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![[Post New]](/templates/default/images/icon_minipost_new.gif) 10 May 2007 01:24:59 IST
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where does g become 1/2 of value of g at surface..( at whta height)?
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10 May 2007 01:49:07 IST
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At h= R (i.e. radius of earth)
Is the answer right. Please tell fast.
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10 May 2007 02:04:44 IST
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it's [ (root2) - 1]R
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10 May 2007 02:06:34 IST
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g/2 = g/(1+h/r)^2
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10 May 2007 02:14:32 IST
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Try the formula given by him26
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I cannot even solve 20% of the problem asked here. But each time i click i learn something new. GOIIT ROCKS ................................................................................................................................................. |
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10 May 2007 02:17:06 IST
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Use this formula :
g' = g(1 - h/2R)
h = height above the earth
g' = gravity at that point.
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Will nip in at times to solve problems :)
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10 May 2007 02:18:32 IST
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ans is h = (root2 - 1)R
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10 May 2007 02:21:04 IST
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kartik u r wrong. the formula given by u is valid only for very small height. so, apply g' = g/(1+h/r)^2
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10 May 2007 02:22:21 IST
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gh= GM upon (R+h)2 ...............1 g = GM upon R2 ...................2 divide one by two gh upon g = R^2 upon (R+H)^2 1/2 = R^2upon (R+H)^2 solve and u will get the answer.
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10 May 2007 04:40:24 IST
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Ahh yes..
got the mistake. Thanks for pointing it out
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Will nip in at times to solve problems :)
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11 May 2007 20:11:03 IST
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Amit singh is right.
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