Physics Vidyapith ✍️

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 $W=GM m \int^{\infty}_{R} \frac{dx}{x^{2}}$

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

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

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}= \

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

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 $(1)$

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 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

Kepler's Laws → Kepler's found important uniformity in the motion of the planets. This uniformity is known as "Kepler's Laws of planetary motion". There are basically three laws → First law (Law of orbits) Second Law (Law of Areal Speed) Third Law (Law of Periods) First law (Law of orbits) → All the planets move around the sun in elliptical path and the sun is at the one foci of the ellipse. Second Law (Law of Areal Speed) → A line joining the any planet to sun sweeps out equal areas in equal interval of of times. i.e The areal speed of the planets remains constant. According to the second law, When the planet is farthest from the sun, then its speed is maximum, and when it is nearest the sun, then its speed is maximum. Planetary Motion From the above figure, A planet is moving around the sun from $A$ to $B$ in a given time interval, and from $C$ to $D$ in the same time inter

Relation between $g$ and $G$ → Let us consider: The mass of earth = $M_{e}$ The radius of earth = $R_{e}$ The mass of the object = $m$ If the object is placed on the surface of the earth then the gravitational force on the object is → $F=G \frac{M_{e} m}{R_{e}^{2}} \qquad(1)$ The force on the object due to gravitational acceleration is → $F=mg \qquad(2)$ From equation $(1)$ and equation $(2)$ $mg=G \frac{M_{e} m}{R_{e}^{2}}$ $g=G \frac{M_{e}}{R_{e}^{2}}$ $G M_{e}=g R_{e}^{2}$

Gravitational Force → Newton's Gravitational Law statement is a combination of three individual statements. These are The force between the two-particle is directly proportional to the product of their masses i.e. $F \propto m_{1} \: m_{2} \qquad(1)$ Where $m_{1}$ & $m_{2}$ are the masses of the particles. The force between the two-particle is inversely proportional to the square of the distance between them i.e. $F \propto \frac{1}{r^{2}} \qquad(2)$ Where $r$ is the distance between the particles. This force always acts between the line joining the masses. From the above the equation $(1)$ and equation $(2)$ $F\propto \frac{m_{1} \: m_{2}}{r^{2}}$ $F=G \frac{m_{1} \: m_{2}}{r^{2}}$ Where $G$ is Newton's gravitaional constant and its experimental value $6.67\times 10^{-11} \frac{N-