Compute for the value of the acceleration due to gravity g of an object at an altitude equal to twice the radius of the Earth? (radius of Earth = 6.4 x 106 m)
"g(0)=G\\frac{M}{R^2}"
"g(h)=G\\frac{M}{(R+h)^2}" . So, we have
"g(h)=g(0)\\cdot (\\frac{R}{R+h})^2=g(0)\\cdot (\\frac{R}{R+R})^2=g(0)\/4=9.18\/4=2.45\\ (m\/s^2)" . Answer
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