A satellite is revolving around a planet of mass 2M in orbit of radius R with time period T₁. Another satellite is revolving around a planet of mass 4M in orbit of radius 2R with time period T₂. Find T₁ /T₂ .

The correct answer is:

$\frac{1}{2}$

Step-by-Step Explanation:

The time period $T$ of a satellite revolving in a circular orbit of radius $r$ around a planet of mass $M$ is given by Kepler's third law, which can be derived by equating the gravitational force to the centripetal force:
$\frac{GMm}{r^2}=\frac{m{v}^2}{r}=m{\left(\frac{2\pi r}{T}\right)}^2\frac{1}{r}$

Solving for $T$, we get the general formula for the time period:
$T=2\pi \sqrt{\frac{r^3}{GM}}$

Now, let us write the expressions for both satellites:

For the first satellite:
Mass of the planet, $M_1=2M$
Orbit radius, $R_1=R$
Time period, $T_1=2\pi \sqrt{\frac{R^3}{G\left(2M\right)}}=2\pi \sqrt{\frac{R^3}{2GM}}$

For the second satellite:
Mass of the planet, $M_2=4M$
Orbit radius, $R_2=2R$
Time period, $T_2=2\pi \sqrt{\frac{{\left(2R\right)}^3}{G\left(4M\right)}}=2\pi \sqrt{\frac{8R^3}{4GM}}=2\pi \sqrt{\frac{2R^3}{GM}}$

To find the ratio of their time periods, we divide $T_1$ by $T_2$:
$\frac{T_1}{T_2}=\frac{2\pi \sqrt{{\displaystyle \frac{R^3}{2GM}}}}{2\pi \sqrt{{\displaystyle \frac{2R^3}{GM}}}}$

Simplifying the fraction inside the square roots:
$\frac{T_1}{T_2}=\sqrt{\frac{\frac{R^3}{2GM}}{\frac{2R^3}{GM}}}=\sqrt{\frac{R^3}{2GM}\times \frac{GM}{2R^3}}$

Canceling the common terms $R^3$ and $GM$:
$\frac{T_1}{T_2}=\sqrt{\frac{1}{4}}=\frac{1}{2}$