{"id":88064,"date":"2025-06-01T07:00:45","date_gmt":"2025-06-01T07:00:45","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=88064"},"modified":"2025-06-01T07:00:45","modified_gmt":"2025-06-01T07:00:45","slug":"escape-speed-from-the-earth-is-close-to-11-2-km-s%e2%81%bb%c2%b9-on-another-planet","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/escape-speed-from-the-earth-is-close-to-11-2-km-s%e2%81%bb%c2%b9-on-another-planet\/","title":{"rendered":"Escape speed from the Earth is close to 11.2 km s\u207b\u00b9. On another planet"},"content":{"rendered":"

Escape speed from the Earth is close to 11.2 km s\u207b\u00b9. On another planet whose radius is half of the Earth’s radius and whose mass density is four times that of the Earth, the escape speed in km s\u207b\u00b9 will be close to :<\/p>\n

[amp_mcq option1=”11.2″ option2=”15.8″ option3=”5.6″ option4=”7.9″ correct=”option1″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC NDA-1 – 2024<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe correct option is A. The escape speed from the new planet will be close to 11.2 km s\u207b\u00b9, the same as Earth’s.
\n<\/section>\n
\nThe escape speed (v_e) from a spherical body of mass M and radius R is given by v_e = sqrt(2GM\/R). The mass M can be expressed in terms of density (rho) and volume (V = 4\/3 * pi * R^3) as M = rho * (4\/3 * pi * R^3). Substituting this into the escape speed formula: v_e = sqrt(2G * (rho * 4\/3 * pi * R^3) \/ R) = sqrt(8\/3 * pi * G * rho * R\u00b2) = R * sqrt(8\/3 * pi * G * rho). This shows that v_e is proportional to R * sqrt(rho).
\n<\/section>\n
\nLet Earth’s radius, density, and escape speed be R_e, rho_e, and v_e_e respectively. So, v_e_e \u221d R_e * sqrt(rho_e) = 11.2 km\/s. For the new planet, R_p = R_e \/ 2 and rho_p = 4 * rho_e. The escape speed from the new planet is v_e_p \u221d R_p * sqrt(rho_p) = (R_e \/ 2) * sqrt(4 * rho_e) = (R_e \/ 2) * 2 * sqrt(rho_e) = R_e * sqrt(rho_e). Thus, v_e_p is proportional to the same value as v_e_e, meaning v_e_p = v_e_e = 11.2 km\/s.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

Escape speed from the Earth is close to 11.2 km s\u207b\u00b9. On another planet whose radius is half of the Earth’s radius and whose mass density is four times that of the Earth, the escape speed in km s\u207b\u00b9 will be close to : [amp_mcq option1=”11.2″ option2=”15.8″ option3=”5.6″ option4=”7.9″ correct=”option1″] This question was previously asked … <\/p>\n

Detailed SolutionEscape speed from the Earth is close to 11.2 km s\u207b\u00b9. On another planet<\/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":[1103,1129,1128],"class_list":["post-88064","post","type-post","status-publish","format-standard","hentry","category-upsc-nda-1","tag-1103","tag-mechanics","tag-physics","no-featured-image-padding"],"yoast_head":"\nEscape speed from the Earth is close to 11.2 km s\u207b\u00b9. On another planet<\/title>\n<meta name=\"description\" content=\"The correct option is A. The escape speed from the new planet will be close to 11.2 km s\u207b\u00b9, the same as Earth's. The escape speed (v_e) from a spherical body of mass M and radius R is given by v_e = sqrt(2GM\/R). The mass M can be expressed in terms of density (rho) and volume (V = 4\/3 * pi * R^3) as M = rho * (4\/3 * pi * R^3). Substituting this into the escape speed formula: v_e = sqrt(2G * (rho * 4\/3 * pi * R^3) \/ R) = sqrt(8\/3 * pi * G * rho * R\u00b2) = R * sqrt(8\/3 * pi * G * rho). This shows that v_e is proportional to R * sqrt(rho).\" \/>\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\/escape-speed-from-the-earth-is-close-to-11-2-km-s\u207b\u00b9-on-another-planet\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Escape speed from the Earth is close to 11.2 km s\u207b\u00b9. On another planet\" \/>\n<meta property=\"og:description\" content=\"The correct option is A. The escape speed from the new planet will be close to 11.2 km s\u207b\u00b9, the same as Earth's. The escape speed (v_e) from a spherical body of mass M and radius R is given by v_e = sqrt(2GM\/R). The mass M can be expressed in terms of density (rho) and volume (V = 4\/3 * pi * R^3) as M = rho * (4\/3 * pi * R^3). Substituting this into the escape speed formula: v_e = sqrt(2G * (rho * 4\/3 * pi * R^3) \/ R) = sqrt(8\/3 * pi * G * rho * R\u00b2) = R * sqrt(8\/3 * pi * G * rho). This shows that v_e is proportional to R * sqrt(rho).\" \/>\n<meta property=\"og:url\" content=\"https:\/\/exam.pscnotes.com\/mcq\/escape-speed-from-the-earth-is-close-to-11-2-km-s\u207b\u00b9-on-another-planet\/\" \/>\n<meta property=\"og:site_name\" content=\"MCQ and Quiz for Exams\" \/>\n<meta property=\"article:published_time\" content=\"2025-06-01T07:00:45+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":"Escape speed from the Earth is close to 11.2 km s\u207b\u00b9. On another planet","description":"The correct option is A. The escape speed from the new planet will be close to 11.2 km s\u207b\u00b9, the same as Earth's. The escape speed (v_e) from a spherical body of mass M and radius R is given by v_e = sqrt(2GM\/R). The mass M can be expressed in terms of density (rho) and volume (V = 4\/3 * pi * R^3) as M = rho * (4\/3 * pi * R^3). Substituting this into the escape speed formula: v_e = sqrt(2G * (rho * 4\/3 * pi * R^3) \/ R) = sqrt(8\/3 * pi * G * rho * R\u00b2) = R * sqrt(8\/3 * pi * G * rho). This shows that v_e is proportional to R * sqrt(rho).","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\/escape-speed-from-the-earth-is-close-to-11-2-km-s\u207b\u00b9-on-another-planet\/","og_locale":"en_US","og_type":"article","og_title":"Escape speed from the Earth is close to 11.2 km s\u207b\u00b9. On another planet","og_description":"The correct option is A. The escape speed from the new planet will be close to 11.2 km s\u207b\u00b9, the same as Earth's. The escape speed (v_e) from a spherical body of mass M and radius R is given by v_e = sqrt(2GM\/R). The mass M can be expressed in terms of density (rho) and volume (V = 4\/3 * pi * R^3) as M = rho * (4\/3 * pi * R^3). Substituting this into the escape speed formula: v_e = sqrt(2G * (rho * 4\/3 * pi * R^3) \/ R) = sqrt(8\/3 * pi * G * rho * R\u00b2) = R * sqrt(8\/3 * pi * G * rho). 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