kg
m2: The mass of another massive body measured in kg
r2: The separation between them measured in kilometres (Km)

## Solved Questions Based on the Class 11 Gravitation Chapter

1. Explain Kepler’s Laws of Planetary Motion.

In astronomy, Kepler’s laws of planetary motion are three laws describing the motion of planets around the sun.

• Kepler’s first law – The law of orbits – According to Kepler’s first law, ”All the planets revolve around the sun in elliptical orbits having the sun at one of the foci”.
• Kepler’s second law – The law of equal areas – Kepler’s second law states that ”The radius vector drawn from the sun to the planet sweeps out equal areas in equal intervals of time”
• Kepler’s third law – The law of periods – Kepler’s law of periods states that ”The square of the time period of revolution of a planet around the sun in an elliptical orbit is directly proportional to the cube of its semi-major axis”.

2. Explain Newton’s Law Of Universal Gravitation.

Newton’s Law of Universal Gravitation states that every particle attracts every other particle in the universe with a force that is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.

where,

• F is the gravitational force between bodies
• m1 is the mass of one of the objects
• m2 is the mass of the second object
• r is the distance between the centres of two objects
• G is the universal gravitational constant

The constant proportionality (G) in the above equation is known as the universal gravitation constant. The value of G is G = 6.673 x 10-11 N m2/kg2.

3. Why doesn’t the moon crash into the earth?

The forces of speed and gravity are what keeps the moon in constant orbit around the Earth. The Moon hovers around in the sky, which is unaffected by gravity. However, the reason the Moon stays in orbit is precisely because of gravity. This is the reason why the moon doesn’t crash into the Earth.

4. What can be the force of gravity acting on an object at the Earth’s surface?

Given:
Mass of Earth (m1) = 5.98 × 1024kg
Mass of object (m2) = 2000kg
The radius of the Earth (r)= 6.38 × 106m
Acceleration due to gravity (g) = 9.8 m/s2
Universal constant (G) = 6.67 x 10-11 N m2 / kg2

Solution:

F = Gm1m2/r2
F = ( 6.67 x 10-11) (5.98 × 1024)(2 x 103)/(6.38 × 106)2
F = (7.978 x 1017)/ (4.07044  × 1013)

F = 1.959 x 104 or F = 19.59 N

5. What is the Centre of Gravity?

The Centre of gravity is a theoretical point in the body where the total weight of the body is thought to be concentrated. It is important to know the centre of gravity because it predicts the behaviour of a moving body when acted on by gravity and it is also useful in designing static structures such as buildings and bridges.

6. How does the Centre of gravity affect the balance?

The centre of gravity affects the stability of objects. The lower the centre of gravity (G) is, the more stable the object. The higher it is the more likely the object is to topple over if it is pushed. Racing cars have really low centres of gravity so that they can corner rapidly without turning over.

7. Let us assume that our galaxy consists of 2.5 × 1011 stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution?

Solution:
Take the diameter of the Milky Way to be 105 ly
Number of stars in our Galaxy (N) = 2.5 × 10¹¹
Mass of each stars = 2 × 10^30kg
So, mass of the stars of the galaxy(M) = 2.5× 10¹¹× 2× 10^30
= 5 × 10⁴¹ Kg
The radius of orbit of a star (r) = 50000 light-years

We know,
1 light-years = 9.46 × 10^15 m
So, r = 50000×9.46×10^15 m
= 5 × 9.46 × 10^19 m

Centripetal force = Gravitational force

mv²/r = GMm/r²
v² = GM/r
(2πr/T)² = GM/r [ v = 2πr/T
4π²r²/T² = GM/r
T² = 4π²r³/GM
Put the values of r, G and M
T = √{ 4×(3.14)²× (5×9.46×10^19)³/6.67×10^-11×5×10⁴¹}
= 111.93 × 10¹⁴ sec
= 111.93 × 10¹⁴/(365×24×3600) yr

= 3.55 × 10^8 yr

8. What does the escape speed of a body from the earth depend on:

1. the mass of the body
2. the location from where it is projected
3. the direction of projection
4. the height of the location from where the body is launched

Solution: B. The location from where it is projected

Explanation:

Escape velocity is independent of the direction of projection and the mass of the body. It depends upon the gravitational potential at the place from where the body is projected. Gravitational potential depends slightly on the altitude and the latitude of the place, thus escape velocity depends slightly upon the location from where it is projected.

9. A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?

Solution:
Weight of the body, W = 63 N

Acceleration due to gravity at height h from the Earth’s surface is given by the relation:
g‘ = g / [1 + ( h / Re) ]2

Where,
g = Acceleration due to gravity on the Earth’s surface
Re = Radius of the Earth
For h = Re / 2
g‘ = g / [(1 + (Re / 2Re) ]2
= g / [1 + (1/2) ]2  =  (4/9)g
The weight of a body of mass m at height h is given as:
W‘ = mg
= m × (4/9)g  =  (4/9)mg
= (4/9)W

= (4/9) × 63  =  28 N.

We hope you liked our blog, where we tried to cover comprehensive information about the class 11 Gravitation chapter. For more informative articles like this keep following Leverage Edu.