{"id":940,"date":"2018-02-20T18:10:09","date_gmt":"2018-02-20T10:10:09","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=940"},"modified":"2018-02-20T18:10:09","modified_gmt":"2018-02-20T10:10:09","slug":"keplers-law-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/keplers-law-problems-and-solutions.htm","title":{"rendered":"Kepler’s law \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

1. The Earth’s distance<\/a> from the Sun is 149.6 x 10<\/span><\/span>6<\/span><\/span><\/sup> km and period of Earth’s revolution is 1 year. Calculate T<\/span><\/span>2<\/span><\/span><\/sup> \/ r<\/span><\/span>3<\/span><\/span><\/sup><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

T = 1 year, r = 149.6 x 10<\/span><\/span>6<\/span><\/span><\/sup> km <\/span><\/span><\/p>\n

Wanted <\/u><\/span><\/span>: T<\/span><\/span>2<\/span><\/span><\/sup> \/ r<\/span><\/span>3<\/span><\/span><\/sup> = … ?<\/span><\/span><\/p>\n

Solution <\/u><\/span><\/span>:<\/span><\/span><\/p>\n

k = T<\/span><\/span>2<\/span><\/span><\/sup> \/ r<\/span><\/span>3 <\/span><\/span><\/sup>= 1<\/span><\/span>2<\/span><\/span><\/sup> \/ (149.6 x 10<\/span><\/span>6<\/span><\/span><\/sup>)<\/span><\/span>3<\/span><\/span><\/sup> = 1 \/ (3348071.9 x 10<\/span><\/span>18<\/span><\/span><\/sup>) = 2.98 x 10<\/span><\/span>-25 <\/span><\/span><\/sup>year<\/span><\/span>2<\/span><\/span><\/sup>\/km<\/span><\/span>3<\/span><\/span><\/sup><\/p>\n

<\/p>\n

2. <\/span><\/span>Universal constant (G) = 6.67 x 10<\/span><\/span>-11<\/span><\/span><\/sup> N.m<\/span><\/span>2<\/span><\/span><\/sup>\/kg<\/span><\/span>2 <\/span><\/span><\/sup>and Sun’s 1.99 x 10<\/span><\/span>30<\/span><\/span><\/sup> kg.<\/span><\/span><\/p>\n

\"Kepler's<\/p>\n

\"Kepler's<\/p>\n

<\/p>\n

3. The mean distance of Earth from the Sun is 149.6 x 10<\/span><\/span>6<\/span><\/span><\/sup> km and the mean distance of Mercury from the Sun is 57.9 x 10<\/span><\/span>6<\/span><\/span><\/sup> km. The period of Earth’s revolutions is 1 year, what is the period of Mercury’s revolution? <\/span><\/span><\/p>\n

Known :<\/u><\/span><\/span><\/p>\n

r of Earth = 149.6 x 10<\/span><\/span>6<\/span><\/span><\/sup> km<\/span><\/span><\/p>\n

r of mercury = 57.9 x 10<\/span><\/span>6<\/span><\/span><\/sup> km<\/span><\/span><\/p>\n

T of Earth = 1 year<\/span><\/span><\/p>\n

Wanted:<\/u><\/span><\/span> T of mercury?<\/span><\/span><\/p>\n

Solution :<\/u><\/span><\/span><\/p>\n

\"Kepler's<\/p>\n

[wpdm_package id=’953′]<\/p>\n

    \n
  1. Newton’s law of universal gravitation problems and solutions<\/a><\/li>\n
  2. Gravitational force, weight problems, and solutions<\/a><\/li>\n
  3. Acceleration due to gravity problems and solutions<\/a><\/li>\n
  4. Geosynchronous satellite problems and solutions<\/a><\/li>\n
  5. Kepler’s law problems and solutions problems and solutions<\/a><\/li>\n<\/ol>\n

    <\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

    1. The Earth’s distance from the Sun is 149.6 x 106 km and period of Earth’s revolution is 1 year. Calculate T2 \/ r3 Known : T = 1 year, r = 149.6 x 106 km Wanted : T2 \/ r3 = … ? Solution : k = T2 \/ r3 = 12 \/ (149.6 … Read more<\/a><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_seopress_titles_title":"","_seopress_titles_desc":"","_seopress_robots_index":"","_seopress_robots_follow":"","_seopress_robots_imageindex":"","_seopress_robots_snippet":"","_seopress_robots_primary_cat":"","_seopress_robots_breadcrumbs":"","_seopress_robots_freeze_modified_date":"","_seopress_robots_custom_modified_date":"","_seopress_robots_canonical":"","_seopress_social_fb_title":"","_seopress_social_fb_desc":"","_seopress_social_fb_img":"","_seopress_social_fb_img_attachment_id":0,"_seopress_social_fb_img_width":0,"_seopress_social_fb_img_height":0,"_seopress_social_twitter_title":"","_seopress_social_twitter_desc":"","_seopress_social_twitter_img":"","_seopress_social_twitter_img_attachment_id":0,"_seopress_social_twitter_img_width":0,"_seopress_social_twitter_img_height":0,"_seopress_redirections_value":"","_seopress_redirections_enabled":"","_seopress_redirections_enabled_regex":"","_seopress_redirections_logged_status":"","_seopress_redirections_param":"","_seopress_redirections_type":0,"_seopress_analysis_target_kw":"Kepler's law \u2013 problems and solutions","_seopress_news_disabled":"","_seopress_video_disabled":"","_seopress_video":[],"_seopress_pro_schemas_manual":[],"_seopress_pro_rich_snippets_disable_all":"","_seopress_pro_rich_snippets_disable":[],"_seopress_pro_schemas":[],"footnotes":""},"categories":[3],"tags":[],"class_list":["post-940","post","type-post","status-publish","format-standard","hentry","category-solved-problems-in-basic-physics"],"gt_translate_keys":[{"key":"link","format":"url"}],"_links":{"self":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/940","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/comments?post=940"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/940\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=940"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=940"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=940"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}