What is geostationary satellite? What are the conditions for sat
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    Gravitation
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    Two satellites are revolving in circular orbits of radius R and 4R around a planet. What is the ratio of the time period?
    Given, two satellites are revolving in a circular orbits or radius R and 4R.

    According to Kepler’s law of planetary motion, 

                            straight T squared proportional to straight R cubed 

    Therefore, the ratio of the time period is given by,

                        straight T subscript 1 squared over straight T subscript 2 squared equals fraction numerator straight R cubed over denominator left parenthesis 4 straight R right parenthesis cubed end fraction equals 1 over 64 

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    i.e., T1: T2 is 1: 8
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    Two satellites are revolving around the earth. One which has time period of revolution 24 hours revolves in orbit of radius R. What is the radius of orbit of second satellite whose time period is 12 hours?
    According to Kepler’s third law,
    T2 ∝ R3
    Let R’ be the radius of the orbit of second satellite.
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    ∴           straight R apostrophe equals open parentheses 1 fourth close parentheses to the power of 1 divided by 3 end exponent space straight R     
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    What is geostationary satellite? What are the conditions for satellite to appear stationary?

    A satellite which appears to be stationary to an observer standing on the earth is known as a geostationary satellite.

    The conditions for satellite to appear stationary are:

    (i) The time-period should be 24 hours.

    (ii) Its orbit should be in the equatorial plane of the earth.

    (iii) Its direction of motion should be the same as that of the earth about its polar axis.

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    What is the height of a geostationary satellite from the surface of the earth?
    The time period of the satellite is given by, 

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    rightwards double arrow          space space space space space space space space space space space space straight T squared space equals space fraction numerator 4 straight pi squared over denominator straight g space straight R squared end fraction left parenthesis straight R plus straight h right parenthesis cubed 

    rightwards double arrow            left parenthesis straight R plus straight h right parenthesis cubed space equals space fraction numerator straight g space straight R squared space straight T squared over denominator 4 straight pi squared end fraction 

    rightwards double arrow                      straight h equals open parentheses fraction numerator straight g space straight R squared space straight T squared over denominator 4 straight pi squared end fraction close parentheses to the power of 1 divided by 3 end exponent space minus space straight R 
    We have,

    Acceleration due to gravity, g = 9.8 m/s2

     Raidus space of space the space Earth comma space straight R space equals space 6.4 cross times 10 to the power of 6 space straight m divided by straight s Time space period comma space straight T space equals space 86400 space straight s 

    Therefore, height of a geostationary satellite from the surface of the earth is, 

    straight h equals open parentheses fraction numerator 9.8 cross times left parenthesis 6.4 cross times 10 to the power of 6 right parenthesis squared space left parenthesis 86400 right parenthesis squared over denominator 4 straight pi squared end fraction close parentheses to the power of 1 divided by 3 end exponent minus left parenthesis 6.4 cross times 10 to the power of 6 right parenthesis 
      equals 4.234 cross times 10 to the power of 6 minus 6.4 cross times 10 to the power of 6 

       equals 3.594 cross times 10 to the power of 7 straight m 
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    Can an artificial satellite be put into orbit in such a way that it will always remain directly above Chandigarh?
    Chandigarh does not lie on the equatorial line.

    Therefore, no satellite can be made to remain just above Chandigarh. 
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