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A planet of mass M moves along a circle ...

A planet of mass M moves along a circle around the Sun with velocity `v=34.9km//s` (relative to the heliocentric reference frame). Find the period of revolution of this planet around the Sun.

Text Solution

Verified by Experts

We have
`(Mv^2)/(r)=(gammaMm_s)/(r^2)` or `r=(gammam_s)/(v^2)`
Thus `omega=v/r=(v)/(gammam_s//v^2)=(v^3)/(gammam_s)`
(Hence `m_s` is the mass of the Sun.)
So `T=(2pigammam_s)/(v^3)=(2pixx6*67xx10^(-11)xx1*97xx10^(30))/((34*9xx10^3)^3)=1*94xx10^7sec=225` days.
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Knowledge Check

  • All the planets move around the sun in

    A
    circular path
    B
    rectangular path
    C
    elongated elliptical path
    D
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  • The planet which completes one revolution in 88 days around the sun is:

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  • A planet is revolving around the sun. its distance from the sun at apogee is r_(A) and that at perigee is r_(p) . The masses of planet and sun are 'm' and M respectively, V_(A) is the velocity of planet at apogee and V_(P) is at perigee respectively and T is the time period of revolution of planet around the sun, then identify the wrong answer.

    A
    `T^(2)=(pi^(2))/(Gm)(r_(A)+r_(P))^(3)`
    B
    `T^(2)=(pi^(2))/(2GM)(r_(A)+r_(p))^(3)`
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