Question

A satellite is in a circular orbit about the earth (ME = 5.98 x 1024 kg). The period of the satellite is 1.04 x 104 s. What is the speed at which the satellite travels

Answer #1

we know the formula

F=GMm/r^{2}................1)

force that is helping in circular motion of the satellite around the earth is centripetal force

so F=mv^{2}/r........2)

equating both the equations we get

GMm/r^{2}=mv^{2}/r

v^{2}=GM/r.........3)

we know that speed=distance/time

here s= circumference=2pir and time=T

v=2pir/T

r=Tv/2pi

substituting the value of r in eqn 3

v^{2}=2piGM/Tv

v^{3}=2piGM/T

v===6.22*10^{3}m/s

so answer is 6.22*10^{3}m/s. ( if this is incorrect try
6223m/s or 6222 m/s or 6.223*10^{3} m/s or
6.222*10^{3}m/s)

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