Question

An artificial satellite is revolving around a planet of mass M and radius R, in a circular orbit period of a satellite around a common central body, square or 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 dimensionless constant and g is acceleration due to gravity,

Answer 1

Given Form Kepler;s III law, 

T∝ r ^{\(\frac{3}{2}\)}

T is also function of g and R

⇒ T∝r^{\(\frac{3}{2}\)}R^yg^x= kr^{\(\frac{3}{2}\)}R^yg^x

[k = dimension less constant of proportionality]

Substituting dimensionless in each term-

∴ [L⁰M⁰T^-1] = k[L^3/2M⁰T⁰][L¹M⁰T^-2]^x [L]^y 

= k[L^x+y+3/2T^-2x]

For L, 0 = \(\frac{3}{2}\)+x+y

For T, 1 = 0 – 2x 

⇒ x = \(-\frac{1}{2}\)

Therefore, 0 = \(\frac{3}{2}\) \(-\frac{1}{2}\)+ y

⇒ y = -1

Thus,

T = kr^3/2g^−1/2 R^-1

^{=\(\frac{k}{R}\sqrt\frac{r^3}{g}\)}