What is the acceleration due to gravity on the top of mouint everest? Mount everest is the highest mountain peak of the world at the height of 8848 m. the value at sea level is $9.80 m s^{- 2}$

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Solution

Let g be th eacceleration due to grvity on muint everest
$g^{'} = g (1 - \frac{2 h}{R}) = 9.8 (1 - \frac{1796}{6400000}) = 9.8 (1 - 0.00276) = 9.0 (0.99724) = 9.773 m / s^{2}$

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What is the acceleration due to gravity on the top of mouint everest? Mount everest is the highest mountain peak of the world at the height of 8848 m. the value at sea level is 9.80ms−2

Let g be th eacceleration due to grvity on muint everest g′=g(1−2hR)=9.8(1−17966400000)=9.8(1−0.00276)=9.0(0.99724)=9.773m/s2

What is the acceleration due to gravity on the top of mouint everest? Mount everest is the highest mountain peak of the world at the height of 8848 m. the value at sea level is $9.80 m s^{- 2}$

Let g be th eacceleration due to grvity on muint everest
$g^{'} = g (1 - \frac{2 h}{R}) = 9.8 (1 - \frac{1796}{6400000}) = 9.8 (1 - 0.00276) = 9.0 (0.99724) = 9.773 m / s^{2}$