{"id":87307,"date":"2025-06-01T06:39:50","date_gmt":"2025-06-01T06:39:50","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=87307"},"modified":"2025-06-01T06:39:50","modified_gmt":"2025-06-01T06:39:50","slug":"the-radius-of-the-moon-is-about-one-fourth-that-of-the-earth-and-accel","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/the-radius-of-the-moon-is-about-one-fourth-that-of-the-earth-and-accel\/","title":{"rendered":"The radius of the Moon is about one-fourth that of the Earth and accel"},"content":{"rendered":"

The radius of the Moon is about one-fourth that of the Earth and acceleration due to gravity on the Moon is about one-sixth that on the Earth. From this, we can conclude that the ratio of the mass of Earth to the mass of the Moon is about<\/p>\n

[amp_mcq option1=”10″ option2=”100″ option3=”1,000″ option4=”10,000″ correct=”option2″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-1 – 2015<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe acceleration due to gravity (g) on a celestial body with mass M and radius R is given by the formula g = GM\/R\u00b2, where G is the gravitational constant. We are given that the radius of the Moon (Rm) is about one-fourth that of the Earth (Re), i.e., Rm \u2248 Re\/4. We are also given that the acceleration due to gravity on the Moon (gm) is about one-sixth that on the Earth (ge), i.e., gm \u2248 ge\/6.
\nFrom g = GM\/R\u00b2, we can write M = gR\u00b2\/G.
\nThe ratio of the mass of Earth (Me) to the mass of the Moon (Mm) is:
\nMe\/Mm = (ge * Re\u00b2 \/ G) \/ (gm * Rm\u00b2 \/ G)
\nMe\/Mm = (ge\/gm) * (Re\/Rm)\u00b2
\nSubstitute the given ratios: ge\/gm \u2248 6 and Re\/Rm \u2248 4.
\nMe\/Mm \u2248 6 * (4)\u00b2 = 6 * 16 = 96.
\nAmong the options, 96 is closest to 100.
\n<\/section>\n
\nThe ratio of masses can be calculated using the relationship between acceleration due to gravity, mass, and radius (g = GM\/R\u00b2).
\n<\/section>\n
\nThe actual ratio of Earth’s mass to Moon’s mass is approximately 81.3, so the given approximate values lead to a result (96) that is closest to 100, indicating the question uses approximate figures typical for simplified calculations.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

The radius of the Moon is about one-fourth that of the Earth and acceleration due to gravity on the Moon is about one-sixth that on the Earth. From this, we can conclude that the ratio of the mass of Earth to the mass of the Moon is about [amp_mcq option1=”10″ option2=”100″ option3=”1,000″ option4=”10,000″ correct=”option2″] This … <\/p>\n

Detailed SolutionThe radius of the Moon is about one-fourth that of the Earth and accel<\/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":[1093],"tags":[1443,1129,1128],"class_list":["post-87307","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-1","tag-1443","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nThe radius of the Moon is about one-fourth that of the Earth and accel<\/title>\n<meta name=\"description\" content=\"The acceleration due to gravity (g) on a celestial body with mass M and radius R is given by the formula g = GM\/R\u00b2, where G is the gravitational constant. We are given that the radius of the Moon (Rm) is about one-fourth that of the Earth (Re), i.e., Rm \u2248 Re\/4. We are also given that the acceleration due to gravity on the Moon (gm) is about one-sixth that on the Earth (ge), i.e., gm \u2248 ge\/6. From g = GM\/R\u00b2, we can write M = gR\u00b2\/G. The ratio of the mass of Earth (Me) to the mass of the Moon (Mm) is: Me\/Mm = (ge * Re\u00b2 \/ G) \/ (gm * Rm\u00b2 \/ G) Me\/Mm = (ge\/gm) * (Re\/Rm)\u00b2 Substitute the given ratios: ge\/gm \u2248 6 and Re\/Rm \u2248 4. Me\/Mm \u2248 6 * (4)\u00b2 = 6 * 16 = 96. Among the options, 96 is closest to 100. The ratio of masses can be calculated using the relationship between acceleration due to gravity, mass, and radius (g = GM\/R\u00b2).\" \/>\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\/the-radius-of-the-moon-is-about-one-fourth-that-of-the-earth-and-accel\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"The radius of the Moon is about one-fourth that of the Earth and accel\" \/>\n<meta property=\"og:description\" content=\"The acceleration due to gravity (g) on a celestial body with mass M and radius R is given by the formula g = GM\/R\u00b2, where G is the gravitational constant. We are given that the radius of the Moon (Rm) is about one-fourth that of the Earth (Re), i.e., Rm \u2248 Re\/4. We are also given that the acceleration due to gravity on the Moon (gm) is about one-sixth that on the Earth (ge), i.e., gm \u2248 ge\/6. From g = GM\/R\u00b2, we can write M = gR\u00b2\/G. The ratio of the mass of Earth (Me) to the mass of the Moon (Mm) is: Me\/Mm = (ge * Re\u00b2 \/ G) \/ (gm * Rm\u00b2 \/ G) Me\/Mm = (ge\/gm) * (Re\/Rm)\u00b2 Substitute the given ratios: ge\/gm \u2248 6 and Re\/Rm \u2248 4. Me\/Mm \u2248 6 * (4)\u00b2 = 6 * 16 = 96. Among the options, 96 is closest to 100. The ratio of masses can be calculated using the relationship between acceleration due to gravity, mass, and radius (g = GM\/R\u00b2).\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/the-radius-of-the-moon-is-about-one-fourth-that-of-the-earth-and-accel\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T06:39:50+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":"The radius of the Moon is about one-fourth that of the Earth and accel","description":"The acceleration due to gravity (g) on a celestial body with mass M and radius R is given by the formula g = GM\/R\u00b2, where G is the gravitational constant. We are given that the radius of the Moon (Rm) is about one-fourth that of the Earth (Re), i.e., Rm \u2248 Re\/4. We are also given that the acceleration due to gravity on the Moon (gm) is about one-sixth that on the Earth (ge), i.e., gm \u2248 ge\/6. From g = GM\/R\u00b2, we can write M = gR\u00b2\/G. The ratio of the mass of Earth (Me) to the mass of the Moon (Mm) is: Me\/Mm = (ge * Re\u00b2 \/ G) \/ (gm * Rm\u00b2 \/ G) Me\/Mm = (ge\/gm) * (Re\/Rm)\u00b2 Substitute the given ratios: ge\/gm \u2248 6 and Re\/Rm \u2248 4. Me\/Mm \u2248 6 * (4)\u00b2 = 6 * 16 = 96. Among the options, 96 is closest to 100. The ratio of masses can be calculated using the relationship between acceleration due to gravity, mass, and radius (g = GM\/R\u00b2).","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\/the-radius-of-the-moon-is-about-one-fourth-that-of-the-earth-and-accel\/","og_locale":"en_US","og_type":"article","og_title":"The radius of the Moon is about one-fourth that of the Earth and accel","og_description":"The acceleration due to gravity (g) on a celestial body with mass M and radius R is given by the formula g = GM\/R\u00b2, where G is the gravitational constant. We are given that the radius of the Moon (Rm) is about one-fourth that of the Earth (Re), i.e., Rm \u2248 Re\/4. We are also given that the acceleration due to gravity on the Moon (gm) is about one-sixth that on the Earth (ge), i.e., gm \u2248 ge\/6. From g = GM\/R\u00b2, we can write M = gR\u00b2\/G. The ratio of the mass of Earth (Me) to the mass of the Moon (Mm) is: Me\/Mm = (ge * Re\u00b2 \/ G) \/ (gm * Rm\u00b2 \/ G) Me\/Mm = (ge\/gm) * (Re\/Rm)\u00b2 Substitute the given ratios: ge\/gm \u2248 6 and Re\/Rm \u2248 4. Me\/Mm \u2248 6 * (4)\u00b2 = 6 * 16 = 96. Among the options, 96 is closest to 100. The ratio of masses can be calculated using the relationship between acceleration due to gravity, mass, and radius (g = GM\/R\u00b2).","og_url":"https:\/\/exam.pscnotes.com\/mcq\/the-radius-of-the-moon-is-about-one-fourth-that-of-the-earth-and-accel\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T06:39:50+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\/the-radius-of-the-moon-is-about-one-fourth-that-of-the-earth-and-accel\/","url":"https:\/\/exam.pscnotes.com\/mcq\/the-radius-of-the-moon-is-about-one-fourth-that-of-the-earth-and-accel\/","name":"The radius of the Moon is about one-fourth that of the Earth and accel","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T06:39:50+00:00","dateModified":"2025-06-01T06:39:50+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The acceleration due to gravity (g) on a celestial body with mass M and radius R is given by the formula g = GM\/R\u00b2, where G is the gravitational constant. We are given that the radius of the Moon (Rm) is about one-fourth that of the Earth (Re), i.e., Rm \u2248 Re\/4. We are also given that the acceleration due to gravity on the Moon (gm) is about one-sixth that on the Earth (ge), i.e., gm \u2248 ge\/6. From g = GM\/R\u00b2, we can write M = gR\u00b2\/G. The ratio of the mass of Earth (Me) to the mass of the Moon (Mm) is: Me\/Mm = (ge * Re\u00b2 \/ G) \/ (gm * Rm\u00b2 \/ G) Me\/Mm = (ge\/gm) * (Re\/Rm)\u00b2 Substitute the given ratios: ge\/gm \u2248 6 and Re\/Rm \u2248 4. Me\/Mm \u2248 6 * (4)\u00b2 = 6 * 16 = 96. Among the options, 96 is closest to 100. The ratio of masses can be calculated using the relationship between acceleration due to gravity, mass, and radius (g = GM\/R\u00b2).","breadcrumb":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/the-radius-of-the-moon-is-about-one-fourth-that-of-the-earth-and-accel\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/exam.pscnotes.com\/mcq\/the-radius-of-the-moon-is-about-one-fourth-that-of-the-earth-and-accel\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/exam.pscnotes.com\/mcq\/the-radius-of-the-moon-is-about-one-fourth-that-of-the-earth-and-accel\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/exam.pscnotes.com\/mcq\/"},{"@type":"ListItem","position":2,"name":"UPSC NDA-1","item":"https:\/\/exam.pscnotes.com\/mcq\/category\/upsc-nda-1\/"},{"@type":"ListItem","position":3,"name":"The radius of the Moon is about one-fourth that of the Earth and accel"}]},{"@type":"WebSite","@id":"https:\/\/exam.pscnotes.com\/mcq\/#website","url":"https:\/\/exam.pscnotes.com\/mcq\/","name":"MCQ and Quiz for Exams","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/exam.pscnotes.com\/mcq\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209","name":"rawan239","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/image\/","url":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/761a7274f9cce048fa5b921221e7934820d74514df93ef195a9d22af0c1c9001?s=96&d=mm&r=g","caption":"rawan239"},"sameAs":["https:\/\/exam.pscnotes.com"],"url":"https:\/\/exam.pscnotes.com\/mcq\/author\/rawan239\/"}]}},"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/87307","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/comments?post=87307"}],"version-history":[{"count":0,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/posts\/87307\/revisions"}],"wp:attachment":[{"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/media?parent=87307"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/categories?post=87307"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/exam.pscnotes.com\/mcq\/wp-json\/wp\/v2\/tags?post=87307"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}