Question

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20. At what height above the surface of earth acceleration due to gravity reduces by 1%?
R
Hint: 9. = 9
Rih
Find the percentage decrease in the acceleration due to gravity...... Show the process​

Answer 1

Answer:

  The new value of gravity at a height h is given as g′=(R+h)2gR2 

As visible from the formula that incresing h results in decresing gravity.

So just put g′=0.99g i.e. 1% less than g where g is gravity at surface. so we get 0.99g=(R+h)2gR2 

Solving the equation we get  h=0.005R

Answer 2

Answer:

Required Height = 1/200 R

Explanation:

The formula we use for finding g at a given height is :

g' = (R+h)×2g×R2

From this formula it is clear that 'h' is inversely proportional to 'g'.

Hence, we put g' = 0.99g

So,

0.99g = (R+h)×2gR2

So, g will cancel out from LHS and RHS

0.99 = (6.4 × 10^6 + h)×2×R2

h = (0.99/2) - 6.4 × 10^6

h = (0.495 - 6.4 × 10^6)R

From this equation,

h = 5/10^3R

h = 5/1000R

h = 0.005 R

Hence, it should be 0.005 R.