calculate the value of accleration due to gravity 3200km above the surface of the earth
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Radius of earth = 6400 km
Acceleration due to gravity at a height “h” is calculated as
g’ = g[R / (R +h)]²
= 9.8 × [6400 / (6400 + 3200)]²
= 4.35 m/s²
Acceleration due to gravity at that height is 4.35 m/s²
Acceleration due to gravity at a height “h” is calculated as
g’ = g[R / (R +h)]²
= 9.8 × [6400 / (6400 + 3200)]²
= 4.35 m/s²
Acceleration due to gravity at that height is 4.35 m/s²
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JunaidMirza:
My answer is correct. Answer should be 4.35 m/s²
Image attached. Have a look it.
I don’t want to waste my time. You should know that……no matter in what way we solve an equation we’ll get same answer.
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as g=GM/R^2 on the surface of earth
at height h from the surface of earth
g(h) = GM/(R+h)^2
g(h) = GM/R^2+h^2+2Rh
as h is small as comparison to radius of earth
hh^2<<<hence we neglect h^3
g(h)=GM/(R^2+2Rh)
now take common R^2 from denominator
g(h)=GM/R^2(1+2h/R)
g(h)=g{1+2h/R}^-1
by binomial throrem
g(h)=g{1-2h/R}
at height h from the surface of earth
g(h) = GM/(R+h)^2
g(h) = GM/R^2+h^2+2Rh
as h is small as comparison to radius of earth
hh^2<<<hence we neglect h^3
g(h)=GM/(R^2+2Rh)
now take common R^2 from denominator
g(h)=GM/R^2(1+2h/R)
g(h)=g{1+2h/R}^-1
by binomial throrem
g(h)=g{1-2h/R}
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