{"id":3766,"date":"2012-10-20T21:00:53","date_gmt":"2012-10-21T00:00:53","guid":{"rendered":"http:\/\/scienceblogs.com.br\/hypercubic\/?p=3766"},"modified":"2012-10-20T21:00:53","modified_gmt":"2012-10-21T00:00:53","slug":"mais-homenzinhos-verdes-devoradores-de-titnio","status":"publish","type":"post","link":"https:\/\/www.blogs.unicamp.br\/hypercubic\/2012\/10\/mais-homenzinhos-verdes-devoradores-de-titnio\/","title":{"rendered":"Mais Homenzinhos Verdes Devoradores de Tit\u00e2nio"},"content":{"rendered":"

Continuando nossa s\u00e9rie sobre os Homenzinhos Verdes Devoradores de Tit\u00e2nio<\/a>: agora a INEXAC precisa saber qual \u00e9 o valor da acelera\u00e7\u00e3o gravitacional no centro da superf\u00edcie plana do hemisf\u00e9rio que sobrou ap\u00f3s os homenzinhos verdes devorarem meio aster\u00f3ide de tit\u00e2nio.<\/p>\n

<\/p>\n

Solu\u00e7\u00e3o \u201cOficial\u201d<\/strong><\/h4>\n
\n

Para calcular a acelera\u00e7\u00e3o da gravidade no centro da plan\u00edcie de nosso aster\u00f3ide hemisf\u00e9rico, vamos imaginar que o astro ainda esteja intacto e que temos uma sonda bem em seu n\u00facleo. Agora, vamos supor que um hemisf\u00e9rio pode ser dividido em muitas camadas\u00a0 \u2014 ou \u2018conchas\u2019 \u2014 conc\u00eantricas de espessuras id\u00eanticas. Qual \u00e9 a for\u00e7a exercida por todas essas \u2018conchas\u2019 sobre a sonda situada no centro da esfera? Desde que a massa de uma concha \u00e9 diretamente proporcional ao quadrado de seu raio e que a for\u00e7a exercida por uma concha de determinada massa sobre a sonda \u00e9 inversamente proporcional ao quadrado de seu raio, a acelera\u00e7\u00e3o gravitacional no centro das diferentes conchas esf\u00e9ricas \u00e9 a mesma:<\/p>\n

\"image\"<\/a><\/p>\n

Se houver n<\/span><\/strong> conchas, ent\u00e3o a massa da concha mais externa \u00e9 2\u1d28R\u00b2(R\/n)\u03c1<\/span><\/strong>, onde R<\/span><\/strong> \u00e9 o raio do asteroide e \u03c1<\/span><\/strong>, sua densidade. O campo gravitacional total do hemisf\u00e9rio \u00e9 n<\/span><\/strong> vezes o de sua camada mais externa, ou seja, \u00e9 como se a superf\u00edcie do hemisf\u00e9rio tivesse massa M = 2\u1d28\u00b3d<\/span><\/strong>. Essa massa \u00e9, de fato, tr\u00eas vezes a massa real do hemisf\u00e9rio.<\/p>\n

Qual \u00e9, ent\u00e3o, a for\u00e7a exercida por essa concha hemisf\u00e9rica sobre a massa M da sonda? Considerando-se apenas sua magnitude, a for\u00e7a \u00e9 a mesma que seria exercida pela sonda na concha. Ou seja, p = G(M\/2\u1d28R\u00b2)(1\/R\u00b2) = G\u03c1\/R<\/span><\/strong> por unidade de superf\u00edcie. Para estender esse efeito para a casca inteira, a situa\u00e7\u00e3o pode ser comparada com a que temos quando \u00e9 preciso encontrar a for\u00e7a excercida por um l\u00edquido de press\u00e3o p<\/span><\/strong> sobre um hemisf\u00e9rio equivalente.<\/p>\n

\"image\"<\/a><\/p>\n

Uma vez que a for\u00e7a resultante de um l\u00edquido que agisse sobre um hemisf\u00e9rio completo seria zero, a for\u00e7a que age sobre sua superf\u00edcie curva deve ser igual \u00e0 de sua superf\u00edcie plana. Assim, g = p\u1d28R\u00b2 = \u03c1G\u1d28R<\/span> <\/strong>[ou Rog\u00ea Pierre, um cara bem afrancesado].<\/p>\n

\"image\"<\/a><\/p>\n

Agora, lembre-se da acelera\u00e7\u00e3o gravitacional inicial,<\/p>\n

\"image\"<\/a><\/p>\n

na superf\u00edcie do aster\u00f3ide. Portanto, o valor acima resulta em g = \u00beg0<\/sub> = 7,36m\/s\u00b2<\/span><\/strong>. Se voc\u00ea souber a densidade do tit\u00e2nio, tamb\u00e9m pode estimar o raio original do asteroide: R \u2248 78km<\/span><\/strong>.<\/p>\n<\/blockquote>\n

Solu\u00e7\u00e3o do Leitor:<\/h3>\n

Apenas um leitor se arriscou dessa vez #foreveralonefeelings #chatiado. Mas talvez o Murilo Varela<\/a> n\u00e3o tenha compreendido bem o enunciado.<\/p>\n

Considerando que 9,81 m\/s\u00b2 \u00e9 a acelera\u00e7\u00e3o para essa massa (a metade da original):
\n9,81=(G*M)\/(3r)\u00b2
\n9,81*9=(G*M)\/r\u00b2
\n88.29=g<\/p>\n

>>88.29 m\/s\u00b2<\/p><\/blockquote>\n

O que ped\u00edamos era a acelera\u00e7\u00e3o gravitacional do hemisf\u00e9rio restante. A acelera\u00e7\u00e3o de 9,81m\/s\u00b2<\/strong> era do asteroide inteiro e n\u00e3o apenas da metade em quest\u00e3o. Evidentemente, o valor de g tinha que ser menor que g0<\/sub>.<\/p>\n

O enigma final, que fecha essa trilogia, sai na pr\u00f3xima segunda-feira. O que posso adiantar \u00e9 que \u00e9 mais pr\u00f3ximo de um problema de engenharia \u2014 e talvez seja o mais simples de todos. Para evitar que o problema caia no esquecimento (como parece ter acontecido com esse), a resposta ser\u00e1 divulgada na pr\u00f3xima quarta-feira.<\/em><\/p>\n","protected":false},"excerpt":{"rendered":"

Continuando nossa s\u00e9rie sobre os Homenzinhos Verdes Devoradores de Tit\u00e2nio: agora a INEXAC precisa saber qual \u00e9 o valor da acelera\u00e7\u00e3o gravitacional no centro da superf\u00edcie plana do hemisf\u00e9rio que sobrou ap\u00f3s os homenzinhos verdes devorarem meio aster\u00f3ide de tit\u00e2nio.<\/p>\n","protected":false},"author":486,"featured_media":3778,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_eb_attr":"","_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"material-hide-sections":[],"pgc_sgb_lightbox_settings":"","_vp_format_video_url":"","_vp_image_focal_point":[],"footnotes":""},"categories":[152],"tags":[],"class_list":["post-3766","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-geral"],"yoast_head":"\nMais Homenzinhos Verdes Devoradores de Tit\u00e2nio - 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