Question

An artificial satellite is moving in a circular path of radius 42250 km. Calculate it's speed if takes 24 hrs to complete one revolution

Answer 1

Answer:

s=1760.3 km/hr

Explanation:

speed= distance/time

s=42250/24

s=1760.4 km/hr

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Answer 2

$\underline{\blacksquare\:\:\:\footnotesize{\red{Artificial\:Satelite}}}$

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$\underline{\blacksquare\:\:\:\footnotesize{\red{SolutioN}}}$

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$\footnotesize{\text{Let , a artificial satelite of mass}\:\red{m}\:\text{is rounding Earth}}$

$\footnotesize{\text{and the radius of the parking orbit is}\:\red{r}\:\text{and orbital }}$

$\footnotesize{\text{velocity is}\:\red{v} .}$

$\footnotesize{\therefore\:\:\dfrac{mv^2}{r}=\dfrac{GMm}{r^2}}$

$\footnotesize{\text{If the radius not Earth is}\:\red{R}\:, \:then\:;\:\red{GM=gR^2}}$

$\footnotesize{\therefore\:\:\dfrac{mv^2}{r}=\dfrac{mgR^2}{r^2}}$

$\footnotesize{\therefore\:v^2=\dfrac{gR^2}{r}}$

$\footnotesize{\text{If the Time period of the satelite is}\:\red{T}\:, \:then\:;}$

$\footnotesize{\red{\:\:v=\dfrac{2\pi r}{T}}}$

$\footnotesize{\therefore\:\:\dfrac{4\pi^2r^2}{T^2}=\dfrac{gR^2}{r}}$

$\footnotesize{\implies\:T^2=\dfrac{4\pi^2r^2}{gR^2}}$

$\footnotesize{\implies\:\bf{\boxed{T=\sqrt{\dfrac{4\pi^2r^2}{gR^2}}}}}$

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$\footnotesize{\text{Here the Using formula is}\:\:,\:\:\red{v=\dfrac{2\pi r}{T}}}$

$\footnotesize{\underline{Given}}$

• $\footnotesize{\red{r=42250\:km}}$

• $\footnotesize{\red{T=24\:hours\:=86400\:seconds}}$

$\footnotesize{\therefore\:v=\dfrac{2\pi \times 42250}{86400}}$

$\footnotesize{\implies\boxed{\:\red{v=3.074\:km.s^{-1}}}}$

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