{"id":3681,"date":"2018-02-13T14:18:42","date_gmt":"2018-02-13T14:18:42","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3681"},"modified":"2024-08-20T23:18:52","modified_gmt":"2024-08-20T23:18:52","slug":"resolucao-comentada-gravitacao-universal-velocidade-de-escape-e-aceleracao-da-gravidade","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/enem\/gravitacao-universal-velocidade-de-escape-e-aceleracao-da-gravidade\/resolucao-comentada-gravitacao-universal-velocidade-de-escape-e-aceleracao-da-gravidade\/","title":{"rendered":"Gravita\u00e7\u00e3o Universal – Resolu\u00e7\u00e3o"},"content":{"rendered":"

Resolu\u00e7\u00e3o Comentada<\/b><\/span><\/span><\/span><\/p>\n

Gravita\u00e7\u00e3o Universal <\/b><\/span><\/span><\/span><\/p>\n

Velocidade de escape e Acelera\u00e7\u00e3o da Gravidade<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

\u00a00<\/b><\/span><\/span><\/span>1-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>As leis de Kepler n\u00e3o justificam a afirma\u00e7\u00e3o do astronauta porque elas versam sobre forma da \u00f3rbita, per\u00edodo da \u00f3rbita e \u00e1rea varrida na \u00f3rbita\u00a0 —\u00a0 a afirma\u00e7\u00e3o explica-se pelo Princ\u00edpio Fundamental da Din\u00e2mica, pois o que est\u00e1 em quest\u00e3o s\u00e3o a massa e o peso do telesc\u00f3pio\u00a0 —\u00a0 como o astronauta e o telesc\u00f3pio est\u00e3o em \u00f3rbita, est\u00e3o sujeitos apenas \u00e0 for\u00e7a peso, e, consequentemente, \u00e0 mesma acelera\u00e7\u00e3o (centr\u00edpeta), que \u00e9 a da gravidade local, tendo peso APARENTE nulo\u00a0 —\u00a0 F<\/b><\/span><\/span><\/span>R<\/b><\/span><\/span><\/sub><\/span>=P\u00a0 —\u00a0 \u00a0m a = m g\u00a0 —\u00a0 a = g.<\/b><\/span><\/span><\/span><\/p>\n

\u00c9 pelo mesmo motivo que os objetos flutuam dentro de uma nave e diz-se nesse caso que os corpos est\u00e3o em estado de \u00a0imponderabilidade (sensa\u00e7\u00e3o de aus\u00eancia de peso)\u00a0 —\u00a0\u00a0todavia, a sensa\u00e7\u00e3o de “levitar” ocorre n\u00e3o por “aus\u00eancia de peso, ou seja, falta de gravidade”, mas, sim, porque a nave e tudo o que se encontra no seu interior caem com a mesma velocidade vetorial e est\u00e3o submetidos \u00e0 mesma acelera\u00e7\u00e3o durante a queda\u00a0 \u00a0 — isto significa que astronauta e nave ficam em repouso um em rela\u00e7\u00e3o ao outro\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- D<\/b><\/span><\/span><\/span><\/p>\n

Observa\u00e7\u00e3o\u00a0 —\u00a0 considerando R = 6.400 km o raio da Terra, \u00e0 altura h = 540 km, o raio da \u00f3rbita do telesc\u00f3pio \u00e9 r = R + h = 6.400 + 540 = 6.940 km\u00a0 —\u00a0 de acordo com a lei de Newton da gravita\u00e7\u00e3o, a intensidade do campo gravitacional num ponto da \u00f3rbita \u00e9 \u00a0\u00a0—\u00a0 g = g<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>.(R\/r)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>, sendo g<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>= 10 m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0(acelera\u00e7\u00e3o da gravidade na superf\u00edcie da Terra)\u00a0 —\u00a0 g=10.(6.400\/6.940)2<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0—\u00a0 g=8,5m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 ou seja, o peso REAL do telesc\u00f3pio na \u00f3rbita n\u00e3o \u00e9 pequeno, \u00e9 85% do seu peso na superf\u00edcie terrestre<\/b><\/span><\/span><\/span>.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A massa de qualquer corpo \u00e9 a mesma na Terra, na Lua ou em qualquer outro planeta\u00a0 —\u00a0 peso do corpo na Lua\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>Lua<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>corpo<\/b><\/span><\/span><\/sub><\/span>.g<\/b><\/span><\/span><\/span>Lua<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 a intensidade da for\u00e7a de atra\u00e7\u00e3o gravitacional, na superf\u00edcie da Lua de raio R, entre esse corpo e a Lua \u00e9 fornecida por F<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>=G.M<\/b><\/span><\/span><\/span>Lua<\/b><\/span><\/span><\/sub><\/span>.m<\/b><\/span><\/span><\/span>corpo<\/b><\/span><\/span><\/sub><\/span>\/R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 P=F<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 m<\/b><\/span><\/span><\/span>corpo<\/b><\/span><\/span><\/sub><\/span>.g<\/b><\/span><\/span><\/span>Lua <\/b><\/span><\/span><\/sub><\/span>= G.M<\/b><\/span><\/span><\/span>Lua<\/b><\/span><\/span><\/sub><\/span>.m<\/b><\/span><\/span><\/span>corpo<\/b><\/span><\/span><\/sub><\/span>\/R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 g<\/b><\/span><\/span><\/span>Lua<\/b><\/span><\/span><\/sub><\/span>=G.M<\/b><\/span><\/span><\/span>Lua<\/b><\/span><\/span><\/sub><\/span>\/R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=7.10<\/b><\/span><\/span><\/span>-11<\/b><\/span><\/span><\/sup><\/span>.7.10<\/b><\/span><\/span><\/span>22<\/b><\/span><\/span><\/sup><\/span>\/(2.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 g<\/b><\/span><\/span><\/span>Lua<\/b><\/span><\/span><\/sub><\/span>=1,2m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 — peso na Lua\u00a0 —\u00a0\u00a0 P<\/b><\/span><\/span><\/span>Lua<\/b><\/span><\/span><\/sub><\/span>=m<\/b><\/span><\/span><\/span>corpo<\/b><\/span><\/span><\/sub><\/span>.g<\/b><\/span><\/span><\/span>Lua<\/b><\/span><\/span><\/sub><\/span>=80.1,2\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>astronauta-Lua<\/b><\/span><\/span><\/sub><\/span>= 96N\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Na superf\u00edcie da Terra\u00a0 —\u00a0 g<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=GM\/R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 na altura h=2R + r=3R\u00a0 —\u00a0 g<\/b><\/span><\/span><\/span>h<\/b><\/span><\/span><\/sub><\/span>=GM\/(3R)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 gh=GM\/9R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0g<\/b><\/span><\/span><\/span>h<\/b><\/span><\/span><\/sub><\/span>=g<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/9\u00a0<\/b><\/span><\/span><\/span><\/p>\n

R- E\u00a0<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Velocidade de escape V<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>\u00a0 — velocidade m\u00ednima necess\u00e1ria para que um objeto,\u00a0<\/b><\/span><\/span><\/span>sem propuls\u00e3o pr\u00f3pria,<\/b><\/span><\/span><\/span>\u00a0saia da superf\u00edcie da Terra ou de qualquer planeta, ou ainda, saia da altura R em rela\u00e7\u00e3o ao centro da Terra ou do planeta \u00a0e chegue no infinito com velocidade zero\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=\u221a(2GM\/R)<\/b><\/span><\/span><\/span>\u00a0 —\u00a0 M \u00e9 a massa da Terra ou do planeta e R \u00e9 a dist\u00e2ncia do centro da Terra ou do planeta at\u00e9 o ponto onde ele \u00e9 lan\u00e7ado\u00a0 —\u00a0 substituindo os valores de G, M, e R (para a Terra) que conhecemos, obtemos:V<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>\u224811,3km\/s que \u00e9 a velocidade com que um corpo,\u00a0sem propuls\u00e3o pr\u00f3pria\u00a0deve sair da superf\u00edcie da Terra para \u201clibertar-se\u201d de seu campo gravitacional\u00a0 —\u00a0 o valor da velocidade de escape n\u00e3o depende da massa do corpo que est\u00e1 sendo lan\u00e7ado mas apenas da massa do corpo central, no caso, a Terra e tamb\u00e9m n\u00e3o depende do \u00e2ngulo de lan\u00e7amento\u00a0 — \u00a0\u00a0em fun\u00e7\u00e3o da acelera\u00e7\u00e3o da gravidade g, a velocidade de escape ser\u00e1 dada por:<\/b><\/span><\/span><\/span><\/p>\n

g=GM\/R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00a0\u00a0e\u00a0 V<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=\u221a(2GM\/R)\u00a0 —\u00a0 \u00a0V<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=\u221a(2gR<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/R)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=\u221a(2gR)<\/b><\/span><\/span><\/span>\u00a0\u00a0—\u00a0 V<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=\u221a(2.3.500.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>)=\u221a(30.10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>)\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=\u221a(3.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>)=\u221a(3).10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>m\/s\u00a0 –V<\/b><\/span><\/span><\/span>e<\/b><\/span><\/span><\/sub><\/span>=\u221a3 km\/h\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>R- 5<\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Devido ao movimento de rota\u00e7\u00e3o da Terra e ao fato da Terra ser achatada nos p\u00f3los e dilatada no equador, o valor de g e, consequentemente do peso de um corpo, \u00e9 m\u00e1ximo nos\u00a0 p\u00f3los (9,823m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>) onde n\u00e3o h\u00e1 influ\u00eancia da rota\u00e7\u00e3o da Terra e m\u00ednimo no equador (g=9,789m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>) onde essa influ\u00eancia \u00e9 m\u00e1xima\u00a0 —\u00a0\u00a0 como a massa de um corpo \u00e9 invari\u00e1vel, o mesmo ocorre com o peso. Anote que o peso \u00e9 o mesmo no Polo Norte e no Polo Sul\u00a0 —\u00a0 latitude \u00e9 definida como sendo a medida angular ao longo de um<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

meridiano, do equador at\u00e9 os polos da Terra, na dire\u00e7\u00e3o norte ou sul, variando de latitude zero (no equador) a latitude 90<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0(nos polos)\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Leia atentamente as informa\u00e7\u00f5es abaixo\u00a0 —\u00a0 \u00a0imponderabilidade\u00a0\u00e9 a\u00a0sensa\u00e7\u00e3o de aus\u00eancia de peso\u00a0 —\u00a0 um astronauta dentro da nave espacial e a pr\u00f3pria nave tem a mesma velocidade orbital\u00a0 V\u00a0 e a mesma acelera\u00e7\u00e3o de queda g (queda livre), pois as \u00f3rbitas s\u00e3o de mesmo raio\u00a0 —\u00a0 assim, astronauta, nave, e tudo mais no ve\u00edculo ficam em repouso uns em rela\u00e7\u00e3o aos outros e o <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00e9 apenas sensa\u00e7\u00e3o, pois ali existem\u00a0\"\"\u00a0e\u00a0\"\", caso contr\u00e1rio eles n\u00e3o estariam em \u00f3rbita circular, mas sim em MRU, fora do campo gravitacional terrestre\u00a0 —\u00a0 observe pela figura abaixo que eles <\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

(astronauta e nave) est\u00e3o sempre caindo sobre a Terra\u00a0 mas n\u00e3o a atingem por causa da velocidade tangencial\u00a0\"\".<\/b><\/span><\/span><\/span><\/p>\n

R- C<\/b><\/span><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Do enunciado\u00a0 —\u00a0 M<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>=3M<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 r<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>=0,2r<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 T=37 dias\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>=0,5V<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 r (raio da \u00f3rbita)\u00a0 —\u00a0 R (raio do planeta)<\/b><\/span><\/span><\/span><\/p>\n

I. Falsa\u00a0 —\u00a0 voc\u00ea deve calcular a acelera\u00e7\u00e3o da gravidade de cada planeta e depois iguala-las\u00a0 —\u00a0 g<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>=GM<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>\/R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>G<\/b><\/span><\/span><\/sub><\/span>=G3M<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>\/R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>G<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 g<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>=GM<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>\/R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>T<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 g<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>=g\u00ad<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 3GM<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>\/R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>G<\/b><\/span><\/span><\/sub><\/span>=GM<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>\/R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>T<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 3\/R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>G<\/b><\/span><\/span><\/sub><\/span>=1\/R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>T<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 R<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>=\u221a(3.R<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0R<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>=1,73R<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>\u00a0(deve ser 1,73 vezes maior).<\/b><\/span><\/span><\/span><\/p>\n

II. Verdadeira\u00a0 —\u00a0 velocidade orbital\u00a0 —\u00a0 V=\u221a(GM\/r)\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=GM\/r\u00a0 —\u00a0 r<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 dist\u00e2ncia de Gleese 581g \u00e0 estrela\u00a0 —\u00a0 r<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 dist\u00e2ncia da Terra ao Sol\u00a0 —\u00a0 velocidade orbital de Gleese 581g\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>G<\/b><\/span><\/span><\/sub><\/span>=GM<\/b><\/span><\/span><\/span>estrela<\/b><\/span><\/span><\/sub><\/span>\/r<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 (0,5V<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>)2=GM<\/b><\/span><\/span><\/span>estrela<\/b><\/span><\/span><\/sub><\/span>\/0,2r<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 0,25VT2=GM<\/b><\/span><\/span><\/span>estrela<\/b><\/span><\/span><\/sub><\/span>\/0,2r<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>T<\/b><\/span><\/span><\/sub><\/span>=GM<\/b><\/span><\/span><\/span>estrela<\/b><\/span><\/span><\/sub><\/span>\/0,05r<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>\u00a0(I)\u00a0 — velocidade orbital da Terra\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>T<\/b><\/span><\/span><\/sub><\/span>=GM<\/b><\/span><\/span><\/span>Sol<\/b><\/span><\/span><\/sub><\/span>\/r<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>\u00a0(II)\u00a0 —\u00a0 igualando (I) com (II)\u00a0 —\u00a0 GM<\/b><\/span><\/span><\/span>estrela<\/b><\/span><\/span><\/sub><\/span>\/0,05r<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>=GM<\/b><\/span><\/span><\/span>Sol<\/b><\/span><\/span><\/sub><\/span>\/r<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 M<\/b><\/span><\/span><\/span>estrela<\/b><\/span><\/span><\/sub><\/span>=0,05M<\/b><\/span><\/span><\/span>Sol<\/b><\/span><\/span><\/sub><\/span>=5%M<\/b><\/span><\/span><\/span>Sol<\/b><\/span><\/span><\/sub><\/span>\u00a0(menos que a metade).<\/b><\/span><\/span><\/span><\/p>\n

(III) Correta\u00a0 —\u00a0 uma face \u00e9 sempre clara e a outra sempre escura\u00a0 —\u00a0 o tempo que Gleese demora para efetuar uma volta completa em torno de si mesma (mesmo per\u00edodo, mesma velocidade angular) \u00e9 o mesmo que Gleese demora para efetuar uma volta completa ao redor de sua estrela.<\/b><\/span><\/span><\/span><\/p>\n

(IV) Falsa\u00a0 —\u00a0 W=V\/r\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>\/r<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>=0,5V<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>\/0,2r<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>=2,5V<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>\/r<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 W<\/b><\/span><\/span><\/span>G<\/b><\/span><\/span><\/sub><\/span>=2,5W<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>\u00a0(\u00e9 maior)<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>R- C<\/b><\/span><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Rela\u00e7\u00e3o entre as alturas\u00a0 —\u00a0 sendo g<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span>\u00a0a acelera\u00e7\u00e3o da gravidade na Lua e g<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>\u00a0a acelera\u00e7\u00e3o da gravidade na Terra\u00a0 —\u00a0 aplicando a equa\u00e7\u00e3o de Torricelli\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2gh\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0 (partem do repouso)\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=2gh\u00a0 —\u00a0 na Lua\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=2g<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span>h\u00a0 —\u00a0 na Terra\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=2g<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>h\u00a0 —\u00a0 observe nessas duas express\u00f5es que, como a altura h \u00e9 a mesma, as velocidades na Lua e na Terra s\u00e3o diferentes e, consequentemente as energias cin\u00e9ticas tamb\u00e9m\u00a0 —\u00a0 sendo g<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span><g<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>, isso implica em que V<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span><V<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 assim, ele chegar\u00e1 com energia cin\u00e9tica menor na Lua\u00a0\u00a0 —\u00a0 rela\u00e7\u00e3o entre os tempos de queda\u00a0 —\u00a0 na queda livre\u00a0 —\u00a0h =gt<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 t=\u221a(2h\/g)\u00a0 —\u00a0 observe que o tempo de queda \u00e9 inversamente proporcional a \u221ag\u00a0 —\u00a0 como g<\/b><\/span><\/span><\/span>L<\/b><\/span><\/span><\/sub><\/span>\u00a0< g<\/b><\/span><\/span><\/span>T<\/b><\/span><\/span><\/sub><\/span>, o tempo de queda na Lua \u00e9 menor do que na Terra\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>I) Falsa \u2013 a for\u00e7a que age sobre ele \u00e9 a for\u00e7a resultante centr\u00edpeta que \u00e9 igual \u00e0 for\u00e7a gravitacional\u00a0 —\u00a0\u00a0\u00a0 II) Correta \u2013 vide I\u00a0 —\u00a0III) Falsa \u2013 exerce sim, pelo princ\u00edpio da a\u00e7\u00e3o e rea\u00e7\u00e3o\u00a0 —\u00a0 \u00a0IV) Verdadeira\u00a0 —\u00a0 V=\u221a(GM\/R) \u2013 V e R s\u00e3o inversamente proporcionais, ou seja, \u00e0 medida que a velocidade V vai diminuindo, a altura R tamb\u00e9m vai diminuindo e ele ir\u00e1 caindo.<\/b><\/span><\/span><\/span><\/p>\n

R- D<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Enunciado da lei da gravita\u00e7\u00e3o universal de Newton\u00a0 —\u00a0 \u201cmat\u00e9ria atrai mat\u00e9ria com for\u00e7as cujas intensidades s\u00e3o diretamente proporcionais ao produto de suas massas e inversamente proporcionais ao quadrado da dist\u00e2ncia que os separa\u201d<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

a) Falsa\u00a0 —\u00a0 essa constante, representada por G, pode ser chamada de constante de gravita\u00e7\u00e3o universal ou de constante de Gauss\u00a0 —\u00a0 G \u00e9 n\u00e3o \u00e9 dependente dos corpos que se atraem, nem da dist\u00e2ncia ou at\u00e9 mesmo do meio interposto entre os dois corpos, sendo assim uma constante universal.
\nb) Falsa\u00a0 —\u00a0 ambos se atraem,\u00a0 mas a intensidade dessa for\u00e7a consegue atrair apenas o objeto e n\u00e3o a Terra por causa da<\/b><\/span><\/span><\/span><\/p>\n

diferen\u00e7a de massa entre eles.<\/b><\/span><\/span><\/span><\/p>\n

c) Falsa\u00a0 —\u00a0 o sat\u00e9lite s\u00f3 fica em \u00f3rbita se estiver dentro da campo gravitacional terrestre e, nesse caso, a for\u00e7a gravitacional \u00e9 a pr\u00f3pria for\u00e7a resultante centr\u00edpeta.<\/b><\/span><\/span><\/span><\/p>\n

d) Falsa\u00a0 —\u00a0 diminui sim, mas n\u00e3o linearmente,pois, \u00e9 inversamente proporcional ao\u00a0<\/b><\/span><\/span><\/span>quadrado<\/b><\/span><\/span><\/span>\u00a0da dist\u00e2ncia (fun\u00e7\u00e3o do segundo grau)<\/b><\/span><\/span><\/span><\/p>\n

e) Correta\u00a0 — o peso de um corpo \u00e9 resultado da for\u00e7a de atra\u00e7\u00e3o entre ele e o centro da Terra e P=m.g<\/b><\/span><\/span><\/span><\/p>\n

R- E<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

Voltar para os exerc\u00edcios<\/span><\/a><\/h3>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o Comentada Gravita\u00e7\u00e3o Universal Velocidade de escape e Acelera\u00e7\u00e3o da Gravidade \u00a0 \u00a001-\u00a0As leis de Kepler n\u00e3o justificam a afirma\u00e7\u00e3o do astronauta porque elas versam sobre forma da \u00f3rbita, per\u00edodo da \u00f3rbita e \u00e1rea varrida na \u00f3rbita\u00a0 —\u00a0 a afirma\u00e7\u00e3o explica-se pelo Princ\u00edpio Fundamental da Din\u00e2mica, pois o que est\u00e1 em quest\u00e3o s\u00e3o a massa e o peso do telesc\u00f3pio\u00a0<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3679,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"class_list":["post-3681","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3681","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/comments?post=3681"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3681\/revisions"}],"predecessor-version":[{"id":10670,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3681\/revisions\/10670"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3679"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3681"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}