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the height above the surface of earth where g decreases to 9% of its maximum value on earth surface is.(radius of earth 6400 km)
Dear anand s g=G Me/(Re +h)2 wher h is hight from earth surface value of g on earth surface is for h=0 g1=G Me/(Re )2 so we need to find g such that g=.09 g1 G Me/(Re +h)2 =.09G Me/(Re )2 or Re =.3 (Re+h) h=7Re/3 Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE & AIEEE preparation. All the best. Regards, Askiitians Experts Badiuddin
Dear anand s
g=G Me/(Re +h)2
wher h is hight from earth surface
value of g on earth surface is for h=0
g1=G Me/(Re )2
so we need to find g such that
g=.09 g1
G Me/(Re +h)2 =.09G Me/(Re )2
or Re =.3 (Re+h)
h=7Re/3
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