Gravitation Class Notes PDF Arjuna JEE 2025

## ARJUNA JEE 2025 - Gravitation **By - Rajwant Singh (RJ)** **Physics Wallah** ### Lecture - 1 ### Topics 1. Newton laws of gravitation 2. Question 3. Gravitational field intensity 3. Questions ### To be Covered: - Vectors - FBD (force) - Circular Motion ### Newton Law's of Gravitation -...

## ARJUNA JEE 2025 - Gravitation **By - Rajwant Singh (RJ)** **Physics Wallah** ### Lecture - 1 ### Topics 1. Newton laws of gravitation 2. Question 3. Gravitational field intensity 3. Questions ### To be Covered: - Vectors - FBD (force) - Circular Motion ### Newton Law's of Gravitation - Two point masses _m_1 & _m_2 are separated by distance _r_. They always attract each other by a force _F_, which is proportional to the product of masses (_m_1_m_2_) and inversely proportional to the square of the distance between them. - **F = Gm₁m₂/r²** - _G_ = universal gravitational constant = 6.67 x 10^-11 _Nm_²/_kg_² ### Important Points: 1. Only applicable for point masses. Distributed mass system is not valid. 2. _G_ = universal gravitational constant = 6.67 x 10^-11 _Nm_²/_kg_² 3. Always attractive force 4. Force is central, meaning it acts along the line joining the center. 5. Action/Reaction pairs ### Gravitational Force is Independent of - The medium between two particles. - Presence of a third particle. ### Fgravitation is a Vector The resultant of force is calculated by using vector laws. ### Examples **a)** - A particle _m_0 is placed at x = 0. - Particles with mass _m_ are placed at x = 1m, 2m, 4m, and 8m etc. - **Find F_net on _m_0 = ?** - **F_net = Gm₀m/1² + Gm₀m/2² + Gm₀m/4² + Gm₀m/6² ... = (4/3)Gm₀m** **b)** - Three masses, _m_, _m_, and _m_0 are placed at the corners of an equilateral triangle of side _l_. - Find _F_net_ on _m_0. - **F_net = √3Gm₀m/l²** **c)** - Four equal masses, _m_, are placed at the corners of a square. - Find _F_net_ on _m_0. - **F_net = (√2 + 1)Gm₀m/l²** **d)** - Four equal masses, _m_, are placed at the corners of rectangle. - Find _F_net_ on _m_0. - **F_net = (2√2 + 5)Gm₀m/l²** **e)** - Four equal masses, _m_, are placed at the corners of a circle, with radius _R_. - Find _F_net_ on _m_0. - **F_net = (√2 + 1)Gm₀m/2R²** **f)** - Six equal masses, _m_, are placed on a circle. - Find _F_net_ on _m_0. - **F_net = (2 + √3)Gm₀m/R²** **g)** - Multiple masses, _m_, are placed on the circumference of a circle at equal angles. - _m_0 is at the center of the circle. - **F_net on _m_0 = 0** **h)** - Three equal masses, _m_, are placed at the corners of a triangle. - Find the force on _m_0. - **F = 3Gm₀m/(2√3)²** **i)** - Four equal masses, _m_, are placed at the corners of a square. - Find the force on _m_0. - **F_net on _m_0 = 0** **j)** - Five equal masses, _m_, are placed at the corners of a regular pentagon. - Find the force on _m_0. - **F_net on _m_0 = 0** **k)** - A regular pentagon with five masses, _m_. - Find the force on _m_0. - **F_net on _m_0 = Gm₀m/a²** - **Concept of Broken Symmetry** - When a particle is removed from the pentagon , the symmetry breaks and there is a net attraction between the particles. **l)** - Twelve equal masses, _m_. - Find the force on _m_0. - **F_net on _m_0 = Gm₀m/R²** **m)** - A mass, _M_, is broken into two parts. - The two parts are separated by a distance _r_. - **Find the force between the two part.** - **The force is maximum when the mass is broken into two equal parts (m = M/2).** ### Null Point Problems **Calculation of Null Point** - Point where force = 0 **Null Vector** - Magnitude: 0 - Direction: not defined - **Null point is obtained between the line joining the masses.** **a)** - Masses _m_ and _m_ are separated by a distance _d_. - A mass, _m_0, is placed at a distance _x_ from _m_ and a distance _d-x_ from _m_. - **Find the value of _x_ so that the net force on _m_0 is 0** - **x = d/2** **Generalised Formula for Null Point** - **x = d√n/(√n + 1)** - **Always towards the smaller mass** **b)** - A mass of 4_m_ is placed at one end of a line segment with a length _d_. - A mass of 5_m_ is placed at the other end of the line segment. - Find the Null point. - **x = d/√5+1 = 2d/√5+2** **c)** - Two equal masses, _m_, are fixed at a distance _d_. - A mass _m_0 is placed at a distance _x_ from the midpoint of the line segment. - Find the net force on _m_0 and find at what value of _x_ the force will be maximum. - **F_net = 2Gm₀m.x/(d² + x²)^3/2** - **The force is maximum at x = d/√2** **HW** - A mass, _m_0, is placed at the centroid of an equilateral triangle, while three equal masses, _m_, are placed at the corner of the triangle. Find the net force on _m_0. - A mass _m_0_ is placed at the center of a square. Four equal masses, _m_, are placed at the corners of the square. Find the net force on _m_0. - A mass, _m_0, is placed at the center of a ring with radius _R_. The mass, _m_, is distributed uniformly across the circumference of the ring. Find the net force on _m_0.