PhysicsKund Physics class 9 to 12 NCERT, JEE - NEET

Ncert Solution CBSE Class 11 Physics Chapter 7 Gravitation 7.1 Answer the following : (a) You can shield a charge from electrical forces by putting it inside a hollow conductor.Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means ? Solution: You cannot shield a body from the gravitational influence of nearby matter by any means. It is because the gravitational force on a body due to nearby matter is not altered due to the presence of other bodies. In other words , gravitational field cannot be shielded by any means. (b) An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity ? Solution: Yes , If the size of the spaceship orbiting earth is very large , an astronaut in it can detect the gravity. (c) If you compare the gravitational force on the earth due to the sun to that due to the ...

Objective: To derive expressions for the Total Energy and Binding Energy of an orbiting satellite. Total Energy of a Satellite The total energy of a satellite in its orbit is the sum of its Potential Energy (P.E.) and Kinetic Energy (K.E.) . $$E = U + K.E.$$ Where, $M$ = Mass of Earth $m$ = Mass of satellite $R$ = Radius of Earth $h$ = Height of satellite above Earth's surface $r = R+h$ = Radius of orbit 1. Potential Energy of Satellite The gravitational potential energy of a satellite revolving around the Earth in a circular orbit is: $$ U = -\frac{GMm}{R+h} $$ or $$U = -\frac{GMm}{r} $$ 2. Kinetic Energy of Satellite The kinetic energy of the satellite is: $$ K.E.=\frac{1}{2}mv_0^2 $$ where $v_0$ is the orbital velocity. Since the satellite moves in a circular orbit, Centripetal Force = Gravitational Force $$ \frac{mv_0^2}{R+h} = \frac{GMm}{(R+h)^2} $$ or $$ mv_0^2=\frac{GMm}{R+h} $$ Substituting in the kinetic energy equation, $$K.E. = \frac{1...

Orbital Velocity of Satellite, Time Period, Height of Satellite and Verification of Kepler's Third Law Orbital Velocity of Satellite The minimum velocity required by a satellite to remain in a stable circular orbit around the Earth is called orbital velocity . Derivation of Orbital Velocity Consider: Mass of Earth = \(M\) Mass of Satellite = \(m\) Orbital Radius = \(r\) Orbital Velocity = \(v_0\) Gravitational Force on Satellite: $$ F_g=\frac{GMm}{r^2} $$ Centripetal Force Required: $$ F_c=\frac{mv_0^2}{r} $$ For stable circular motion, $$ F_g=F_c $$ $$ \frac{GMm}{r^2}=\frac{mv_0^2}{r} $$ Cancelling \(m\), $$ \frac{GM}{r^2}=\frac{v_0^2}{r} $$ $$ v_0^2=\frac{GM}{r} $$ $$ v_0=\sqrt{\frac{GM}{r}} $$ For a satellite, $$ r=R+h $$ where \(R\) is the radius of Earth and \(h\) is the height of the satellite above Earth's surface. Therefore, $$ v_0=\sqrt{\frac{GM}{R+h}} $$ Using $$ g=\frac{GM}{R^2} $$ or $$ GM=gR^2 $$ we get $$ v_0=\sqrt{...

Notes : Earth Satellite it's type and Principle of Launching Satellite - Physicskund  What is a Earth Satellite? A satellite is a small body that revolves around a planet in a definite orbit due to gravitational attraction. Types of Satellites 1. Natural Satellite A satellite that revolves around a planet naturally is called a natural satellite . Example: The Moon is the natural satellite of the Earth. 2. Artificial Satellite A man-made object launched into space to revolve around the Earth or another planet is called an artificial satellite . Difference Between Natural and Artificial Satellites Natural Satellite Artificial Satellite Formed naturally. Made and launched by humans. Revolves around a planet due to natural processes. Placed into orbit for specific purposes. Example: Moon. Example: Aryabhata, INSAT, Rohini. No human control. Can be monitored and controlled from Earth. Uses of Artificial Satellites Communication Weather fo...

Definition of Gravitational Potential Energy Gravitational potential energy of a body at a point is defined as the work done in bringing the body from infinity to that point in the gravitational field of another body (such as Earth). It is denoted by U . Derivation of Gravitational Potential Energy Consider a body of mass m placed at a distance x from the centre of Earth of mass M . According to Newton's law of gravitation, the gravitational force acting on the body is: $$ F=\frac{GMm}{x^2} $$ Here, G = Universal Gravitational Constant M = Mass of Earth m = Mass of the body x = Distance of the body from Earth's centre Let the body be displaced through a very small distance dx . The small work done during this displacement is: $$ dW=F\,dx $$ Substituting the value of gravitational force: $$ dW=\frac{GMm}{x^2}dx $$ To find the total work done in bringing the body from infinity to distance r , integrate both sides: $$ W=\int_{\infty}...

Variation of Acceleration Due to Gravity with Depth (Below the Earth's Surface) When a body is taken to a depth d below the Earth's surface, the acceleration due to gravity decreases. This variation of gravity with depth can be derived by assuming that the Earth is a homogeneous sphere of uniform density. Derivation of Variation of g with Depth The value of acceleration due to gravity on the Earth's surface is: $$ g=\frac{GM}{R^2} $$ where: G = Universal Gravitational Constant M = Mass of Earth R = Radius of Earth For a homogeneous Earth of density ρ, $$ M=\frac{4}{3}\pi R^3\rho $$ Substituting the value of M: $$ g=\frac{G\left(\frac{4}{3}\pi R^3\rho\right)}{R^2} $$ $$ g=\frac{4}{3}\pi GR\rho $$ Let a body be at a depth d below the Earth's surface. The effective radius becomes: $$ r=R-d $$ The mass enclosed within radius (R − d) is: $$ M'=\frac{4}{3}\pi (R-d)^3\rho $$ Therefore, acceleration due to gravity at depth d is: ...

Variation of Acceleration Due to Gravity with Altitude (Height)) Acceleration due to gravity ( g ) decreases as we move away from the Earth's surface. This variation can be explained using Newton's law of gravitation. Acceleration Due to Gravity on the Surface of Earth The acceleration due to gravity at the Earth's surface is given by: $$g=\frac{GM}{R^2}$$ Where: G = Universal Gravitational Constant M = Mass of the Earth R = Radius of the Earth Acceleration Due to Gravity at Height h Consider a body at a height h above the Earth's surface. Distance from the centre of Earth = (R + h) Therefore, acceleration due to gravity at height h is: $$g_h=\frac{GM}{(R+h)^2}$$ Derivation of Relation Between g and g h Dividing the above equation by the surface gravity equation: $$\frac{g_h}{g}=\frac{\frac{GM}{(R+h)^2}}{\frac{GM}{R^2}}$$ $$\frac{g_h}{g}=\frac{R^2}{(R+h)^2}$$ $$\frac{g_h}{g}=\frac{1}{\left(1+\frac{h}{R}\right)^2}$$ Hence, $$g_...

Class 11 Physics – Chapter 7: Gravitation (NCERT Topics) Introduction Kepler's Laws Universal Law of Gravitation Newton's law of Gravitation in Vector form Principle of Superposition : Gravitation force Shell Theorem and Gravitation Shielding The Gravitational Constant Acceleration Due to Gravity of the Earth Mass and Mean Density of Earth Acceleration Due to Gravity Above the Surface of Earth Acceleration Due to Gravity Below the Surface of Earth Gravitational Potential Energy Escape Speed Earth Satellites Orbiting velocity of Satellite Total Energy of Satellite

Using Newton's law of gravitation, we can determine the mass of the Earth and its mean density . These quantities are calculated using the values of acceleration due to gravity, radius of the Earth, and the universal gravitational constant. Mass of the Earth Consider a body of mass \(m\) placed on the surface of the Earth. According to Newton's law of gravitation, the gravitational force acting on the body is: \[ F=\frac{GMm}{R^2} \] Where: \(M\) = Mass of the Earth \(R\) = Radius of the Earth \(G\) = Universal Gravitational Constant \(m\) = Mass of the body The weight of the body is: \[ F=mg \] Equating the two expressions: \[ mg=\frac{GMm}{R^2} \] Cancelling \(m\) from both sides: \[ g=\frac{GM}{R^2} \] Therefore, \[ M=\frac{gR^2}{G} \] Calculation of Earth's Mass Given: \[ g=9.8\ \text{m s}^{-2} \] \[ R=6400\ \text{km}=6.4\times10^6\ \text{m} \] \[ G=6.67\times10^{-11}\ \text{N m}^2\text{kg}^{-2} \] Substituting...

Acceleration Due to Gravity on Earth (g): Derivation, Formula, Factors and Value Acceleration due to gravity ( g ) is one of the most important concepts in Gravitation. Every object near the Earth's surface experiences a force of attraction due to the Earth's gravity. The acceleration produced by this gravitational force is called acceleration due to gravity (g) . What is Acceleration Due to Gravity? Acceleration due to gravity is the acceleration experienced by a body due to the gravitational force of the Earth. Derivation of Acceleration Due to Gravity Consider a body of mass m lying on the surface of the Earth of mass M and radius R . According to the shell theorem, the gravitational force acting on the body of mass m due to the Earth is equal to the gravitational force between mass m and the mass of Earth concentrated at the centre of the Earth. Therefore, the gravitational force acting on the body is: F = GMm / R² Where: G = Universal Gravitationa...

Determination of Gravitational Constant (G) – Cavendish Experiment The value of the universal gravitational constant ( G ) entering Newton's law of universal gravitation was first determined experimentally by Henry Cavendish in 1798 . The experiment demonstrated that gravitational forces between ordinary laboratory objects can be measured directly. Cavendish Experiment Principle The experiment is based on the principle that the tiny gravitational attraction between two pairs of masses produces a measurable torque on a suspended rod. This gravitational torque twists the suspension wire until it is balanced by the restoring torque of the wire. By measuring the angle of twist, the value of the gravitational constant G can be determined. Experimental Arrangement The apparatus consists of a light horizontal rod AB carrying two small lead spheres, each of mass m , at its ends. The rod is suspended from its centre by a fine torsion wire attached to a rigid support, allowin...