Question

The orbital speeds of $2$ satellites having the ratio of their time period as $64:1$ are: (Consider $V_{01}$ and $V_{02}$ are the orbital speed of satellites)

• A. $\frac{V_{o2}}{V_{o1}}=\frac{1}{2}$
• B. $\frac{V_{o1}}{V_{o2}}=\frac{1}{3}$
• C. $\frac{V_{o2}}{V_{o1}}=\frac{1}{16}$
• D. $\frac{V_{o1}}{V_{o2}}=\frac{1}{4}$

Answer 1

The correct option is D $\frac{V_{o1}}{V_{o2}}=\frac{1}{4}$

Orbital speed of a satellite around a planet of mass $M$ at distance $r$ from the centre of the planet is given by,
$$V_o=\sqrt{\frac{GM}{r}}$$If $r_1$ and $r_2$ are the centre distanace for the satellites, then

$V_{o1}=\sqrt{\frac{GM}{r_1}}$ and $V_{o2}=\sqrt{\frac{GM}{r_2}}$

$\therefore \frac{V_{o1}}{V_{o}2}=\sqrt{\frac{GM}{r_1}}\times \sqrt{\frac{r_2}{GM}}=\sqrt{\frac{r_2}{r_1}}.....\left(1\right)$

Time period of revolution of a satellite is given by$$T^2\propto r^3$$$\Rightarrow {\left(\frac{T_1}{T_2}\right)}^2={\left(\frac{r_1}{r_2}\right)}^3$

As per problem, $\frac{T_1}{T_2}=\frac{64}{1}$,

$\Rightarrow {\left(\frac{64}{1}\right)}^2={\left(\frac{r_1}{r_2}\right)}^3$

$\Rightarrow {\left(\frac{r_1}{r_2}\right)}^3=(64{)}^2$

$\Rightarrow \frac{r_2}{r_1}={\left[{\left(\frac{1}{64}\right)}^{\frac{1}{3}}\right]}^2={\left(\frac{1}{4}\right)}^2.....\left(2\right)$

Using equation (1) and (2), we get

$\therefore \frac{V_{o1}}{V_{o2}}=\sqrt{{\left(\frac{1}{4}\right)}^2}$

$\Rightarrow \frac{V_{o1}}{V_{o2}}=\frac{1}{4}$

Hence, option (d) is the correct answer.

Why this Question? To understand the relation between the orbial speed and time period of revolution.