Question

At what height the acceleration due to gravity decreases by 51% of its value on the surface of the earth​

Answer 1

Solution :

We know that the acceleration due to gravity from the surface of the earth at height 'h' is given as,

$\huge{\boxed{\sf{g' = g [\frac{1- 2h}{R}]}}}$ - - - - (1)

Where 'h' is the height from the surface of the earth.

We know that:

The radius of the Earth (R) = 6400 km

We have to calculate the height from the surface of the earth at which the value of g is reduced by 51% from the value at the surface of earth.

That is:

$\huge{\boxed{\sf{g' = 51\%\:of\:g}}}$ - - - - (2)

Comparing equations (1) and (2), we get:

$\implies$ 51% of g = g [1- 2h/R] 

$\implies$ (51/100) g = g [1- 2h/R] 

$\implies$ 0.51 = 1 - 2h/R

$\implies$ 2h/R = 1 - 0.51

$\implies$ 2h/R = 0.49

So:

$\implies$ 2h = (0.49) x R

$\implies$ h = (0.49)/2 x R

$\implies$ h = 0.74 x R

$\implies$ h = 0.74 x 6400

Hence:

$\implies$ h = 4736 km

Therefore:

The required height at which the value of 'g' is 51% from the value at the surface of earth is 4736 km.

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Answered by: Niki Swar, Goa❤️

Answer 2

Explanation:

refer to the attachment