At what height the acceleration due to gravity will be reduced to 64% of its value on the surface of the earth?
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gravity is inversely proportional to the square of distance...
let g, a be the accelerations on earth surface and the required surface respectively. let r , h be the radii of earth surface and required surface respectively.
now,
g=k/(r^2)
a=k/(h^2)
where k is proportionality constant equals to GM.
Now according to question
a/g={64/100}
g/a=(100/64)
(h^2)/(r^2)=100/64
h/r=10/8
hence h=10r/8
where r = 6400 km approx
let g, a be the accelerations on earth surface and the required surface respectively. let r , h be the radii of earth surface and required surface respectively.
now,
g=k/(r^2)
a=k/(h^2)
where k is proportionality constant equals to GM.
Now according to question
a/g={64/100}
g/a=(100/64)
(h^2)/(r^2)=100/64
h/r=10/8
hence h=10r/8
where r = 6400 km approx
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