A satellite is in a circular orbit around the earth. The period of the satellite is 3.3 hours. What is the speed at which the satellite travels in units of km/s?. Assume the mass of the earth is 5.98x10^24kg
Centripetal acceleration,
a=G×Mr2=v2rr=G×Mv2v=2πrTv=(2πT)×(GMv2)v3=2πGMTv=2πGMT3T=3.3×60×60=11880 sv=[2π×(6.67×10−11)×(5.98×1024)(11880)]1/3=5951 m/s=5.951 km/sa = \frac{G \times M}{r^2} = \frac{v^{2}}{r} \\ r = \frac{G \times M}{v^{2}} v = \frac{2\pi r}{T} \\ v = (\frac{2\pi }{T}) \times (\frac{GM}{v^{2}} ) \\ v^{3} = \frac{2\pi GM}{T} \\ v = \sqrt[3]{\frac{2\pi GM}{T}} \\ T = 3.3 \times 60 \times 60 = 11880 \;s \\ v = [\frac{2π \times (6.67 \times 10^{ -11} ) \times (5.98 \times 10^{24} ) }{ (11880)}]^{1/3} \\ = 5951 \; m/s \\ = 5.951 \; km/sa=r2G×M=rv2r=v2G×Mv=T2πrv=(T2π)×(v2GM)v3=T2πGMv=3T2πGMT=3.3×60×60=11880sv=[(11880)2π×(6.67×10−11)×(5.98×1024)]1/3=5951m/s=5.951km/s
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