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Units and Measurements physics ncert solutions class 11th Physics Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Two

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= $\frac{k}{R} \sqrt[]{\frac{r^{3}}{g}}$ ,

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

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Payal Gupta | Contributor-Level 10

4 months ago

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

According to kepler’s law T² $\propto$ r³

T $\propto r^{3 / 2}$

T $\propto r^{\frac{3}{2}} R^{a} g^{b}$

T $\propto k r^{\frac{3}{2}} R^{a} g^{b}$

[M⁰L⁰T]=K [L]^3/2 [L]^a [LT^-2]^b

= k [L^a+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=kr^3/2R^-1g^-1/2

T= $\frac{k}{R} \sqrt[]{\frac{r^{3}}{g}}$

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