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Definition and Expression of escape velocity of an object on the planet

Definition of Escape Velocity: The minimum velocity, by which an object is thrown vertically in an upward direction and that object goes out from the gravitation field of the planet and does not come back, is called the escape velocity. Deduction Escape Velocity Expression: Let us consider the following: The mass of the planet =$M$ The radius of the planet = $R$ The mass of the object = $m$ The gravitational force on an object at position $P$ which is a distance $x$ from the surface of the planet $F=G\frac{M m}{x^{2}} \qquad(1)$ The work done by the force to move the object a very small distance $dx$ from position $A$ to $B$ $W=F.dx$ $dw=G\frac{M m}{x^{2}}dx$ The total work done to move the object from the surface of the planet to infinity $\int^{W}_{0}dw= \int^{\infty}_{R}G\frac{M m}{x^{2}}dx$ $\left[ w \right]^{W}_{0}=G M m \int^{\infty}_{R} \frac{dx}{x^{2}}$ On solving the above equation $W=GM m \int^{\infty}_{R} \frac{dx}{x^{2}}$

Numerical Problems and Solutions of Centripetal Force and Centripetal Acceleration

Solutions to Numerical Problems of the Centripetal Force and Centripetal Acceleration Numerical Problems and Solutions Q.1 An object of mass $0.20 \: Kg$ is being revolved in a circular path of diameter $2.0 \: m$ on a frictionless horizontal plane by means of a string. It performs $10$ revolutions in $3.14 \: s$. Find the centripetal force acting on an object. Solution: Given that: Mass of body $(m)=0.20 \: Kg$ Diameter of Circular path $(d)=2.0: m$ Number of revolutions $(n)=10$ Time taken to complete $(10)$ revolution$(t)=3.14 s$ The centripetal force acting on a body $(F)=?$ Now the centripetal force: $F=m r \omega^2$ $F=m r \left( \frac{2 \pi n}{t} \right)^2$ Now Substitute the given values in the centripetal force formula: $F=0.20 \times 1 \left( \frac{2 \times 3.14 \times 10}{3.14} \right)^2$ $F=0.8 N$ Check Solution Q.2 A car of mass $1200 Kg$ is moving with a speed of $10.5 ms^{-1}$ on a

Variation of Acceleration Due to Gravity

Variation in Gravitational Acceleration: The value of $g$ varies above and below the surface of the earth. It also changes with the latitude at different places and due to the rotation of the earth about its axis. The earth is not a perfect sphere but is approximately an ellipsoid. The radius of the earth is more at the equator and lesser at the poles. The acceleration due to gravity also changes due to the shape of the earth. 1.) Effect of Altitude on Acceleration due to Gravity 2.) Effect of depth below the earth's surface on Acceleration due to Gravity 3.) Effect of the shape of the earth 4.) Effect of rotation about its own axis 1.) Effect of Altitude on Acceleration due to Gravity: Let us consider: The mass of the earth = $M_{e}$ The radius of the earth = $R_{e}$ A satellite is moving around the earth from the surface of the earth at height= $h$ The gravitational acceleration on the surface of the earth at point $A$ $g=\frac{GM_{e}}{R

Total energy of an orbiting Satellite and its Binding Energy

Definition: The total energy of a satellite revolving around the planet is the sum of kinetic energy (i.e. due to orbital motion) and potential energy (i.e. due to the gravitational potential energy of the satellite). Derivation of the total energy of the satellite revolving around the planet: Let us consider The mass of the satellite = $m$ The mass of the planet = $M$ The satellite revolving around a planet at distance from the center of the planet = $r$ The radius of planet =$R$ The potential energy of the satellite is $U=-\frac{G \: M \: m}{r} \qquad(1)$ The kinetic energy of the satellite is $K=\frac{1}{2}m v_{e}^{2}$ $K=\frac{1}{2} m \left( \sqrt{\frac{G \: M}{r}} \right)^{2} \qquad \left( \because v_{e}^{2}=\sqrt{\frac{G \: M}{r}} \right)$ $K=\frac{1}{2} \left( \frac{G \: M \: m}{r} \right) \qquad(2)$ The total energy of the satellite is $E= K+U$ Now substitute the value of the kinetic energy and potential energy in the above equat

Expression for Orbital velocity of Satellite and Time Period

Orbital Velocity of Satellite: When any satellite moves about the planet in a particular orbit then the velocity of the satellite is called the orbital velocity of the satellite. Expression for Orbital Velocity of Satellite: Let us consider: The mass of the satellite = $m$ The mass of planet= $M$ The radius of Planet =$R$ The satellite is moving about the planet at height=$h$ The satellite is moving about the planet with orbital velocity=$v_{\circ}$ The distance from the center of the planet to satellite=$r$ The force of gravitation between the planet and the satellite $F=G \frac{M \: m}{r^{2}} \qquad(1)$ This force work act as a centripetal force to revolve the satellite around the planet i.e. $F=\frac{m v_{\circ}^{2}}{r} \qquad(2)$ From the above equation $(1)$ and equation $(2)$, we get $\frac{m v_{\circ}^{2}}{r} = G \frac{M \: m}{r^{2}}$ $v_{\circ}= \sqrt{\frac{G \: M }{r}}$ Where $r=R+h$ then $v_{\circ}= \

Derivation of Gravitational Potential Energy due to Point mass and on the Earth

Definition of Gravitational Potential Energy: When an object is brought from infinity to a point in the gravitational field then work done acquired by the gravitational force is stored in the form of potential energy which is called gravitational potential energy. Let us consider, An object of mass $m$ brought from infinity to a point in the gravitational field. If work done acquired by force is $W$ then gravitational potential energy $U=W_{\infty \rightarrow r}$ Derivation of Gravitational Potential energy due to a Point mass: Let us consider, The mass of the point object (i.e point mass)=$m$ The mass of the object that produces the gravitational field = $M$ If the point mass $m$ is at a distance $x$ then the gravitational force between the objects is $F=G \frac{M \: m}{x^{2}} \qquad(1)$ If the point mass moves a very small distance element $dx$ that is at distance $x$ from point $O$ then the work done to move the point object from point

Derivation of Gravitational Potential due to Point mass and on the Earth

Definition of Gravitational Potential: When an object is brought from infinity to a point in the gravitational field then work done acquired by the gravitational force is called gravitational potential. Let us consider, An object of mass $m$ brought from infinity to a point in the gravitational field. If work done acquired by force is $W$ then gravitational potential $V=-\frac{W}{m}$ Derivation of Gravitational Potential due to a Point mass: Let us consider, The mass of the point object (i.e point mass)=$m$ The mass of the object that produces the gravitational field = $M$ If the point mass $m$ is at a distance $x$ then the gravitational force between the objects is $F=G \frac{M \: m}{x^{2}} \qquad(1)$ If the point mass moves a very small distance element $dx$ that is at distance $x$ from point $O$ then the work done to move the point object from point $B$ to $A$ $dw=F.dx$ Now substitute the value of $F$ from equation $(1)$

Gravitational field, Intensity of Gravitational field and its expression

Definition of Gravitational Field: The region around an object in which another object experiences a gravitational force then the region of that object is called the gravitational field. Gravitational Field Intensity: The force applied per unit mass of an object that is placed in the gravitational field is called the intensity of the gravitational field. $\overrightarrow{E}=- \frac{G \: M}{r^{2}} \hat{r}$ The Expression for gravitational field intensity: Let us consider The mass of a lighter object that experience the force = $m$ The mass of a heavy object that produces the gravitational field= $M$ The distance between the objects = $r$ The gravitational force between the objects is $F=G\frac{M\:m}{r^{2}} \qquad(1)$ Now the force per unit mass i.e Gravitational field intensity $E=-\frac{F}{m} \qquad (2)$ Here the negative indicates that the direction of force is opposite to $\hat{r}$ The vector form of the gravitationa

Deduction of Newton's Law of gravitation from Kepler's Law

Deduction of Newton's Law of gravitational force from Kepler's Law: Let us consider: The mass of a planet = $m$ The radius of the circular path of a planet=$r$ The mass of the sun = $M$ The velocity of the planet = $v$ The time of the revolution=$T$ The attraction force between the planet and the sun is achieved by the centripetal force i.e $F=\frac{m v^{2}}{r} \qquad(1)$ The orbital velocity of the planet: $v=\frac{Circumference \: of \: the \: circular \: path}{Time \: period}$ $v=\frac{2\pi r}{T} \qquad(2)$ From equation $(1)$ and equation $(2)$, we get $F=\frac{m}{r} \left( \frac{2\pi r}{T} \right)^{2}$ $F=\frac{4 \pi^{2} mr}{T^{2}} \qquad(3)$ According to Kepler's third law i.e $T^{2}=Kr^{3} \qquad(4)$ From equation$(3)$ and equation$(4)$, we get $F=\frac{4 \pi^{2}mr}{Kr^{3}}$ $F=\frac{4 \pi^{2}}{K}\frac{mr}{r^{3}}$ $F=\frac{4 \pi^{2}}{K}\frac{m}{r^{2}} \qquad(5)$ The source of this force is

Force of Friction | Expression for the acceleration and the work done on inclined rough surface| Advantage and Disadvantage

Definition of Friction: When an object slides over the surface of another object then each surface applies a parallel force to the other. This force is called friction force. The direction of the force of friction on every object with respect to the other is always opposite to the motion of the second object. $F=\mu R$ Properties of Friction Force: The force of friction always acts in the opposite direction of the motion of the object. The force of friction always opposes the motion of the object. So it does not help in motion. Friction force always helps an object to be in the rest of the position. i.e If an object is in the rest of the position that is due to only frictional force. The frictional force is act on both states of the object i.e Rest and motion. Modern Concept of Friction: It is now believed that frictional forces arise because of a fundamental force occurring in nature commonly known as an electric force. Every object has surface irr