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A satellite revolves in an orbit close t...

A satellite revolves in an orbit close to the surface of a planet of mean density `5.51 xx 10^(3) kg m^(-3)`. Calculate the time period of satellite.
Given `G = 6.67 xx 10^(-11) Nm^(2) kg^(-2)`.

Text Solution

Verified by Experts

`T = (2pi R)/(upsilon_(0)) = (2pi R)/(sqrt(GM//R)) = (2pi R)/(sqrt((G)/(R ) xx (4)/(3) pi R^(2) rho))`
`= 2 sqrt((3pi)/(4G rho))`
`:. T = 2 sqrt((3 xx 22)/(7 xx 4 xx 6.67 xx 10^(-11) xx 5.5 xx 10^(3)))`
`= 5065 s`
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Knowledge Check

  • A spherical planet has uniform density pi/2xx10^(4)kg//m^(3) . Find out the minimum period for a satellite in a circular orbit around it in seconds (Use G=20/3xx10^(-11) (N-m^(2))/(kg^(2)) ).

    A
    `7500`
    B
    `3000`
    C
    `4500`
    D
    `6000`

  • A satellite is orbiting very close to a planet. Its periodic time depends only on

    A
    density of the planet
    B
    mass of the planet
    C
    radius of the planet
    D
    mass of the satellite

  • Determine the gravitational potential on the surface of earth, given that radius of the earth is 6.4 xx 10^(6) m : its mean density is 5.5 xx 10^(3)kg m^(-3) , G = 6.67 xx 10^(-11) Nm^(2) kg^(-2) .

    A
    `-6.297 xx 10^(7) J kg^(-1)`
    B
    `3.11 xx 10^(11) J kg^(-1)`
    C
    `4 xx 10^(8) J kg^(-1)`
    D
    `-2 xx 10^(9) J kg^(-1)`