At what altitude the value of g would become one-fourth that on the surface of the Earth? (One Earth’s radius)
Difficulty: Hard
Solution:
Mass of Earth = $M_{e}$ = $6.0\times 10^{24}$ kg
Radius of Earth = $R_{e}$ = $6.4\times 10^{6}$ m
Gravitational acceleration = gh =$\frac{1}{4}$ g =$\frac{1}{4}\times 10$ $ms^{-2}$ = $2.5ms^{-2}$
Altitude above Earth’s surface = h =?
gh = $\frac{〖GM〗_{e}}{(R + h)^{2}}$
or $(R + h)^{2}$ = $\frac{〖GM〗_{e}}{g_{h}}$
Taking square root on both sides
$√16*10^{13}m^{2}– 6.4*10^{6}$
or$√((R + h)^{2})$=$\frac{√(〖GM〗_{e}}{g_{h})}$
orR + h = $\frac{√(G〖GM〗_{e}}{g_{h})}$
orh =$\frac{√(G〖 GM〗_{e}}{g_{g} -R )}$
orh = $√((\frac{6.673 \times 10^{-11}\times 6.0 \times 10^{24})}{2.5)- 6.4 \times 10^{6}}$
= $√((\frac{40.038\times 10^{13})}{2.5 )-6.4\times 10^{6}}$
= $√(16 \times 10^{13} m^{2})- 6.4\times 10^{6}$ = $√(0.16 \times 10^{12})- 6.4 \times 10^{6}$
= $0.4 \times 10^{6}- 6.4 \times 10^{6}$
= $-6.0 \times 10^{6}$ m
As height is always taken as positive, therefore
h = $6.0 \times 10^{6}$ m = One Earth’s radius
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