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At what height from surface of earth the value of acceleration due to gravity will fall to half that on the surface of the earth?
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The correct answer is:
\(2625 \mathrm{~km}\)
If \(g\) be the gravitational acceleration on the surface of earth, then gravitational acceleration at height \(h\) given as
\(\begin{aligned}
& g_h=\frac{g}{\left(1+\frac{h}{R_e}\right)^2} \\
& \text{But, } g_h=\frac{g}{2} \\
& \therefore \quad \frac{g}{2}=\frac{g}{\left(1+\frac{h}{R_e}\right)^2} \\
& \Rightarrow \quad\left(1+\frac{h}{R_e}\right)^2=2 \\
& \Rightarrow \quad 1+\frac{h}{R_e}=\sqrt{2} \\
& \Rightarrow \quad h=(\sqrt{2}-1) R_e \\
& =(1.414-1) 6400 \\
& =2649.6 \mathrm{~km} \\
& \simeq 2625 \mathrm{~km} \\
\end{aligned}\)
\(\begin{aligned}
& g_h=\frac{g}{\left(1+\frac{h}{R_e}\right)^2} \\
& \text{But, } g_h=\frac{g}{2} \\
& \therefore \quad \frac{g}{2}=\frac{g}{\left(1+\frac{h}{R_e}\right)^2} \\
& \Rightarrow \quad\left(1+\frac{h}{R_e}\right)^2=2 \\
& \Rightarrow \quad 1+\frac{h}{R_e}=\sqrt{2} \\
& \Rightarrow \quad h=(\sqrt{2}-1) R_e \\
& =(1.414-1) 6400 \\
& =2649.6 \mathrm{~km} \\
& \simeq 2625 \mathrm{~km} \\
\end{aligned}\)
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