Question

two satellites a and b are revolving around a planet .the period of revolution of these satellites is ta/tb=1/8.calculate the ratio of the orbital speed if satellites​

Answer 1

Answer:

$\boxed{\mathfrak{Ratio \ of \ orbital \ speed \ of \ satellite = 2:1}}$

Explanation:

Period of revolution of satellite is inversely proportional to cube of orbital speed of satellite i.e.

$\boxed{ \bold{T \propto \frac{1}{{v_o}^3}}}$

According to the question,

Two satellite a & b are revolving around a planet and there ratio of time period of revolution is given as:

$\rm \dfrac{T_a}{T_b} = \dfrac{1}{8}$

So,

Ratio of orbital speed of satellite will be:

$\rm \implies { (\dfrac{{{v}_{o}}_{a}}{{{v}_{o}}_{b}} )}^{3} = \dfrac{T_b}{T_a} \\ \\ \rm \implies { (\dfrac{{{v}_{o}}_{a}}{{{v}_{o}}_{b}} )}^{3} = \dfrac{8}{1} \\ \\ \rm \implies \dfrac{{{v}_{o}}_{a}}{{{v}_{o}}_{b}} = {( \dfrac{2^3}{ 1} )}^{ \frac{1}{3} } \\ \\ \rm \implies \dfrac{{{v}_{o}}_{a}}{{{v}_{o}}_{b}} = \frac{2}{1}$

Answer 2

Answer:

2:1... steps in attachment