Question

An artificial satellite is revolving around a planet of mass M and radius R in a certain orbit of radius r. Using dimensional analysis show that the period of satellite $T=\frac {K} {R}$ . $\sqrt({\frac {r^{3}}{R}})$ where K is a dimensionless constant and G is acceleration due to gravity.

Answer 1

we know from Kepler's 3rd law, The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.
if T is orbital period of satellite and r is the semi major axis of its orbit then, $T^2\propto r^3$

Now suppose , $T^2\propto r^3R^xg^y$

or, $T^2=Ar^3R^xg^y$, where A is proportionality constant

or, [T²] = A[L³] [L]^x [LT^-2]^y

or, [T²] = A[L^(3 + x + y) T^-2y]

compare both sides,

-2y = 2 => y = -1
3 + x + y = 0 => 3 + x - 1 = 0=> x = -2

so, T² = Ar³R^-2g^-1

or, T² = Ar³/R²g

taking square root both sides,

$T=\kappa\sqrt{\frac{r^3}{R^2g}}$, k = √A

so, $T=\frac{\kappa}{R}\sqrt{\frac{r^3}{g}}$

Answer 2

Explanation:

Woh pet help you please follow me