Question

Drive an expression for time period of satellite

Answer 1

Answer:

$\sf T\:=\:\dfrac{Circumference\:of\:the\:orbit}{Orbital\:velocity}$

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$\sf T\:=\:\dfrac{2\pi(R_e+h)}{V_0}$

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but, $\sf v_0\:=\:\sqrt{\dfrac{GM_e}{R_e+h}}$

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$\sf T\:=\: \dfrac{2\pi(R_e+h)}{\sqrt{\dfrac{GM_e}{R_e+h}}}$

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$\sf = \dfrac{2\pi(R_e+h)\sqrt{R_e+h}}{\sqrt{GM_e}}$

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$\sf T\:=\:\dfrac{2\pi\sqrt{(R_e+h)^2\times (R_e+h)}}{\sqrt{GM_e}}$

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$\sf = \dfrac{2\pi\sqrt{(R_e+h)^3}}{\sqrt{GM_e}}$

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but $\sf gR_e^L\:=\:GM_e$

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$\sf T\:=\: \dfrac{2\pi\sqrt{(R_e+h)^3}}{\sqrt{gR_e^2}}$

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$\sf T\:=\: 2\pi\sqrt{\dfrac{(R_e+h)^3}{gR_e^2}}$

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Hence $\sf T\:=\: \pi\sqrt{\dfrac{R_e}{g}}$

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If the satellite is very close to earth, h = 0,

$\sf T\:=\: 2\pi\sqrt{\dfrac{R_e^3}{gR_e^L}}$

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Hence $\boxed{\sf{T\:=\:\pi\sqrt{\dfrac{R_e}{g}}}}$

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⊱⋅ ──────────── ⋅⊰

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$\Large{\underline{\sf{\blue{Extra\:Info:}}}}$

✶ $\sf{\red{Time\:period}}$ of a satellite is the time taken by a satellite to complete one revoultion around earth.