An artificial satellite is revolving around a planet of mass M and radius R in a circularorbit 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) sqrt((r^(3))/(g)) where k is dimensionless RV g constant and g is acceleration due to gravity.

An artificial satellite is revolving around a planet of mass M and rad

1.	An artificial satellite is revolving around a planet of mass M and radius R in a circularorbit 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) sqrt((r^(3))/(g)) where k is dimensionless RV g constant and g is acceleration due to gravity.
Answer» Solution :According to Kepler.s third law `T^(2) prop r^(2) rArr T prop r^((3)/(2))` We know that T is a function of R and G, Let `T prop r^((3)/(2)) R^(a)g^(b)` `:.T=kr^((3)/(2))R^(a)g^(b) ""...(i)` where K is a DIMENSIONLESS constant of proportionality. Substituting the dimensions of each term in equ. (i) we get `[M^(0)L^(0)T]=[L]^((3)/(2))[L]^(a)+[LT^(-2)]^(b)` `=[L^(a+b+(3)/(2))T^(-2b)]` By comparing the powers `a+2b+(3)/(2)=0 ""...(II)` `=-2b=1 rArr b=-(1)/(2) ""...(iii)` USING equ. (ii) `a-(1)/(2)+(3)/(2)=0 rArr a=-1` By substituting the values of a and b in equ (i) `T=kr^((3)/(2))R^(-1)g^((1)/(2))` `:.T=(k)/(R) sqrt((r^(3))/(g))`