{"id":259032,"date":"2025-07-18T16:34:15","date_gmt":"2025-07-18T16:34:15","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=259032"},"modified":"2025-07-18T17:49:13","modified_gmt":"2025-07-18T17:49:13","slug":"the-gravitational-force-between-the-sun-mass-1-99-x-1030-kg-and-mercury-mass-3-30-x-1023-kg-is-8-99-x-1021-n","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/18\/the-gravitational-force-between-the-sun-mass-1-99-x-1030-kg-and-mercury-mass-3-30-x-1023-kg-is-8-99-x-1021-n\/","title":{"rendered":"The gravitational force between the Sun (mass = 1.99 \u00d7 1030 kg) and Mercury (mass = 3.30 \u00d7 1023 kg) is 8.99 \u00d7 1021 N"},"content":{"rendered":"\n<p>The gravitational force between the Sun (mass = 1.99 \u00d7 1030 kg) and Mercury (mass = 3.30 \u00d7 1023 kg) is 8.99 \u00d7 1021 N. How far is Mercury from the Sun? 6.98 \u00d7 1010 km 6.98 \u00d7 107 km 4.87 \u00d7 1013 km 4.87 \u00d7 1024 km<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To solve for the distance between Mercury and the Sun, we can use <strong>Newton&#8217;s Law of Universal Gravitation<\/strong>, which is given by:F=Gm1m2r2F = G \\frac{m_1 m_2}{r^2}F=Gr2m1\u200bm2\u200b\u200b<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>FFF is the gravitational force between the two objects (in this case, the Sun and Mercury),<\/li>\n\n\n\n<li>GGG is the gravitational constant, which is approximately 6.674\u00d710\u221211\u2009N\u22c5m2\/kg26.674 \\times 10^{-11} \\, \\text{N} \\cdot \\text{m}^2\/\\text{kg}^26.674\u00d710\u221211N\u22c5m2\/kg2,<\/li>\n\n\n\n<li>m1m_1m1\u200b is the mass of the first object (the Sun),<\/li>\n\n\n\n<li>m2m_2m2\u200b is the mass of the second object (Mercury),<\/li>\n\n\n\n<li>rrr is the distance between the two objects.<\/li>\n<\/ul>\n\n\n\n<p>From the problem, we know the following values:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The gravitational force, F=8.99\u00d71021\u2009NF = 8.99 \\times 10^{21} \\, \\text{N}F=8.99\u00d71021N,<\/li>\n\n\n\n<li>The mass of the Sun, m1=1.99\u00d71030\u2009kgm_1 = 1.99 \\times 10^{30} \\, \\text{kg}m1\u200b=1.99\u00d71030kg,<\/li>\n\n\n\n<li>The mass of Mercury, m2=3.30\u00d71023\u2009kgm_2 = 3.30 \\times 10^{23} \\, \\text{kg}m2\u200b=3.30\u00d71023kg.<\/li>\n<\/ul>\n\n\n\n<p>We need to solve for rrr, the distance between Mercury and the Sun.<\/p>\n\n\n\n<p>Rearranging the formula to solve for rrr:r=Gm1m2Fr = \\sqrt{ \\frac{G m_1 m_2}{F} }r=FGm1\u200bm2\u200b\u200b\u200b<\/p>\n\n\n\n<p>Substituting the known values:r=(6.674\u00d710\u221211)\u00d7(1.99\u00d71030)\u00d7(3.30\u00d71023)8.99\u00d71021r = \\sqrt{ \\frac{(6.674 \\times 10^{-11}) \\times (1.99 \\times 10^{30}) \\times (3.30 \\times 10^{23})}{8.99 \\times 10^{21}} }r=8.99\u00d71021(6.674\u00d710\u221211)\u00d7(1.99\u00d71030)\u00d7(3.30\u00d71023)\u200b\u200b<\/p>\n\n\n\n<p>First, calculate the numerator:(6.674\u00d710\u221211)\u00d7(1.99\u00d71030)=1.327\u00d71020(6.674 \\times 10^{-11}) \\times (1.99 \\times 10^{30}) = 1.327 \\times 10^{20}(6.674\u00d710\u221211)\u00d7(1.99\u00d71030)=1.327\u00d71020<\/p>\n\n\n\n<p>Then,(1.327\u00d71020)\u00d7(3.30\u00d71023)=4.379\u00d71043(1.327 \\times 10^{20}) \\times (3.30 \\times 10^{23}) = 4.379 \\times 10^{43}(1.327\u00d71020)\u00d7(3.30\u00d71023)=4.379\u00d71043<\/p>\n\n\n\n<p>Now divide by the gravitational force:4.379\u00d710438.99\u00d71021=4.87\u00d71022\\frac{4.379 \\times 10^{43}}{8.99 \\times 10^{21}} = 4.87 \\times 10^{22}8.99\u00d710214.379\u00d71043\u200b=4.87\u00d71022<\/p>\n\n\n\n<p>Finally, take the square root:r=4.87\u00d71022\u22486.98\u00d71010\u2009mr = \\sqrt{4.87 \\times 10^{22}} \\approx 6.98 \\times 10^{10} \\, \\text{m}r=4.87\u00d71022\u200b\u22486.98\u00d71010m<\/p>\n\n\n\n<p>To convert meters to kilometers, divide by 100010001000:r\u22486.98\u00d7107\u2009kmr \\approx 6.98 \\times 10^7 \\, \\text{km}r\u22486.98\u00d7107km<\/p>\n\n\n\n<p>Thus, the distance between Mercury and the Sun is approximately <strong>6.98 \u00d7 10^7 km<\/strong>.<\/p>\n\n\n\n<p>So, the correct answer is: <strong>6.98 \u00d7 10^7 km<\/strong>.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner6-913.jpeg\" alt=\"\" class=\"wp-image-259033\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>The gravitational force between the Sun (mass = 1.99 \u00d7 1030 kg) and Mercury (mass = 3.30 \u00d7 1023 kg) is 8.99 \u00d7 1021 N. How far is Mercury from the Sun? 6.98 \u00d7 1010 km 6.98 \u00d7 107 km 4.87 \u00d7 1013 km 4.87 \u00d7 1024 km The Correct Answer and Explanation is: To [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"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":"","ast-disable-related-posts":"","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":[25],"tags":[],"class_list":["post-259032","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/259032","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=259032"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/259032\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=259032"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=259032"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=259032"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}