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Calculate the value of g at a height of 3600 km above the surface of the Earth. (4.0 $ms^{-2}$)

Difficulty: Medium

Solution:

Height = h = 3600 km = $3600 \times 1000$ m = $3.6 \times 10^{6}$ m
        Mass of Earth = $M_{e}$ = $6.0 \times 10^{24}$ kg
        Gravitational acceleration = $g_{h}$ =?


        $g_{h}$ =$\frac{〖 GM〗_{e}}{(R + h)^{2}}$
        $g_{h}$ = $\frac{6.673 \times 10^{-11} \times (6.0 \times 10^{24})}{((6.4 \times 10^{6} + 3.6 \times 10^{6})^{2})}$
              = $\frac{6.673 \times 10^{-11}\times (6.0 \times 10^{24})}{((10.0 \times 10^{6})^{2})}$
              =$\frac{6.673 \times10^{-11} \times (6.0 \times 10^{24})}{(100 \times10^{12})}$


= $6.673 \times 10^{-11}\times 6.0 \times 10^{10}$ = $40\times 10^{-1}$ = $4.0 ms^{-2}$

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