The solution for the acceleration of gravity is given as
- [tex]g_{1}=9.789 \mathrm{~m} / \mathrm{s}^{2}$[/tex]
- [tex]g_2=1.955 \mathrm{~m} / \mathrm{s}^{2}$[/tex]
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 [tex]$M=5.97 \times 10^{24} \mathrm{~kg}$[/tex]
Radius of earth [tex]$R=6371 \mathrm{~km}$[/tex]
Gravitational const. [tex]$G=6.67 \times 10^{-11} \mathrm{Nm}^{2} \mathrm{~kg}^{-2}$[/tex]
height [tex]$h_{1}=7900 \mathrm{~m}=7.9 \mathrm{~km}$$[/tex]
[tex]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}[/tex]
In conclusion, acceleration due to gravity at this point will be
[tex]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}$[/tex]
for [tex]$h_{2}=7900 \mathrm{~km}$[/tex]
[tex]R+h_{2}=6371+7900\\\\R+h_{2}=14271 \mathrm{~km}\\\\$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}$[/tex]
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