Science NCERT Grade 9, Chapter 10, Gravitation as the name suggests, talks about gravitation and universal law of gravitation.The motion of the object under the influence of the gravitational force of the earth is explained in the chapter, Gravitation.The main discussion of the chapter starts by explaining the concept of gravitation by taking the example of the motion of the moon around the earth. Introduce the universal gravitation constant. Share on Facebook Share on Twitter. Gravitation is simply a phenomenon that occurs in nature, where things with energy or mass are attracted to each other. Tx = $\frac{1}{2}$mVx2 = $\frac{1}{2}$mgR. = $\frac{{{\rm{g}}{{\rm{R}}^2}}}{{\rm{r}}}$, = $\frac{{2{\rm{\pi r}}}}{{{\rm{R}}\sqrt {\frac{{\rm{g}}}{{\rm{R}}}} }}$, Or, T2 = $\frac{{4{{\rm{\pi }}^2}{{\rm{r}}^2}}}{{{{\rm{R}}^2}}}$.$\frac{{\rm{r}}}{{\rm{g}}}$ = $\frac{{4{{\rm{\pi }}^2}{{\rm{r}}^3}}}{{{{\rm{R}}^2}{\rm{g}}}}$, Or, g = $\frac{{4{{\rm{\pi }}^2}{{\rm{r}}^3}}}{{{{\rm{T}}^2}{{\rm{R}}^3}}}$, = $\frac{{4{{\rm{\pi }}^2}{\rm{*}}{{\left( {60.1} \right)}^3}{{\rm{R}}^3}}}{{{{\rm{T}}^2}{{\rm{R}}^2}}}$, = $\frac{{4{{\rm{\pi }}^2}{\rm{*}}{{\left( {60.1} \right)}^3}{\rm{*}}6.36{\rm{*}}{{10}^6}}}{{2361600}}$. The revision notes help you revise the whole chapter in minutes. In orbit, the energy is given by: E2 = $ - \frac{{{\rm{G}}{{\rm{M}}_{\rm{e}}}{\rm{m}}}}{{2{\rm{r}}}}$, = $\frac{{ - 6.67{\rm{*}}{{10}^{ - 11}}{\rm{*}}5.97{\rm{*}}{{10}^{24}}{\rm{*}}1000}}{{2\left( {6.68{\rm{*}}{{10}^6}} \right)}}$. Or, g’ = ${\left( {1 + \frac{{\rm{h}}}{{\rm{R}}}} \right)^{ - 2}}$g……(i). Solve Numericals. If T is the time period of revolution of the satellite then, T = $\frac{{2{\rm{\pi r}}}}{{\rm{v}}}$. Hindi Gravitation. Distance of satellite from the centre of the earth d = 2R. T = $\frac{{2{\rm{\pi }}}}{{\rm{w}}}$ = $\frac{{2{\rm{\pi }}}}{{\frac{{\rm{v}}}{{\rm{r}}}}}$ = $\frac{{2{\rm{\pi r}}}}{{\rm{v}}}$, = $\frac{{2{\rm{\pi *}}6.559{\rm{*}}{{10}^6}}}{{7852}}$. Complete Physics Course - Class 11 OFFERED PRICE: Rs. 2 ] What is the gravitational force between the Sun and the Earth? 01.Physical World; ... Lect 05:Gravitation Potential. So, work done against the attraction of moon (i.e.) = 6.67 * 10-11 * 6.42 * 1023 * 3 * 103 * $\frac{1}{2}$ * $\left( {\frac{1}{{5.4{\rm{*}}{{10}^{ - 6}}}} - \frac{1}{{7.4{\rm{*}}{{10}^{ - 6}}}}} \right)$. Derive its mathematical formula. Or, g’ = $\left( {1 - \frac{{\rm{d}}}{{\rm{R}}}} \right)$g, W = $\frac{{2{\rm{\pi }}}}{{\rm{T}}}$ = $\frac{{2{\rm{\pi }}}}{{24{\rm{*}}60{\rm{*}}60}}$. Hence, the escape speed for the given asteroid is 177 m/s. or, $\frac{{{\rm{g'}}}}{{\rm{g}}}$ = ${\left( {1 + \frac{{\rm{h}}}{{\rm{R}}}} \right)^{ - 2}}$. In what source is ‘G’ universal? Learn the concepts of Class 11 Physics Gravitation with Videos and Stories. T = $\frac{{2{\rm{\pi r}}}}{{\rm{V}}}$ = $\frac{{20{\rm{T}}\left( {6.68{\rm{*}}{{10}^6}} \right)}}{{7720}}$ = 5436.18sec. And we know, T = $\frac{{2{\rm{\pi r}}}}{{\rm{v}}}$ and v2 = $\frac{{{\rm{GM}}}}{{\rm{r}}}$, Or, v = $\sqrt {\frac{{{\rm{Gm}}}}{{\rm{r}}}} $ = $\left( {\frac{{\sqrt {\rm{G}} {\rm{m}}}}{{\sqrt {\rm{r}} }}} \right)$, So, T = $\frac{{2{\rm{\pi }}}}{{{\rm{Gm}}}}$. From the above two relation it is clear that Tx> Ty. 3. CBSE CLASS 11 Physics Notes. Or, 10 = $\frac{{{\rm{GM}}}}{{{{\rm{R}}^2}}}$ …(i), Also, F’ = $\frac{{{\rm{GMm}}}}{{{{\rm{r}}^2}}}$ = $\frac{{{\rm{GMm}}}}{{{{\left( {\frac{{3{\rm{R}}}}{2}} \right)}^2}}}$, = $\frac{{{\rm{GM}}}}{{{{\rm{R}}^2}}}$ * $\frac{{4{\rm{m}}}}{9}$, Equilateral radius of earth Req = 6.378 * 106m, gp = $\frac{{{\rm{GM}}}}{{{\rm{R}}{{\rm{p}}^2}}}$, or, gp = $\frac{{6.67{\rm{*}}{{10}^{ - 11}}{\rm{*}}5.957{\rm{*}}{{10}^{24}}}}{{{{\left( {6.357{\rm{*}}{{10}^6}{\rm{\: }}} \right)}^2}}}$. Download Numericals- Chapter 2; Download Numericals- Chapter 3; Download Numericals- Chapter 4; Download Numericals- Chapter 5; Download Numericals- Chapter 6 ; Download Numericals- Chapter 7; Download Numericals- Chapter 8; Download Numericals- Chapter 9; Download Numericals- Chapter 10; Download Numericals- Chapter 11; Download Numericals- Chapter … Given acceleration due to gravity at the surface of the earth g = 9.8m/s2. So, here is the Class 11 Physics Gravitation Notes for IIT JEE, NEET & Board Exam Preparation. Ended on Nov 21, 2020 • 7 lessons. Radius of earth R = 6.4 * 106m, If r be the radius of orbit of satellite. eSaral have already come up with an amazing revision series of Physics where you can easily revise your chapter within minutes with all the important formulae and key points. Then, T = $\frac{{2{\rm{\pi r}}}}{{\rm{V}}}$ = $\frac{{2{\rm{\pi r}}}}{{\sqrt {\frac{{{\rm{GM}}}}{{\rm{r}}}} }}$, = 2πr $\sqrt {\frac{{\rm{r}}}{{{\rm{GM}}}}} $, = 2π * 1 * 107$\sqrt {\frac{{{{10}^7}}}{{6.7{\rm{*}}{{10}^{ - 11}}{\rm{*}}6{\rm{*}}{{10}^{24}}}}} $, Radius of the earth R = 6400km = 6.4 * 106m. So, energy required E = $\frac{1}{2}$mve2. These are (i) the gravitational force (ii) the electromagnetic force Important questions, guess papers, most expected questions and best questions from 11th physics chapter 08 Gravitation have CBSE chapter wise important questions with solution for free download in PDF format. So, here is the Class 11 Physics Gravitation Notes for IIT JEE, NEET & Board Exam Preparation. % different in weight = $\left( {\frac{{{\rm{mg}} - {\rm{mg'}}}}{{{\rm{mg}}}}} \right)$ * 100% = $\frac{{{\rm{g}} - {\rm{g'}}}}{{\rm{g}}}$ * 100%. Forces of gravity are directed along the line joining the interacting particles and are, therefore, called central forces, which is conservative. Here, Density of the earth δ = 5500 kgm-3. Again. V = R $\sqrt {\frac{{\rm{g}}}{{{\rm{R}} + {\rm{h}}}}} $, = 6.4 * 106$\sqrt {\frac{{9.8}}{{71.8{\rm{*}}{{10}^5}}}} $. 17–Thermal properties of matter. or, v2 = $\frac{{{\rm{GM}}}}{{\rm{r}}}$. Reading Time: 9min read 0. Solve Numericals. Download CBSE Important Questions for CBSE Class 11 Physics Gravitation Kepler's laws of planetary motion, universal law of gravitation. Or, t1 = $\frac{1}{{{{\rm{g}}_{\rm{e}}}}}$ …(i). If you're seeing this message, it means we're having trouble loading external resources on our website. If you have any query regarding NCERT Solutions for Class 11 Physics Chapter 8 Gravitation, drop a comment below and we will get back to you at the earliest. (iii) For satellite to escape to infinity, the total mechanical energy must be zero. $\frac{{{\rm{GMm}}}}{{{{\rm{r}}^2}}}$ = $\frac{{{\rm{m}}{{\rm{v}}^2}}}{{\rm{r}}}$. Equatorial radius of the earth R = 6.4 * 106 m. Acceleration of free fall at the poles of the earth g = 9.8 m/s2. Prepared by teachers of the best CBSE schools in India. CBSE class 11 Physics notes with derivations are best notes by our expert team. Nootan Solutions Gravitation Planets and Satellites ISC Class-11 Physics Nageen Prakashan Chapter-12 Numericals of latest edition. Here find Physics Notes, assignments, concept maps and lots of study material for easy learning and understanding. They are as follows: (i) Law of orbits. If T be the time period of revolution of a satellite moving in a circular orbit round the earth. Physics problems with pseudo force and solutions – inclined plane/wedge Try your concepts on pseudo force. Prcatical Center is one of the biggest Coaching Centre in Karachi . Numericals from Physics, Chapter No.6 (Gravitation) for Class 11th, XI, HSC Part 1, 1st Year. For the satellite to be in circular orbit. 13–Elasticity. Solve Numericals. Courses. Our notes has covered all topics which are in NCERT syllabus plus other topics which are required for Board Exams. i.e. Course on Physics for Science Final Exams. or own an. Acceleration due to gravity g = 0.278 m/s2. Or, $\frac{{{\rm{GM}}}}{{{{\rm{R}}^2}}}$ = 10 …(i). April 22, 2019. in CBSE. So, the required value of g from the motion of the moon is 9.8 m/s2. Where, R = 6.4 * 106m is the radius of earth. Before starting with Gravitation Class 11, it is crucial to understand that gravity and gravitation are not similar. Find video courses for class 11 and 12 physics. 1800-212-7858 / 9372462318. 14– Fluid Pressure. The gravitational force between the earth and satellite is given by: F2 = $\frac{{{\rm{GM}}{{\rm{m}}_{\rm{s}}}}}{{{{\rm{d}}^2}}}$ = $\frac{{{\rm{GM}}{{\rm{m}}_{\rm{s}}}}}{{{{\left( {2{\rm{R}}} \right)}^2}}}$, = $\frac{1}{4}$$\left( {\frac{{{\rm{GM}}}}{{{{\rm{R}}^2}}}} \right)$ * 100, = $\frac{{{\rm{GM}}}}{{{{\rm{R}}^2}}}$ * 25 ….(ii). Let acceleration due to gravity at the height h be g’ = 0.98m/s2. Introduce the universal gravitation constant. The physics formulas for class 11 PDF is provided here so that students can understand the subject more effectively. So, Gravitational force F = $\frac{{{\rm{GMm}}}}{{{{\left( {{\rm{R}} + {\rm{h}}} \right)}^2}}}$, = $\frac{{6.67{\rm{*}}{{10}^{ - 11}}{\rm{*}}6.025{\rm{*}}{{10}^{24}}{\rm{*}}2150}}{{{{\left( {7.18{\rm{*}}{{10}^6}} \right)}^2}}}$, Let g’ be the acceleration due to gravity at height h = 780 km, Or, $\frac{{{\rm{g'}}}}{{\rm{g}}}$ = $\left( {1 - \frac{{2{\rm{h}}}}{{\rm{R}}}} \right)$, = $\left( {1 - \frac{{\left( {2{\rm{*}}7.8{\rm{*}}{{10}^5}} \right)}}{{6.4{\rm{*}}{{10}^6}}}} \right)$, = $\left( {1 - \frac{{15.6}}{{64}}} \right)$, Or, Fraction of weight = $\frac{{{\rm{mg'}}}}{{{\rm{mg}}}}$ = $\frac{{{\rm{g'}}}}{{\rm{g}}}$ = 0.8. Some Basic Concepts of Chemistry; ... 11 Chemistry Quiz; 11 Physics Quiz; 12 Chemistry Quiz; 12 Physics Quiz; Study Material. But the potential energy gained must be the same. An amount of work equal to 2.99 * 1010J would have to be done. We have, T = $\frac{{2{\rm{\pi }}}}{{\rm{R}}}$$\sqrt {\frac{{{{\rm{d}}^3}}}{{\rm{g}}}} $. 1. 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