KVPY SX SB Exam 05-Nov-2017 Question Paper

Question : 35 of 120

Marks: +1, -0

Two satellites

S_1 \and S_2

are revolving around a planet in the opposite sense in coplanar circular concentric orbits. At time t = 0, the satellites are farthest apart. The periods of revolution of

S_1 \and S_2

are 3 h and 24 h respectively. The radius of the orbit of S1 is 3 × 10

^{4 }

km. Then the orbital speed of

S_2

as observed from

the planet is $4π × 10^4 \km \h^{-1}$ when $S_2$ is closest from $S_1$ .
the planet is $2π× 10^4 \km \h^{–1}$ when $S_2$ is closest from $S_1$ .
$S_1$ is $π× 10^4 \km \h^{–1}$ when $S_2$ is closest from $S_1$ .
$S_1$ is $3π× 10^4 \km \h^{–1}$ when $S_2$ is closest from $S_1$ .

Validate

Solution:

T^ 2 ∝ r^ 3

9/24^2=({3×10^4}/R^3)^3

{3×3}/{24×24}=({3×10^4}/R )^3

1/4={3×10^4}/R

r = 12 ×10^4

orbital speed of

S_2

seen from planet =

ω_2 r

={2π}/24×12×10^4

=π×10^4\km \h^{–1}

V_1 = ω_1r_1 ={2π}/3×3×10^4

⇒

2π× 10^4 \km \h^{–1}

({2π}/3+{2π}/24)T=π

{9T}/24=1/2

t=12/9

hr
Angle rotate by both satellite

θ_1={2π}/3×12/9⇒{8π}/9

θ_1={2π}/24×12/9⇒{ π}/9

velocity of S2 seen from

S_1 = V_1 + V_2

= 3π × 10^4 \km \h^{–1}

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