Two satellites S1 and S2 are revolving in circular orbits around a planet with radius R1 = 3200km and R2 = 800km respectively. The ratio of speed of satellite S1 to the speed of satellite S2 in their respective orbits would be 1 x where x =…………..

M1 ->Mass of satellite (1) M2 -> Mass of satellite (2) MP -> Mass of planet Now NLM (2) on S1 G M 1 M P R 1 2 = m 1 v 1 2 R 1 Similarly v2 = G M P R 2 v 1 v 2 = R 2 R 1 = 8 0 0 3 2 0 0 = 1 2 = 1 x

Two satellites S₁ and S₂ are revolving in circular orbits around a planet with radius R₁ = 3200km and R₂ = 800km respectively. The ratio of speed of satellite S₁ to the speed of satellite S₂ in their respective orbits would be $\frac{1}{x}$ where x =…………..

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Alok Kumar Singh | Contributor-Level 10

8 months ago

M₁ ->Mass of satellite (1)

M₂ -> Mass of satellite (2)

M_P -> Mass of planet

Now NLM (2) on S₁

_{$\frac{G M_{1} M_{P}}{R_{1}^{2}} = \frac{m_{1} v_{1}^{2}}{R_{1}}$}

Similarly v₂ = $\sqrt[]{\frac{G M_{P}}{R_{2}}}$

$\frac{v_{1}}{v_{2}} = \sqrt[]{\frac{R_{2}}{R_{1}}} = \sqrt[]{\frac{800}{3200}} = \frac{1}{2} = \frac{1}{x}$

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The satellites revolve around a planet in coplanar circular orbits in anticlockwise direction. Their period of revolutions are 1 hour and 8 hour respectively. The radius of the orbit of nearer satellite is 2 × 10³ km. The angular speed of the farther satellite as observed from the nearer satellite at the instant when both the satellites are closest is $\frac{π}{x} r a d h^{- 1}$ where x is__________.

Alok Kumar Singh

For A satellite T₁ = 1hour

$S o ω_{1} = 2 π r e d / h o u r$

for B satellite T₂ = 8 hour

given R₁ = 2 × 10³ Km

Relative $ω = \frac{V_{1} - V_{2}}{R_{2} - R_{1}} = \frac{2 π \times 1 0^{3}}{6 \times 1 0^{3}}$

$\Rightarrow \frac{π}{3}$ rad/hour

x = 3

Alok Kumar Singh

For circular motion F_{net $= \frac{M V^{2}}{R}$}

_{$\sqrt[]{2 F} + F' = \frac{M V^{2}}{R}$}

$\sqrt[]{2} \frac{G M M}{{(\sqrt[]{2} R)}^{2}} + \frac{G M M}{{(2 R)}^{2}} = \frac{M V^{2}}{R}$

$\frac{G M}{R} [\frac{1}{\sqrt[]{2}} + \frac{1}{4}] = V^{2}$

$V = \frac{1}{2} \sqrt[]{\frac{G M}{R} (2 \sqrt[]{2} + 1)}$

Alok Kumar Singh

$V_{A} = - \frac{G M_{1}}{r} - \frac{G M_{2}}{R} = - \frac{50 G}{25} - \frac{100 G}{50} = - 4 G$

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