An artificial satellite is revolving around a planet of mass M and radius R, in a circular orbit of radius r. From Kepler’s Third law about the period of a satellite around a common central body, square of the period of revolution T is proportional to the cube of the radius of the orbit r. Show using dimensional analysis, that

T= k R r 3 g ,

where k is a dimensionless constant and g is acceleration due to gravity.

0 2 Views | Posted 3 months ago
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    Answered by

    Payal Gupta | Contributor-Level 10

    3 months ago

    This is a long answer type question as classified in NCERT Exemplar

    According to kepler’s law T2  r3

    r3/2

    r32Ragb

    kr32Ragb

    [M0L0T]=K [L]3/2 [L]a [LT-2]b

     = k [La+b+3/2T-2b]

    On comparing the powers of same terms we get

    So a+b+3/2=0

    So b=-1/2 and a=-1

    T=kr3/2R-1g-1/2

    T= k R r 3 g

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