{"id":2089,"date":"2022-11-22T16:00:03","date_gmt":"2022-11-22T10:30:03","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=2089"},"modified":"2022-11-22T16:49:02","modified_gmt":"2022-11-22T11:19:02","slug":"earths-gravitational-force","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/earths-gravitational-force\/","title":{"rendered":"What Is Earth&#8217;s Gravitational Force"},"content":{"rendered":"<h2><strong>Earth&#8217;s Gravitational Force<\/strong><\/h2>\n<ul>\n<li>The <a href=\"https:\/\/www.aplustopper.com\/force\/\">force<\/a> which earth exerts on a body is called &#8216;force of gravity&#8217;. i.e.<br \/>\n\\(F=\\frac{GMm}{{{R}^{2}}}\\)<br \/>\nWhere M = mass of the earth, R = radius of the earth.<\/li>\n<li>Due to this force, a body released from some height on the earth&#8217;s surface falls towards the earth with its <a href=\"https:\/\/www.aplustopper.com\/difference-between-speed-and-velocity\/\">velocity<\/a> increasing at a constant rate.<\/li>\n<\/ul>\n<p><strong>Acceleration due to Gravity:<\/strong><\/p>\n<ul>\n<li>The acceleration produced in a body due to attraction of earth is called the <a href=\"https:\/\/www.aplustopper.com\/acceleration-and-types-of-acceleration\/\">acceleration<\/a> due to gravity and is denoted by &#8216;g&#8217;.<\/li>\n<li>\\( g=\\frac{GM}{{{R}^{2}}}=9.8\\text{ m\/}{{\\text{s}}^{\\text{2}}} \\\\ \\)<br \/>\nnear the earth surface<\/li>\n<li>\\( \\text{g on moon }\\!\\!~\\!\\!\\text{ }\\!\\!~\\!\\!\\text{ }\\approx \\frac{{{g}_{e}}}{6}=\\frac{9.8}{6}\\text{m\/}{{\\text{s}}^{\\text{2}}} \\)<\/li>\n<li>A body moving upwards with some initial velocity, experiences a retardation of\u00a09.8 m\/s<sup>2<\/sup> &amp; its velocity decreases continuously unless it becomes zero.<\/li>\n<li>After this, it again starts falling towards the earth with the same acceleration of 9.8 m\/s<sup>2<\/sup>.<\/li>\n<li>The value of g is minimum at equator and maximum at poles.<\/li>\n<li>The value of g does not depend upon the mass of the body falling towards the earth.<\/li>\n<\/ul>\n<h3><strong>Variation In The Value Of Gravitational Acceleration (g)<\/strong><\/h3>\n<p><strong>(A) Variation with altitude or height:<\/strong><\/p>\n<ul>\n<li>When a body moves above the earth&#8217;s surface the distance of the body from the centre of earth increases there by decreasing the force of attraction.<\/li>\n<li>At the earth&#8217;s surface<br \/>\n\\( \\text{g}=\\frac{\\text{GM}}{{{\\text{R}}^{\\text{2}}}} \\)<\/li>\n<li>At a height h above the earth&#8217;s surface<br \/>\n\\( \\text{g}=\\frac{\\text{GM}}{{{\\text{(R + h)}}^{\\text{2}}}} \\)<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-2101 size-full\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/Variation-with-altitude-or-height.png\" alt=\"Variation-with-altitude-or-height\" width=\"110\" height=\"181\" \/><\/li>\n<li>As we go above the earth&#8217;s surface the value of g goes on decreasing.<\/li>\n<\/ul>\n<p><strong>(B) Variation with depth d:<\/strong><\/p>\n<ul>\n<li>As we go deeper inside the earth, the body gets attracted by the core of the earth which is smaller in mass.<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-2102\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/09\/Variation-with-depth-d.png\" alt=\"Variation-with-depth-d\" width=\"160\" height=\"156\" \/><\/li>\n<li>As we go inside the earth, the value of g decreases.<\/li>\n<li>Force of attraction decreases and thus decreasing the value of g and becoming zero at the centre.<\/li>\n<\/ul>\n<p><strong>(C) Variation due to rotation of the earth:<\/strong><br \/>\nDue to the rotation of the earth, the weight of a body is maximum at the poles and minimum at the equator.<\/p>\n<h3><strong>Difference Between Mass and Weight<\/strong><\/h3>\n<table style=\"height: 572px;\" border=\"2\" width=\"662\">\n<tbody>\n<tr>\n<td width=\"44\"><strong>\u00a0<\/strong><\/td>\n<td width=\"214\">\n<p style=\"text-align: center;\"><strong>Mass<\/strong><\/p>\n<\/td>\n<td width=\"248\">\n<p style=\"text-align: center;\"><strong>Weight<\/strong><\/p>\n<\/td>\n<\/tr>\n<tr>\n<td width=\"44\">\n<p style=\"text-align: center;\"><strong>1.<\/strong><\/p>\n<\/td>\n<td width=\"214\">Mass of a body is defined as the quantity of matter contained in it.<\/td>\n<td width=\"248\">Weight of a body is the force with which it is attracted towards the centre of the earth.<br \/>\nW = mg<\/td>\n<\/tr>\n<tr>\n<td width=\"44\">\n<p style=\"text-align: center;\"><strong>2.<\/strong><\/p>\n<\/td>\n<td width=\"214\">Mass of a body\u00a0 remains constant and does not change from place to place.<\/td>\n<td width=\"248\">Weight of a body changes from place to place. It depends upon the value of <em>g<\/em>. Weight of a body on another planet will be different.<\/td>\n<\/tr>\n<tr>\n<td width=\"44\">\n<p style=\"text-align: center;\"><strong>3.<\/strong><\/p>\n<\/td>\n<td width=\"214\">Mass is measured by a pan balance.<\/td>\n<td width=\"248\">Weight is measured by a spring balance.<\/td>\n<\/tr>\n<tr>\n<td width=\"44\">\n<p style=\"text-align: center;\"><strong>4.<\/strong><\/p>\n<\/td>\n<td width=\"214\">Unit of mass is kg.<\/td>\n<td width=\"248\">Unit of weight is newton or kg-wt.<\/td>\n<\/tr>\n<tr>\n<td width=\"44\">\n<p style=\"text-align: center;\"><strong>5.<\/strong><\/p>\n<\/td>\n<td width=\"214\">Mass of a body cannot be zero.<\/td>\n<td width=\"248\">Weight of a body can be zero.<\/p>\n<p>Ex. astronauts experience weightlessness in\u00a0spaceships.<\/td>\n<\/tr>\n<tr>\n<td width=\"44\">\n<p style=\"text-align: center;\"><strong>6.<\/strong><\/p>\n<\/td>\n<td width=\"214\">Mass is a scalar quantity.<\/td>\n<td width=\"248\">Weight is a vector quantity.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2><strong>Earth&#8217;s Gravitational Force Example Problems With Solutions<\/strong><\/h2>\n<p><strong>Example 1:<\/strong> \u00a0 \u00a0Given mass of earth is 6 \u00d7 10<sup>24<\/sup> kg and mean radius of earth is 6.4 \u00d7 10<sup>6<\/sup> m. Calculate the value of acceleration due to gravity (g) on the surface of the earth.<br \/>\n<strong>Solution:<\/strong>\u00a0 \u00a0 The formula for the acceleration due to gravity is given by<br \/>\n\\( \\text{g}=\\frac{\\text{GM}}{{{\\text{R}}^{\\text{2}}}} \\)<br \/>\nHere, G = 6.67 \u00d7 10\u201311 Nm<sup>2<\/sup>\/kg<sup>2<\/sup>;<br \/>\nM = mass of earth = 6 \u00d7 10<sup>24<\/sup> kg;<br \/>\nR = radius of earth = 6.4 \u00d7 10<sup>6<\/sup> m<br \/>\n\\(g=\\frac{6.67\\times {{10}^{-11}}\\times 6\\times {{10}^{24}}}{{{(6.4\\times {{10}^{6}})}^{2}}}\\)<br \/>\ng = 9.8 m\/s<sup>2<\/sup><\/p>\n<p><strong>Example 2:<\/strong> \u00a0 \u00a0Calculate the value of acceleration due to gravity on a planet whose mass is 4 times as that of the earth and radius is 3 times as that of the earth.<br \/>\n<strong>Solution:<\/strong>\u00a0 \u00a0 If M is the mass of the earth and R is the radius of earth, the value of acceleration due to gravity on the earth (g<sub>e<\/sub>) is given by<br \/>\n\\( {{g}_{e}}=\\frac{GM}{{{R}^{2}}}\\text{ }&#8230;..\\text{ (1)} \\)<br \/>\nLet us consider a planet such that mass of the planet is equal to 4 times the mass of earth. Mp = 4M<br \/>\nRadius of the planet is equal to 3 times the radius of earth.<br \/>\nR<sub>e<\/sub> = 3R<br \/>\nThen, acceleration due to gravity on this planet(g<sub>p<\/sub>) is<br \/>\n\\( {{g}_{e}}=\\frac{G\\times (4M)}{{{(3R)}^{2}}}=\\frac{4}{9}.\\frac{GM}{{{R}^{2}}}\\text{ }&#8230;..\\text{ (2)} \\)<br \/>\nDividing equation (2) by equation (1), we get<br \/>\n\\( \\frac{{{g}_{p}}}{{{g}_{e}}}=\\frac{\\frac{4}{9}\\times \\frac{GM}{{{R}^{2}}}}{\\frac{GM}{{{R}^{2}}}} \\)<br \/>\n\\( \\frac{{{g}_{p}}}{{{g}_{e}}}=\\frac{4}{9} \\)<br \/>\n\\( {{g}_{p}}=\\frac{4}{9}{{g}_{e}} \\)<br \/>\nSince g<sub>e<\/sub> = 9.8 m\/s<sup>2<\/sup><br \/>\n\\( {{g}_{p}}=\\frac{4}{5}\\times 9.8=4.35\\text{ m\/}{{\\text{s}}^{\\text{2}}} \\)<br \/>\nThus, acceleration due to gravity on the given planet is 4.35 m\/s<sup>2<\/sup>.<\/p>\n<p><strong>Example 3:<\/strong> \u00a0 \u00a0Given the mass of the moon = 7.35 \u00d7 10<sup>22<\/sup> kg and the radius of the moon = 1740 km. Calculate the acceleration experienced by a particle on the surface of the moon due to the gravitational force of the moon. Find the ratio of this acceleration to that experienced by the same particle on the surface of the earth.<br \/>\n<strong>Solution:<\/strong>\u00a0 \u00a0 If Mm is the mass of the moon and Rm is its radius, then the acceleration experienced by a body on its surface is given by<br \/>\n\\( a=\\frac{G{{M}_{m}}}{R_{m}^{2}} \\)<br \/>\nHere, M<sub>m<\/sub> = 7.3 \u00d7 10<sup>22<\/sup> kg;<br \/>\nR<sub>m<\/sub> = 1740 km = 1.74 \u00d7 10<sup>6<\/sup> m<br \/>\n\\( \\therefore a=\\frac{6.67\\times {{10}^{-11}}\\times 7.3\\times {{10}^{22}}}{{{(1.74\\times {{10}^{6}})}^{2}}}=1.57\\text{ m\/}{{\\text{s}}^{\\text{2}}} \\)<br \/>\nWhile the acceleration due to gravity on the surface of the earth, is given by<br \/>\n\\( {{g}_{e}}=\\frac{G{{M}_{e}}}{R_{e}^{2}}=\\frac{6.67\\times {{10}^{-11}}\\times 6\\times {{10}^{24}}}{{{(6.4\\,\\times \\,{{10}^{6}})}^{2}}}=9.8\\text{ }m\/{{s}^{2}} \\)<br \/>\nComparing acceleration due to gravity on moon to that on the earth is<br \/>\n\\( \\frac{a}{g}=\\frac{1.57}{9.8}=0.16 \\)<\/p>\n<p><strong>Example 4:\u00a0<\/strong> \u00a0 At what height above the earth&#8217;s surface the value of g will be half of that on the earth&#8217;s surface ?<br \/>\n<strong>Solution:<\/strong>\u00a0 \u00a0 We know that the value of g at earth&#8217;s surface is<br \/>\n\\( g=\\frac{GM}{{{R}^{2}}}\\text{ }&#8230;..\\text{ (1)} \\)<br \/>\nWhile the value of g at a height h above the earth&#8217;s surface is given by<br \/>\n\\( g\\acute{\\ }=\\frac{GM}{{{(R+h)}^{2}}}\\text{ }&#8230;..\\text{ (2)} \\)<br \/>\nDividing equation (2) by equation (1), we get<br \/>\n\\( \\frac{g&#8217;}{g}={{\\left( \\frac{R}{R+h} \\right)}^{2}}\\text{ or }g\\acute{\\ }=g{{\\left( \\frac{R}{R+h} \\right)}^{2}} \\)<br \/>\n\\( \\text{Here},\\text{ }g\\acute{\\ }=~\\frac{g}{2} \\)<br \/>\n\\( \\therefore \\text{ }\\frac{g}{2}=g{{\\left( \\frac{R}{R+h} \\right)}^{2}} \\)<br \/>\n\\( \\frac{R+h}{R}=\\sqrt{2}R \\)<br \/>\nor \u00a0 \u00a0R + h =\u00a0\u221a2 R<br \/>\nor \u00a0 \u00a0h = ( \u221a2 \u2013 1)R<br \/>\nor\u00a0 \u00a0\u00a0h = (1.41 \u2013 1) \u00d7 6400 = 0.41 \u00d7 6400<br \/>\n= 2624 km<\/p>\n<p><strong>Example 5:<\/strong> \u00a0 \u00a0Given mass of the planet Mars is 6 \u00d7 10<sup>23<\/sup> kg and radius is 4.3 \u00d7 10<sup>6<\/sup> m. Calculate the weight of a man whose weight on earth is 600 N. (Given g on earth = 10 m\/s<sup>2<\/sup>)<br \/>\n<strong>Solution:<\/strong>\u00a0 \u00a0 Weight of the man on earth, W = mg<br \/>\nor 600 = m \u00d7 10 or m = 60 kg<br \/>\nSo the mass of the man is 60 kg which will remain the same everywhere.<br \/>\nNow acceleration due to gravity on Mars,<br \/>\n\\( {{g}_{m}}=\\frac{G{{M}_{m}}}{R_{m}^{2}} \\)<br \/>\nHere, M<sub>m<\/sub> = 6 \u00d7 10<sup>23<\/sup> kg; R<sub>m<\/sub> = 4.3 \u00d7 10<sup>6<\/sup>m<br \/>\n\\( \\therefore \\text{ }{{g}_{m}}=\\frac{6.67\\times {{10}^{-11}}\\times 6\\times {{10}^{23}}}{{{(4.3\\times {{10}^{6}})}^{2}}} \\)<br \/>\ng<sub>m<\/sub> = 2.17 m\/s<sup>2<\/sup><br \/>\nNow, weight of the man on Mars will be<br \/>\nW<sub>m<\/sub> = m \u00d7 g<sub>m<\/sub> = 60 \u00d7 2.17 = 130.2 N<\/p>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Earth&#8217;s Gravitational Force The force which earth exerts on a body is called &#8216;force of gravity&#8217;. i.e. Where M = mass of the earth, R = radius of the earth. Due to this force, a body released from some height on the earth&#8217;s surface falls towards the earth with its velocity increasing at a constant [&hellip;]<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[404],"tags":[857,856,838,858,860,861,859],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v17.8 (Yoast SEO v17.8) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>What Is Earth&#039;s Gravitational Force - A Plus Topper<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.aplustopper.com\/earths-gravitational-force\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"What Is Earth&#039;s Gravitational Force\" \/>\n<meta property=\"og:description\" content=\"Earth&#8217;s Gravitational Force The force which earth exerts on a body is called &#8216;force of gravity&#8217;. i.e. Where M = mass of the earth, R = radius of the earth. 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