{"id":88562,"date":"2025-06-01T07:14:53","date_gmt":"2025-06-01T07:14:53","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=88562"},"modified":"2025-06-01T07:14:53","modified_gmt":"2025-06-01T07:14:53","slug":"two-bodies-of-mass-m-each-are-placed-r-distance-apart-in-another-syst","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/two-bodies-of-mass-m-each-are-placed-r-distance-apart-in-another-syst\/","title":{"rendered":"Two bodies of mass M each are placed R distance apart. In another syst"},"content":{"rendered":"

Two bodies of mass M each are placed R distance apart. In another system, two bodies of mass 2M each are placed $\\frac{R}{2}$ distance apart. If F be the gravitational force between the bodies in the first system, then the gravitational force between the bodies in the second system will be<\/p>\n

[amp_mcq option1=”16 F” option2=”1 F” option3=”4 F” option4=”None of the above” correct=”option1″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-2 – 2019<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe correct answer is A) 16 F.
\n<\/section>\n
\nThe gravitational force between two bodies of masses $m_1$ and $m_2$ separated by a distance R is given by Newton’s Law of Gravitation: $F = G \\frac{m_1 m_2}{R^2}$, where G is the gravitational constant.
\nIn the first system: $m_1 = M$, $m_2 = M$, $R_1 = R$. The force is $F_1 = G \\frac{M \\times M}{R^2} = G \\frac{M^2}{R^2}$. This force is given as F. So, $F = G \\frac{M^2}{R^2}$.
\nIn the second system: $m_1′ = 2M$, $m_2′ = 2M$, $R_2 = R\/2$. The force is $F_2 = G \\frac{(2M) \\times (2M)}{(R\/2)^2}$.
\nCalculate $F_2$: $F_2 = G \\frac{4M^2}{R^2\/4} = G \\frac{4M^2}{R^2} \\times 4 = 16 G \\frac{M^2}{R^2}$.
\nSubstitute the expression for F: $F_2 = 16 F$.
\n<\/section>\n
\nThe gravitational force is directly proportional to the product of the masses and inversely proportional to the square of the distance between their centers. Doubling the masses quadruples the product of masses ($2M \\times 2M = 4M^2$). Halving the distance quarters the squared distance ($(R\/2)^2 = R^2\/4$), meaning the force is multiplied by 4 due to the inverse square law ($1 \/ (1\/4) = 4$). The combined effect is a multiplication by $4 \\times 4 = 16$.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

Two bodies of mass M each are placed R distance apart. In another system, two bodies of mass 2M each are placed $\\frac{R}{2}$ distance apart. If F be the gravitational force between the bodies in the first system, then the gravitational force between the bodies in the second system will be [amp_mcq option1=”16 F” option2=”1 … <\/p>\n

Detailed SolutionTwo bodies of mass M each are placed R distance apart. In another syst<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1094],"tags":[1119,1129,1128],"class_list":["post-88562","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-2","tag-1119","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nTwo bodies of mass M each are placed R distance apart. In another syst<\/title>\n<meta name=\"description\" content=\"The correct answer is A) 16 F. The gravitational force between two bodies of masses $m_1$ and $m_2$ separated by a distance R is given by Newton's Law of Gravitation: $F = G frac{m_1 m_2}{R^2}$, where G is the gravitational constant. In the first system: $m_1 = M$, $m_2 = M$, $R_1 = R$. The force is $F_1 = G frac{M times M}{R^2} = G frac{M^2}{R^2}$. This force is given as F. So, $F = G frac{M^2}{R^2}$. In the second system: $m_1' = 2M$, $m_2' = 2M$, $R_2 = R\/2$. The force is $F_2 = G frac{(2M) times (2M)}{(R\/2)^2}$. Calculate $F_2$: $F_2 = G frac{4M^2}{R^2\/4} = G frac{4M^2}{R^2} times 4 = 16 G frac{M^2}{R^2}$. Substitute the expression for F: $F_2 = 16 F$.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/two-bodies-of-mass-m-each-are-placed-r-distance-apart-in-another-syst\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Two bodies of mass M each are placed R distance apart. In another syst\" \/>\n<meta property=\"og:description\" content=\"The correct answer is A) 16 F. The gravitational force between two bodies of masses $m_1$ and $m_2$ separated by a distance R is given by Newton's Law of Gravitation: $F = G frac{m_1 m_2}{R^2}$, where G is the gravitational constant. In the first system: $m_1 = M$, $m_2 = M$, $R_1 = R$. The force is $F_1 = G frac{M times M}{R^2} = G frac{M^2}{R^2}$. This force is given as F. So, $F = G frac{M^2}{R^2}$. In the second system: $m_1' = 2M$, $m_2' = 2M$, $R_2 = R\/2$. The force is $F_2 = G frac{(2M) times (2M)}{(R\/2)^2}$. Calculate $F_2$: $F_2 = G frac{4M^2}{R^2\/4} = G frac{4M^2}{R^2} times 4 = 16 G frac{M^2}{R^2}$. Substitute the expression for F: $F_2 = 16 F$.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/two-bodies-of-mass-m-each-are-placed-r-distance-apart-in-another-syst\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T07:14:53+00:00\" \/>\n<meta name=\"author\" content=\"rawan239\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"rawan239\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"1 minute\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"Two bodies of mass M each are placed R distance apart. In another syst","description":"The correct answer is A) 16 F. The gravitational force between two bodies of masses $m_1$ and $m_2$ separated by a distance R is given by Newton's Law of Gravitation: $F = G frac{m_1 m_2}{R^2}$, where G is the gravitational constant. In the first system: $m_1 = M$, $m_2 = M$, $R_1 = R$. The force is $F_1 = G frac{M times M}{R^2} = G frac{M^2}{R^2}$. This force is given as F. So, $F = G frac{M^2}{R^2}$. In the second system: $m_1' = 2M$, $m_2' = 2M$, $R_2 = R\/2$. The force is $F_2 = G frac{(2M) times (2M)}{(R\/2)^2}$. Calculate $F_2$: $F_2 = G frac{4M^2}{R^2\/4} = G frac{4M^2}{R^2} times 4 = 16 G frac{M^2}{R^2}$. Substitute the expression for F: $F_2 = 16 F$.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/exam.pscnotes.com\/mcq\/two-bodies-of-mass-m-each-are-placed-r-distance-apart-in-another-syst\/","og_locale":"en_US","og_type":"article","og_title":"Two bodies of mass M each are placed R distance apart. In another syst","og_description":"The correct answer is A) 16 F. The gravitational force between two bodies of masses $m_1$ and $m_2$ separated by a distance R is given by Newton's Law of Gravitation: $F = G frac{m_1 m_2}{R^2}$, where G is the gravitational constant. In the first system: $m_1 = M$, $m_2 = M$, $R_1 = R$. The force is $F_1 = G frac{M times M}{R^2} = G frac{M^2}{R^2}$. This force is given as F. So, $F = G frac{M^2}{R^2}$. In the second system: $m_1' = 2M$, $m_2' = 2M$, $R_2 = R\/2$. The force is $F_2 = G frac{(2M) times (2M)}{(R\/2)^2}$. Calculate $F_2$: $F_2 = G frac{4M^2}{R^2\/4} = G frac{4M^2}{R^2} times 4 = 16 G frac{M^2}{R^2}$. Substitute the expression for F: $F_2 = 16 F$.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/two-bodies-of-mass-m-each-are-placed-r-distance-apart-in-another-syst\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T07:14:53+00:00","author":"rawan239","twitter_card":"summary_large_image","twitter_misc":{"Written by":"rawan239","Est. reading time":"1 minute"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/exam.pscnotes.com\/mcq\/two-bodies-of-mass-m-each-are-placed-r-distance-apart-in-another-syst\/","url":"https:\/\/exam.pscnotes.com\/mcq\/two-bodies-of-mass-m-each-are-placed-r-distance-apart-in-another-syst\/","name":"Two bodies of mass M each are placed R distance apart. In another syst","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T07:14:53+00:00","dateModified":"2025-06-01T07:14:53+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The correct answer is A) 16 F. The gravitational force between two bodies of masses $m_1$ and $m_2$ separated by a distance R is given by Newton's Law of Gravitation: $F = G \\frac{m_1 m_2}{R^2}$, where G is the gravitational constant. In the first system: $m_1 = M$, $m_2 = M$, $R_1 = R$. The force is $F_1 = G \\frac{M \\times M}{R^2} = G \\frac{M^2}{R^2}$. This force is given as F. So, $F = G \\frac{M^2}{R^2}$. In the second system: $m_1' = 2M$, $m_2' = 2M$, $R_2 = R\/2$. The force is $F_2 = G \\frac{(2M) \\times (2M)}{(R\/2)^2}$. Calculate $F_2$: $F_2 = G \\frac{4M^2}{R^2\/4} = G \\frac{4M^2}{R^2} \\times 4 = 16 G \\frac{M^2}{R^2}$. Substitute the expression for F: $F_2 = 16 F$.","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/two-bodies-of-mass-m-each-are-placed-r-distance-apart-in-another-syst\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/two-bodies-of-mass-m-each-are-placed-r-distance-apart-in-another-syst\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/two-bodies-of-mass-m-each-are-placed-r-distance-apart-in-another-syst\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC NDA-2","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-nda-2\/"},{"@type":"ListItem","position":3,"name":"Two bodies of mass M each are placed R distance apart. 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