# How much above the surface of earth does the acceleration due to gravity reduces by 36% of its value on the suraface of earth. Radius of earth = 6400 km.

Hi Siddhartha,

Since, the acceleration due to gravity reduces by 36%, then the value of acceleration due to gravity

there is = 100 - 36 = 64%. It means, g' = 64/100 g. If h is the height of location above the suraface of earth,

then g' = g R2/(R + h)2 or 64/100g = g R2/(R + h)2       or       8/10 = R/R + h

or  8 R + 8 h = 10 R

or  h = 2 R/8 = R/4 = 6.4 x 106/4 = 1.6 x 106 m....!!@@!!

Hopes this answer will help u out sid....!!@@!! :-)

Cheerrzzzz........!!@@!! :-)

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Thank u soo much Olivea.....!! :)))))

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Att
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thanks for this solution it was very useful

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yapp
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noooooo
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do yourself
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Since, the acceleration due to gravity reduces by 36%, then the value of acceleration due to gravity?

there is = 100 - 36 = 64%. It means, g' = 64/100 g. If h is the height of location above the suraface of earth,

then g' = g R2/(R + h)2?or 64/100g = g R2/(R + h)2?? ? ? or ? ? ? 8/10 = R/R + h

or? 8 R + 8 h = 10 R

or? h = 2 R/8 = R/4 = 6.4 x 106/4 = 1.6 x 106
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4/3Gm
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