{"id":2442,"date":"2014-10-08T08:28:53","date_gmt":"2014-10-08T02:58:53","guid":{"rendered":"http:\/\/physicscatalyst.com\/article\/?p=2442"},"modified":"2022-11-04T12:06:32","modified_gmt":"2022-11-04T06:36:32","slug":"ncert-book-solutions-class-9-gravitation-part-1","status":"publish","type":"post","link":"https:\/\/physicscatalyst.com\/article\/ncert-book-solutions-class-9-gravitation-part-1\/","title":{"rendered":"NCERT book Solutions class-9 Gravitation (Part 1)"},"content":{"rendered":"<p>This page contains NCERT book Solutions class-9 Gravitation starting from page 143 at the end of the chapter. I have also been writing the notes for this chapter to learn more about notes you can follow this<\/p>\n<p>https:\/\/physicscatalyst.com\/Class9\/gravitation.php<\/p>\n<blockquote class=\"wp-embedded-content\" data-secret=\"49s4t57JB2\"><p><a href=\"https:\/\/physicscatalyst.com\/article\/ncert-solutions-class9-gravitation\/\">NCERT book Solutions class-9 Gravitation (In Text Questions)<\/a><\/p><\/blockquote>\n<p><iframe loading=\"lazy\" class=\"wp-embedded-content\" sandbox=\"allow-scripts\" security=\"restricted\" style=\"position: absolute; clip: rect(1px, 1px, 1px, 1px);\" title=\"&#8220;NCERT book Solutions class-9 Gravitation (In Text Questions)&#8221; &#8212; physicscatalyst&#039;s Blog\" src=\"https:\/\/physicscatalyst.com\/article\/ncert-solutions-class9-gravitation\/embed\/#?secret=etj4wOpKdm#?secret=49s4t57JB2\" data-secret=\"49s4t57JB2\" width=\"500\" height=\"282\" frameborder=\"0\" marginwidth=\"0\" marginheight=\"0\" scrolling=\"no\"><\/iframe><\/p>\n<blockquote class=\"wp-embedded-content\" data-secret=\"0rWtWHUZYN\"><p><a href=\"https:\/\/physicscatalyst.com\/article\/ncert-book-solutions-class-9-gravitation-part-2\/\">NCERT book Solutions class-9 Gravitation (Part 2)<\/a><\/p><\/blockquote>\n<p><iframe loading=\"lazy\" class=\"wp-embedded-content\" sandbox=\"allow-scripts\" security=\"restricted\" style=\"position: absolute; clip: rect(1px, 1px, 1px, 1px);\" title=\"&#8220;NCERT book Solutions class-9 Gravitation (Part 2)&#8221; &#8212; physicscatalyst&#039;s Blog\" src=\"https:\/\/physicscatalyst.com\/article\/ncert-book-solutions-class-9-gravitation-part-2\/embed\/#?secret=ivIqzpoQ91#?secret=0rWtWHUZYN\" data-secret=\"0rWtWHUZYN\" width=\"500\" height=\"282\" frameborder=\"0\" marginwidth=\"0\" marginheight=\"0\" scrolling=\"no\"><\/iframe><\/p>\n<p>https:\/\/physicscatalyst.com\/Class9\/gravitation_questions.php<\/p>\n<p><span style=\"color: #000000;\"><strong>Question 1<\/strong><\/span><\/p>\n<p>How does the force of gravitation between two objects change\u00a0when the distance between them is reduced to half ?<\/p>\n<p><span style=\"color: #000000;\"><strong>Answer<\/strong><\/span><\/p>\n<p>According to Universal Law of gravitation , the gravitational force of attrection between any two objects of mass<br \/>\n$M$ and $m$ is proportional to the product of their masses<br \/>\nand inversly proportional to the square of distance $r$ between them<br \/>\nSo, force $F$ is given by<br \/>\n$F = G\\frac{{M \\times m}}{{{r^2}}}$<br \/>\nNow when the distance $r$ is reduced to half then force between two masses becomes<br \/>\n$F&#8217;=G\\frac{M\\times m}{\\left ( \\frac{r}{2} \\right )^2}$<br \/>\nor,<br \/>\n$F&#8217;=4F$<br \/>\nClearly , if distance between two objects is reduced to half then the gravitational force becomes four times larger than its previous value.<\/p>\n<p><span style=\"color: #000000;\"><strong>Question 2<\/strong><\/span><\/p>\n<p>Gravitational force acts on all objects in proportion to their\u00a0masses. Why then, a heavy object does not fall faster than a\u00a0light object<\/p>\n<p><span style=\"color: #000000;\"><strong>Answer<\/strong><\/span><\/p>\n<p>Weight of any object of mass $m$ on the surface of earth is ,<br \/>\n$W=mg$<br \/>\nwhere $g$ is the acceleration due to gravity.<br \/>\nNow from Universal Law of gravitation , force acting on any object due o gravitational pull of eart is<br \/>\n$F = G\\frac{{M \\times m}}{{{R^2}}}$<br \/>\nwhere $M$ is the mass of the earth and $R$ is the radius of the earth<br \/>\nNow,<br \/>\nWeight of object on surface of earth is equal to gravitatinal force acting on it.<br \/>\nSo,<br \/>\n&#40;mg = G\\frac{{M \\times m}}{{{R^2}}}&#41;<br \/>\nRearranging above equation we get<br \/>\n&#40;g = G\\frac{M}{{{R^2}}}&#41;<br \/>\nFrom this expression for acceleration due to gravity $g$ it is clear that $g$ does not depend on the mass of the objects. It is constant and is same for both heavy and light masses.<br \/>\nHence heavy objects do not fall faster then light objects<\/p>\n<p><span style=\"color: #000000;\"><strong>Question 3<\/strong><\/span><\/p>\n<p>What is the magnitude of the gravitational force between the\u00a0earth and a 1 kg object on its surface? (Mass of the earth is $6 \\times\u00a0\u00a010^{24} kg$ and radius of the earth is $6.4 \\times\u00a0\u00a010^{6} m$.)<\/p>\n<p><span style=\"color: #000000;\"><strong>Answer<\/strong><\/span><\/p>\n<p>From Universal law of gravitation , force exerted on an object of mass $m$ by earth is given by<br \/>\n$F = G\\frac{{M \\times m}}{{{R^2}}}$ (1)<br \/>\nwhere,<br \/>\nMass of earth, $M=6 \\times 10^{24}Kg$<br \/>\nMass of the object, $m=1Kg$<br \/>\nGravitational constant $G=6.7 \\times 10^{-11}Nm^{2}Kg^{-2}$<br \/>\nHere radius of rart $R$ is the distance between the objects as object under consideration is on the urface of earth. So,<br \/>\n$R=6.4 \/times 10^{6}m$<br \/>\nPutting all these values in equation 1 we get<br \/>\n&#40;F = \\frac{{6.7 \\times {{10}^{ &#8211; 11}} \\times 6 \\times {{10}^{24}} \\times 1}}{{{{\\left( {6.4 \\times {{10}^6}} \\right)}^2}}} = 9.8N&#41;<\/p>\n<p><span style=\"color: #000000;\"><strong>Question 4<\/strong><\/span><\/p>\n<p>The earth and the moon are attracted to each other by\u00a0gravitational force. Does the earth attract the moon with a force\u00a0that is greater or smaller or the same as the force with which\u00a0the moon attracts the earth? Why?<\/p>\n<p><span style=\"color: #000000;\"><strong>Answer<\/strong><\/span><\/p>\n<p>According to universal law of gravitation tow objects attract each other with equal force but in opposite directions. The Earth and the moon both attracts each other with equal force. Which means that the magnitude of force with which earth attracts moon is equal to the magnitude of force with which moon attracts earth but their directions are opposite to each other,<\/p>\n<p><span style=\"color: #000000;\"><strong>Question 5<\/strong><\/span><\/p>\n<p>If the moon attracts the earth, why does the earth not move\u00a0towards the moon?<\/p>\n<p><span style=\"color: #000000;\"><strong>Answer<\/strong><\/span><\/p>\n<p>Earth and the moon both exerts equal gravitational force on each other. The earth does not moves towards moon as a result of this force of atraction because the mass of earth is much larger then that of moon due to which acceleration experienced by earth due to gravitational pull of moon is very small in comparison to that experienced by moon due to earth. This is the reason earth does not moves towards moon.<\/p>\n<p><span style=\"color: #000000;\"><strong>Question 6<\/strong><\/span><\/p>\n<p>What happens to the force between two objects, if<br \/>\n(i) the mass of one object is doubled?<br \/>\n(ii) the distance between the objects is doubled and tripled?<br \/>\n(iii) the masses of both objects are doubled?<\/p>\n<p><span style=\"color: #000000;\"><strong>Answer<\/strong><\/span><\/p>\n<p>From Universal law of gravitation , force exerted on an object of mass $m$ by earth is given by<br \/>\n$F = G\\frac{{M \\times m}}{{{R^2}}}$ (1)<br \/>\n(i) When mass of the object say $m$ is doubled than<br \/>\n$F&#8217; = G\\frac{{M \\times 2m}}{{{R^2}}}=2F$<br \/>\nSo as the mass of any one of the object is doubled the force is also doubled<br \/>\n(ii) The force $F$ is inversely proportional to the distance between the objects. So if the distance between two objects is doubled then the gravitational force of attraction between them is reduced to one fourth of its original value. Similarly f the distance between two objects is tripled , then the gravitational force of attraction becomes one ninth of its original value.<br \/>\n(iii) Again fron Universal law of attraction from equation (1) force $F$ is directly proportional to the product of both the masses. So if both the masses are doubled then the gravitational force of attraction becomes four times the original value.<\/p>\n<p><span style=\"color: #000000;\"><strong>Question 7<\/strong><\/span><\/p>\n<p>What is the importance of universal law of gravitation?<\/p>\n<p><span style=\"color: #000000;\"><strong>Answer<\/strong><\/span><\/p>\n<p>Universal law of Gravitation is important because it it tells us about<\/p>\n<ul>\n<li>the force that is responsible for binding us to Earth.<\/li>\n<li>the motion of moon around the earth<\/li>\n<li>the motion of planets around the sun<\/li>\n<li>the tides formed by rising and falling of water level in the ocean are due to the gravitational force exerted by both sun and moon on the earth.<\/li>\n<\/ul>\n<p><span style=\"color: #000000;\"><strong>Question 8<\/strong><\/span><\/p>\n<p>What is the acceleration of free fall?<\/p>\n<p><span style=\"color: #000000;\"><strong>Answer<\/strong><\/span><\/p>\n<p>When objects falls towards the Earth from a certain height it falls freely towards the earth under the effect of gravitational force alone. So acceleration of free fall is the acceleration experienced by any object when it falls freely under the influence of gravitational force alone.<br \/>\nIt is denoted by $g$ and its value on the surface of earth is $9.8ms^{-2}$.<\/p>\n<p><span style=\"color: #000000;\"><strong>Question 9<\/strong><\/span><\/p>\n<p>What do we call the gravitational force between the earth and\u00a0an object?<\/p>\n<p><span style=\"color: #000000;\"><strong>Answer<\/strong><\/span><\/p>\n<p>Gravitational force between earth and an object is known as the weight of the object.<\/p>\n<p><span style=\"color: #000000;\"><strong>Question 10<\/strong><\/span><\/p>\n<p>Amit buys few grams of gold at the poles as per the instruction\u00a0of one of his friends. He hands over the same when he meets\u00a0him at the equator. Will the friend agree with the weight of gold\u00a0bought? If not, why? [Hint: The value of g is greater at the\u00a0poles than at the equator.]<\/p>\n<p><span style=\"color: #000000;\"><strong>Answer<\/strong><\/span><\/p>\n<p>The value of acceleration due to gravity varies at different places on the earth . It decreases from poles to equator.<br \/>\nNow, Weight of body on earth is given by<br \/>\n$W=mg$<br \/>\nwhere $g$ is the acceleration due to gravity and $m$ is the mass of the body.<br \/>\nSince the value of $g$ is greater at poles then it is at equator. This is the reason that weight of gold on equator is less than the weight of gold on poles. Hence, Amit&#8217;s friend will not agree with the weight of the gold bought.<\/p>\n<p><span style=\"color: #000000;\"><strong>Question 11<\/strong><\/span><\/p>\n<p>Why will a sheet of paper fall slower than one that is crumpled\u00a0into a ball?<\/p>\n<p><span style=\"color: #000000;\"><strong>Answer<\/strong><\/span><\/p>\n<p>When we crumble a sheet of paper into a ball then its surface area becomes much lesser than the surface area of a plain flat sheet of paper. \u00a0The plain sheet of paper experience\u00a0more resistance due to air than the crumpled ball and it will fall slower than the crumpled ball even if they both experience same force of gravity.<\/p>\n<h4><span style=\"color: #993300;\"><em><span style=\"text-decoration: underline;\">If you are interested in other links containing class 9 notes in science and maths along with free download links consider following the links given below<\/span><\/em><\/span><\/h4>\n<p>[standout-css3-button href=&#8221;https:\/\/physicscatalyst.com\/class9.php&#8221;]Class 9 science study material[\/standout-css3-button] [standout-css3-button href=&#8221;https:\/\/physicscatalyst.com\/class9_maths.php&#8221;]Class 9 maths study material[\/standout-css3-button] [standout-css3-button href=&#8221;https:\/\/physicscatalyst.com\/download9.php&#8221;]Class 9 downloads[\/standout-css3-button]<\/p>\n","protected":false},"excerpt":{"rendered":"<p>This page contains NCERT book Solutions class -9 Gravitation starting from page 143 at the end of the chapter. I have also been writing the notes for this chapter to learn more about notes you can follow this link and to get the in-text questions solution please follow this link<\/p>\n<p>Question 1<\/p>\n<p>How does the force of gravitation between two objects change when the distance between them is reduced to half ?<\/p>\n<p>Answer<\/p>\n<p>According to Universal Law of gravitation , the gravitational force of attrection between any two objects of mass<\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_uag_custom_page_level_css":"","site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[24,279],"tags":[],"class_list":["post-2442","post","type-post","status-publish","format-standard","hentry","category-general","category-ncert-solutions"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.3 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>NCERT book Solutions class-9 Gravitation (Part 1)<\/title>\n<meta name=\"description\" content=\"This page contains NCERT book Solutions class-9 Gravitation starting from page 143 at the end of the chapter. 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I have also been writing the notes for this chapter to learn more about notes you can follow this link and to get the in-text questions solution please follow this link Question 1 How does the&hellip;","_links":{"self":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts\/2442","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/users\/8"}],"replies":[{"embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/comments?post=2442"}],"version-history":[{"count":1,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts\/2442\/revisions"}],"predecessor-version":[{"id":7320,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/posts\/2442\/revisions\/7320"}],"wp:attachment":[{"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/media?parent=2442"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/categories?post=2442"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/physicscatalyst.com\/article\/wp-json\/wp\/v2\/tags?post=2442"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}