Question

A)Calculate the effective value of g, the acceleration of gravity, at 7900 m above the Earth's surface.

Answer 1

The solution for the acceleration of gravity is given as

• $g_{1}=9.789 \mathrm{~m} / \mathrm{s}^{2}$

• $g_2=1.955 \mathrm{~m} / \mathrm{s}^{2}$

This is further explained below.

What is the effective value of g, the acceleration of gravity, at 7900 km above the Earth's surface.?

Generally,

Mass of earth $M=5.97 \times 10^{24} \mathrm{~kg}$

Radius of earth $R=6371 \mathrm{~km}$

Gravitational const. $G=6.67 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{~kg}^{-2}$

height $h_{1}=7900 \mathrm{~m}=7.9 \mathrm{~km}$

$R+h_{1}=6371+7.9 &\\\\R+h_{1}=6378.9 \mathrm{~km} \\\\&R+h_{1}=6378.9 \times 10^{3} \mathrm{~m}\end{aligned}$

In conclusion, acceleration due to gravity at this point will be

$g_{1}=\frac{G M}{\left(\bar{R}+\overline{h_{1}}\right)^{2}} \\\\ g_{1}=\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{\left(6378.9 \times 10^{3}\right)^{2}} \\\\ g_{1}=9.789 \mathrm{~m} / \mathrm{s}^{2}$

for $h_{2}=7900 \mathrm{~km}$

$R+h_{2}=6371+7900\\\\R+h_{2}=14271 \mathrm{~km}\\\\ g_{2}=\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{\left(14271 \times 10^{3}\right)^{2}} \\\\ g_2=1.955 \mathrm{~m} / \mathrm{s}^{2}$

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