{"id":89402,"date":"2025-06-01T10:03:50","date_gmt":"2025-06-01T10:03:50","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=89402"},"modified":"2025-06-01T10:03:50","modified_gmt":"2025-06-01T10:03:50","slug":"the-acceleration-due-to-gravity-on-the-surface-of-the-earth-is-maximum","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/the-acceleration-due-to-gravity-on-the-surface-of-the-earth-is-maximum\/","title":{"rendered":"The acceleration due to gravity on the surface of the Earth is maximum"},"content":{"rendered":"

The acceleration due to gravity on the surface of the Earth is maximum and it<\/p>\n

[amp_mcq option1=”increases as we go up” option2=”decreases as we go up or down” option3=”increases as we go down” option4=”neither increases nor decreases as we go up or down” correct=”option2″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC CAPF – 2011<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe acceleration due to gravity on the surface of the Earth is maximum, and it decreases as we go up or down.
\n<\/section>\n
\nThe acceleration due to gravity (g) at a distance r from the center of the Earth (mass M) is given by g = GM\/r\u00b2. On the surface, r is the Earth’s radius (R), so g_surface = GM\/R\u00b2. As we go up, r increases (r = R + altitude), so g decreases according to the inverse square law. As we go down into the Earth, the mass (M) pulling us decreases (only the mass within the sphere of radius r contributes to the gravitational force at radius r, assuming uniform density for simplicity, though density varies in reality), while the distance from the center (r) decreases. The effect of decreasing mass outweighs the effect of decreasing distance, causing gravity to decrease linearly towards the center (g = GM_r\/r\u00b2, where M_r is the mass within radius r). Gravity is zero at the Earth’s center.
\n<\/section>\n
\nTherefore, gravity is maximum at the surface and decreases both above and below the surface. Minor variations exist due to altitude, latitude (Earth’s bulge), and local geological variations, but the general trend is decrease away from the surface.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

The acceleration due to gravity on the surface of the Earth is maximum and it [amp_mcq option1=”increases as we go up” option2=”decreases as we go up or down” option3=”increases as we go down” option4=”neither increases nor decreases as we go up or down” correct=”option2″] This question was previously asked in UPSC CAPF – 2011 Download … <\/p>\n

Detailed SolutionThe acceleration due to gravity on the surface of the Earth is maximum<\/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":[1085],"tags":[1465,1421,1128],"class_list":["post-89402","post","type-post","status-publish","format-standard","hentry","category-upsc-capf","tag-1465","tag-motion-under-gravity","tag-physics","no-featured-image-padding"],"yoast_head":"\nThe acceleration due to gravity on the surface of the Earth is maximum<\/title>\n<meta name=\"description\" content=\"The acceleration due to gravity on the surface of the Earth is maximum, and it decreases as we go up or down. The acceleration due to gravity (g) at a distance r from the center of the Earth (mass M) is given by g = GM\/r\u00b2. On the surface, r is the Earth's radius (R), so g_surface = GM\/R\u00b2. As we go up, r increases (r = R + altitude), so g decreases according to the inverse square law. As we go down into the Earth, the mass (M) pulling us decreases (only the mass within the sphere of radius r contributes to the gravitational force at radius r, assuming uniform density for simplicity, though density varies in reality), while the distance from the center (r) decreases. The effect of decreasing mass outweighs the effect of decreasing distance, causing gravity to decrease linearly towards the center (g = GM_r\/r\u00b2, where M_r is the mass within radius r). Gravity is zero at the Earth's center.\" \/>\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-acceleration-due-to-gravity-on-the-surface-of-the-earth-is-maximum\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"The acceleration due to gravity on the surface of the Earth is maximum\" \/>\n<meta property=\"og:description\" content=\"The acceleration due to gravity on the surface of the Earth is maximum, and it decreases as we go up or down. The acceleration due to gravity (g) at a distance r from the center of the Earth (mass M) is given by g = GM\/r\u00b2. On the surface, r is the Earth's radius (R), so g_surface = GM\/R\u00b2. As we go up, r increases (r = R + altitude), so g decreases according to the inverse square law. As we go down into the Earth, the mass (M) pulling us decreases (only the mass within the sphere of radius r contributes to the gravitational force at radius r, assuming uniform density for simplicity, though density varies in reality), while the distance from the center (r) decreases. The effect of decreasing mass outweighs the effect of decreasing distance, causing gravity to decrease linearly towards the center (g = GM_r\/r\u00b2, where M_r is the mass within radius r). Gravity is zero at the Earth's center.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/the-acceleration-due-to-gravity-on-the-surface-of-the-earth-is-maximum\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T10:03:50+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 acceleration due to gravity on the surface of the Earth is maximum","description":"The acceleration due to gravity on the surface of the Earth is maximum, and it decreases as we go up or down. The acceleration due to gravity (g) at a distance r from the center of the Earth (mass M) is given by g = GM\/r\u00b2. On the surface, r is the Earth's radius (R), so g_surface = GM\/R\u00b2. As we go up, r increases (r = R + altitude), so g decreases according to the inverse square law. As we go down into the Earth, the mass (M) pulling us decreases (only the mass within the sphere of radius r contributes to the gravitational force at radius r, assuming uniform density for simplicity, though density varies in reality), while the distance from the center (r) decreases. The effect of decreasing mass outweighs the effect of decreasing distance, causing gravity to decrease linearly towards the center (g = GM_r\/r\u00b2, where M_r is the mass within radius r). Gravity is zero at the Earth's center.","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-acceleration-due-to-gravity-on-the-surface-of-the-earth-is-maximum\/","og_locale":"en_US","og_type":"article","og_title":"The acceleration due to gravity on the surface of the Earth is maximum","og_description":"The acceleration due to gravity on the surface of the Earth is maximum, and it decreases as we go up or down. The acceleration due to gravity (g) at a distance r from the center of the Earth (mass M) is given by g = GM\/r\u00b2. On the surface, r is the Earth's radius (R), so g_surface = GM\/R\u00b2. As we go up, r increases (r = R + altitude), so g decreases according to the inverse square law. As we go down into the Earth, the mass (M) pulling us decreases (only the mass within the sphere of radius r contributes to the gravitational force at radius r, assuming uniform density for simplicity, though density varies in reality), while the distance from the center (r) decreases. The effect of decreasing mass outweighs the effect of decreasing distance, causing gravity to decrease linearly towards the center (g = GM_r\/r\u00b2, where M_r is the mass within radius r). Gravity is zero at the Earth's center.","og_url":"https:\/\/exam.pscnotes.com\/mcq\/the-acceleration-due-to-gravity-on-the-surface-of-the-earth-is-maximum\/","og_site_name":"MCQ and Quiz for Exams","article_published_time":"2025-06-01T10:03:50+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-acceleration-due-to-gravity-on-the-surface-of-the-earth-is-maximum\/","url":"https:\/\/exam.pscnotes.com\/mcq\/the-acceleration-due-to-gravity-on-the-surface-of-the-earth-is-maximum\/","name":"The acceleration due to gravity on the surface of the Earth is maximum","isPartOf":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#website"},"datePublished":"2025-06-01T10:03:50+00:00","dateModified":"2025-06-01T10:03:50+00:00","author":{"@id":"https:\/\/exam.pscnotes.com\/mcq\/#\/schema\/person\/5807dafeb27d2ec82344d6cbd6c3d209"},"description":"The acceleration due to gravity on the surface of the Earth is maximum, and it decreases as we go up or down. The acceleration due to gravity (g) at a distance r from the center of the Earth (mass M) is given by g = GM\/r\u00b2. On the surface, r is the Earth's radius (R), so g_surface = GM\/R\u00b2. As we go up, r increases (r = R + altitude), so g decreases according to the inverse square law. As we go down into the Earth, the mass (M) pulling us decreases (only the mass within the sphere of radius r contributes to the gravitational force at radius r, assuming uniform density for simplicity, though density varies in reality), while the distance from the center (r) decreases. The effect of decreasing mass outweighs the effect of decreasing distance, causing gravity to decrease linearly towards the center (g = GM_r\/r\u00b2, where M_r is the mass within radius r). 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