The gravitational force of interaction with earth of mass M will be
"F_g=\\frac{GMm}{R^2}=45\\text{ N}." Since at equilibrium of forces
"m\\frac{v^2}{R}=mg," where
"mg=G\\frac{mM}{R^2},\\\\\\space\\\\\nv=\\sqrt{gR}=\\sqrt{\\frac{GM}{R}}=3077\\text{ m\/s}." The period is
"T=\\frac{d}{v}=\\frac{2\\pi R}{v}=24\\text{ h}."
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