{"id":36001,"date":"2022-11-25T16:57:18","date_gmt":"2022-11-25T11:27:18","guid":{"rendered":"https:\/\/tsboardsolutions.com\/?p=36001"},"modified":"2022-11-25T16:57:18","modified_gmt":"2022-11-25T11:27:18","slug":"ts-inter-1st-year-physics-notes-chapter-9","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.com\/ts-inter-1st-year-physics-notes-chapter-9\/","title":{"rendered":"TS Inter 1st Year Physics Notes Chapter 9 Gravitation"},"content":{"rendered":"
Here students can locate TS Inter 1st Year Physics Notes<\/a> 9th Lesson Gravitation to prepare for their exam.<\/p>\n \u2192 Kepler’s Laws : \u2192 Law of areas (2nd law) : The line joining the planet to the sun sweeps equal areas in equal intervals of time, i.e., \\(\\) = constant. \u2192 Law of periods (3rd law) : The square of time period of revolution of a planet is proportional to the cube of the semi major axis of the ellipse traced out by the planet. \u2192 Newton’s law of gravitation (OR) Universal law of gravitation: Every body in universe attracts other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. <\/p>\n \u2192 Central force : A central force is that force which acts along the line joining the sun and the planet or along the line joining the two mass particles.<\/p>\n \u2192 Conservative force : For a conservative force work done is independent of the path. Work done depends only on initial and final positions only.<\/p>\n \u2192 Gravitational potential energy : Potential energy arising out of gravitational force is called gravitational potential energy. \u2192 Gravitational potential : Gravitational potential due to gravitational force of earth is defined as the “potential energy of a particle of unit mass at that point”. \u2192 Acceleration due to gravity (g) : \u2192 Acceleration due to gravity below and above surface of earth : 2) For a point inside earth at a depth’d’ below the ground mass of earth (Ms<\/sub>) with radius (RE <\/sub>– d) is considered. That mass seems to be at centre of earth. \u2192 Escape speed (v1<\/sub>)min<\/sub> : The minimum initial velocity on surface of earth to overcome gravitational potential energy is defined as “escape speed ve<\/sub>” \u2192 Orbital velocity: Velocity of a body revolving in the orbit is called orbital velocity. Note:<\/p>\n \u2192 Time period of the orbit (T) : Time taken by a satellite to complete one rotation in the orbit is called “time period of rotation”. \u2192 Geostationary orbit : For a geostationary orbit in equatorial plane its time period of rotation is 24 hours, i.e., angular velocity of satellite in that orbit is equal to angular velocity of rotation of earth. <\/p>\n \u2192 Geostationary satellite : Geostationary satellite will revolve above earth in geostationary orbit along the direction of rotation of earth. So it always seems to be stationary w.r.t earth. \u2192 Polar satellites: Polar satellites are low attitude satellites with an altitude of 500 km to 800 km. They will revolve in north-south direction of earth. \u2192 Weightlessness: Fora freely falling body its weight seems to be zero. Weight of a body falling downwards with acceleration ‘a’ is w’ = mg’ = m(g – a). When a = g the body is said to be under free fall and it seems to be weightless.<\/p>\n \u2192 Force between two mass particles, F = \\(\\frac{\\mathrm{Gm}_1 \\mathrm{~m}_2}{\\mathrm{r}^2}\\)<\/p>\n \u2192 Universal gravitational constant, G = \\(\\frac{\\mathrm{Fr}^2}{\\mathrm{~m}_1 \\mathrm{~m}_2}\\) \u2192 Relation between g and G is, \u2192 Variation ofg with depth, gd<\/sub> = g(1 – \\(\\frac{d}{R}\\))<\/p>\n \u2192 Variation ofg with height, gh<\/sub> = g(1 – \\(\\frac{2h}{R}\\)) (a) Gravitational potential, U = –\\(\\frac{\\mathrm{GMm}}{\\mathrm{R}}\\) \u2192 Orbital velocity, V0<\/sub> = \\(\\sqrt{\\frac{\\mathrm{GM}}{\\mathrm{R}}}=\\sqrt{\\mathrm{gR}}\\)<\/p>\n \u2192 Orbital angular velocity, \u03c90<\/sub> = \\(\\sqrt{\\frac{\\mathrm{GM}}{\\mathrm{R}^3}}=\\sqrt{\\frac{\\mathrm{g}}{\\mathrm{R}}}\\)<\/p>\n \u2192 Escape velocity, Ve<\/sub> = \\(\\sqrt{\\frac{2 \\mathrm{GM}}{\\mathrm{R}}}=\\sqrt{2 \\mathrm{gR}}\\)<\/p>\n \u2192 Time period of geostationary orbit = 24 hours.<\/p>\n \u2192 Angular velocity of earth’s rotation Here students can locate TS Inter 1st Year Physics Notes 9th Lesson Gravitation to prepare for their exam. TS Inter 1st Year Physics Notes 9th Lesson Gravitation \u2192 Kepler’s Laws : Law of orbits (1st law) ‘.All planets move in an elliptical orbit with the sun is at one of its foci. \u2192 Law of … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":[],"categories":[27],"tags":[],"yoast_head":"\nTS Inter 1st Year Physics Notes 9th Lesson Gravitation<\/h2>\n
\nLaw of orbits (1st law) ‘.All planets move in an elliptical orbit with the sun is at one of its foci.<\/p>\n
\ni. e., planets will appear to move slowly when they are away from sun, and they will move fast when they are nearer to the sun.<\/p>\n
\ni.e., T2<\/sup> \u221d R3<\/sup> \u21d2 \\(\\frac{\\mathrm{T}^2}{\\mathrm{R}^3}\\) = constant<\/p>\n
\nF \u221d m1<\/sup>m2<\/sup>, F \u221d \\(\\frac{1}{\\mathrm{r}^2}\\) \u21d2 F = G\\(\\frac{\\mathrm{m}_1 \\mathrm{~m}_2}{\\mathrm{r}^2}\\)<\/p>\n
\nSince gravitational force is a conservative force gravitational potential depends on position of object.
\nV = \\(-\\frac{\\mathrm{Gm}_1 \\mathrm{~m}_2}{\\mathrm{r}}\\)<\/p>\n
\nGravitational potential V = \\(\\frac{G M}{r}\\)
\n(r = distance from centre of earth)<\/p>\n
\nAcceleration due to gravity ‘g’ = \\(\\)<\/p>\n
\n1) For points above earth total mass of earth seems to be concentrated at centre of earth.
\nFor a height ‘h’ above earth
\ng(h) = \\(\\frac{\\mathrm{GM}_{\\mathrm{E}}}{\\left(\\mathrm{R}_{\\mathrm{E}}+\\mathrm{h}\\right)^2}\\)
\nwhere h << RE
\ng(h) = g\\(\\left(1+\\frac{h}{R_E}\\right)^{-2}\\) = g\\(\\left(1-\\frac{2 \\mathrm{~h}}{\\mathrm{R}_{\\mathrm{E}}}\\right)\\)<\/p>\n
\ng’ = g\\(\\left(1-\\frac{d}{R}\\right)\\)<\/p>\n
\nve<\/sub> = \\(\\sqrt{2 \\mathrm{gR}}=\\sqrt{\\frac{2 \\mathrm{GM}}{\\mathrm{R}}}\\)<\/p>\n
\nOrbital velocity V0<\/sub> = \\(\\sqrt{\\frac{\\mathrm{GM}}{\\mathrm{R}}}\\) \u21d2 V0<\/sub> = \\(\\sqrt{\\mathrm{gR}}\\)<\/p>\n\n
\nT = 2\u03c0\\(\\frac{\\left(\\mathrm{R}_{\\mathrm{E}}+\\mathrm{h}\\right)^{3 \/ 2}}{\\sqrt{\\mathrm{GM}_{\\mathrm{E}}}}\\)<\/p>\n
\nGeostationary orbit is at a height of 35800 km from earth.<\/p>\n
\nTime period of geostationary satellite is 24 hours. It rotates in equatorial plane in west to east direction.<\/p>\n
\nTime period of polar satellites is nearly 100 minutes.<\/p>\n
\nG = 6.67 \u00d7 10-11<\/sup> Nm2<\/sup> \/ Kg2<\/sup> D.F.: M-1<\/sup> L3<\/sup> T-2<\/sup><\/p>\n
\ng = \\(\\frac{\\mathrm{GM}}{\\mathrm{R}^2}=\\frac{4}{3}\\)\u03c0\u03c1G.R<\/p>\n
\nFor small values of h i.e., h < < R then
\ngh<\/sub> = g(1 – \\(\\frac{2h}{R}\\))<\/p>\n
\n(b) If a body is taken to a height h’ above the ground then
\nGravitational potential, Uh = –\\(\\frac{\\mathrm{GMm}}{(\\mathrm{R}+\\mathrm{h})}\\)<\/p>\n
\n(\u03c9) = \\(\\frac{2 \\pi}{24 \\times 60 \\times 60}\\) = 0 072 \u00d7 10-3<\/sup> rad\/sec<\/p>\n","protected":false},"excerpt":{"rendered":"