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 R ^{2}/(R + h)^{2} or 64/100g = g R^{2}/(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 10 ^{6}/4 = 1.6 x 10^{6} m....!!@@!!**

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

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