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  • 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

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{r3}{g}} where k is a dimensionless constant and g is acceleration due to gravity.

 

Answers (1)

According to Kepler’s 3rd law of planetary motion,

T^{2}\; \alpha \; \; r^{3}

T\; \alpha \; \; r^{\frac{3}{2}}

Now, R & g also affects T

Thus,

T \; \; \; \alpha\; \; \; g^{a}R^{b}r^{\frac{3}{2}}

T = K[LT^{-2}]^{a}[L^{b}][L]^{\frac{3}{2}}

[T^{1}] = K[L^{a+b+\frac{3}{2}}T^{-2a}]

Now, comparing M, L & T according to the principle of homogeneity

a+b+\frac{3}{2} = 0 \; \; \; \; \; \; ..... (i)

-2a = 1 \; \; \; \; \; \; ..... (ii)

a=\frac{-1}{2}

Thus, \frac{-1}{2} + b + \frac{3}{2} = 0


b+1=0

B=-1

-2a=1

T=Kg^{\frac{-1}{2}}R^{-1}r^{\frac{3}{2}}

i.e., T=\frac{K}{R}\sqrt{\frac{r^{3}}{g}}

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