Question

if the acceleration due to gravity at the heght h from the surface of the earth is 96% less than its value on the surface , then h here is

Answer 1

HEYA!!
HERE IS YOUR ANSWER. .

g=GM/r^2

g on earth's surface=9.8m/s^2

So half it's value is 4.9m/s^2

By putting the values in the formula.

The final ans is approx 9038km from centre
So above the earth's surface it's is

9038–6400=2638km(approx)

So accelaration due to gravity decreases by half at an height of 2638km.(approx)

Answer 2

then, g' = g - 96 % of g

then, g' = g - 96 % of g g' = 0.04g....... (1)

then, g' = g - 96 % of g g' = 0.04g....... (1)we know, acceleration due to gravity at a height h from the surface of earth is given by

then, g' = g - 96 % of g g' = 0.04g....... (1)we know, acceleration due to gravity at a height h from the surface of earth is given by so, g' = g/(1 + h/R)²

then, g' = g - 96 % of g g' = 0.04g....... (1)we know, acceleration due to gravity at a height h from the surface of earth is given by so, g' = g/(1 + h/R)² 0.04g = g/(1 + h/R)² [ from equation (1), ]

then, g' = g - 96 % of g g' = 0.04g....... (1)we know, acceleration due to gravity at a height h from the surface of earth is given by so, g' = g/(1 + h/R)² 0.04g = g/(1 + h/R)² [ from equation (1), ](0.2)² = 1/(1 + h/R)²

then, g' = g - 96 % of g g' = 0.04g....... (1)we know, acceleration due to gravity at a height h from the surface of earth is given by so, g' = g/(1 + h/R)² 0.04g = g/(1 + h/R)² [ from equation (1), ](0.2)² = 1/(1 + h/R)² 0.2 = 1/(1 + h/R)²

then, g' = g - 96 % of g g' = 0.04g....... (1)we know, acceleration due to gravity at a height h from the surface of earth is given by so, g' = g/(1 + h/R)² 0.04g = g/(1 + h/R)² [ from equation (1), ](0.2)² = 1/(1 + h/R)² 0.2 = 1/(1 + h/R)² (1 + h/R) = 1/0.2

then, g' = g - 96 % of g g' = 0.04g....... (1)we know, acceleration due to gravity at a height h from the surface of earth is given by so, g' = g/(1 + h/R)² 0.04g = g/(1 + h/R)² [ from equation (1), ](0.2)² = 1/(1 + h/R)² 0.2 = 1/(1 + h/R)² (1 + h/R) = 1/0.2 1 + h/R = 5

then, g' = g - 96 % of g g' = 0.04g....... (1)we know, acceleration due to gravity at a height h from the surface of earth is given by so, g' = g/(1 + h/R)² 0.04g = g/(1 + h/R)² [ from equation (1), ](0.2)² = 1/(1 + h/R)² 0.2 = 1/(1 + h/R)² (1 + h/R) = 1/0.2 1 + h/R = 5 h/R = 4

then, g' = g - 96 % of g g' = 0.04g....... (1)we know, acceleration due to gravity at a height h from the surface of earth is given by so, g' = g/(1 + h/R)² 0.04g = g/(1 + h/R)² [ from equation (1), ](0.2)² = 1/(1 + h/R)² 0.2 = 1/(1 + h/R)² (1 + h/R) = 1/0.2 1 + h/R = 5 h/R = 4h = 4R

then, g' = g - 96 % of g g' = 0.04g....... (1)we know, acceleration due to gravity at a height h from the surface of earth is given by so, g' = g/(1 + h/R)² 0.04g = g/(1 + h/R)² [ from equation (1), ](0.2)² = 1/(1 + h/R)² 0.2 = 1/(1 + h/R)² (1 + h/R) = 1/0.2 1 + h/R = 5 h/R = 4h = 4Rhence, h = 4R