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Acceleration due to gravity
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The correct answer is:
decreases with altitude
Acceleration due to gravity above the surface (at altitude) is given as
\(\begin{aligned}
& g^{\prime}=\frac{g}{\left(1+\frac{h}{R_e}\right)^2}=g\left(1+\frac{h}{R_e}\right)^{-2} \\
& g^{\prime}=g\left(1-\frac{2 h}{R_e}\right) \quad \ldots (i)
\end{aligned}\)
[By binomial theorem \((1+x)^n=1+n x\), when \(x < 1\)] From Eq. (i), it is clear that acceleration due to gravity decreases with altitude.
\(\begin{aligned}
& g^{\prime}=\frac{g}{\left(1+\frac{h}{R_e}\right)^2}=g\left(1+\frac{h}{R_e}\right)^{-2} \\
& g^{\prime}=g\left(1-\frac{2 h}{R_e}\right) \quad \ldots (i)
\end{aligned}\)
[By binomial theorem \((1+x)^n=1+n x\), when \(x < 1\)] From Eq. (i), it is clear that acceleration due to gravity decreases with altitude.
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