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Two satellites S(1) and S(2) are revolvi...

Two satellites `S_(1)` and `S_(2)` are revolving round a planet in coplanar and concentric circular orbit of radii `R_(1)` and `R_(2)` in te same direction respectively. Their respective periods of revolution are 1 hr and 8 hr. the radius of the orbit of satellite `S_(1)` is equal to `10^(4)`km. Find the relative speed in kmph when they are closest.

A

`(pi)/2xx10^(4)`

B

`pixx10^(4)`

C

`2pixx10^(4)`

D

`4pixx10^(4)`

Text Solution

Verified by Experts

The correct Answer is:
B

`T^(2)alpha R^(3), V_(0)=(2pi R)/T`, Rel. velocity `=V_(01)-V_(02)`
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Knowledge Check

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    the ratio of period of revolution `s_(1) & s_(2)` is `1:8`
    B
    their velocities are in the ratio `2:1`
    C
    their angular momentum about the planet are in the ratio `2:1`
    D
    the ratio of angular velocities of `s_(2)` w.r.t. `s_(1)` when all three are in the same line is `9:5`.
  • 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 S_(1) is 3xx10^(4) km. Then the orbital speed of S_(2) as observed from

    A
    the planet is `4pixx10^(4)` km `h^(-1)` when `S_(2)` is closest from `S_(1)`.
    B
    the planet is `2pixx10^(4)` km `h^(-1)` when `S_(2)` is closest from `S_(1)`.
    C
    `S_(1)` is `pi xx10^(4)` km `h^(-1)` when `S_(2)` is closest from `S_(1)`
    D
    `S_(1)` is `3pi xx10^(4)` km `h^(-1)` when `S_(2)` is closest from `S_(1)`
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    A
    `8xx10^4` Km
    B
    `2xx10^4` Km
    C
    `16xx10^4` Km
    D
    `4xx10^4` km
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