A satellite A of mass m is at a distance of r from the centre of the earth. Another satellite B of mass 2m is at distance of 2r from the earth’s centre. Their time periods are in the ratio of

Correct Option: The correct option is 1:2√(2).

Step-by-Step Explanation:

According to Kepler's Third Law of Planetary Motion, the square of the time period ($T$) of a satellite in a circular orbit around a central body is directly proportional to the cube of its orbital radius ($r$). This relationship is expressed as:
$T^2\propto r^3$

Taking the square root on both sides gives:
$T\propto r^{3/2}$

Note that the time period of a satellite depends only on the mass of the planet it orbits (in this case, the Earth) and the orbital radius. It is completely independent of the satellite's own mass. Therefore, the masses of satellite A ($m$) and satellite B ($2m$) do not affect their orbital time periods.

Let the orbital radius and time period of satellite A be $r_A=r$ and $T_A$ respectively.
Let the orbital radius and time period of satellite B be $r_B=2r$ and $T_B$ respectively.

Now, let's write the ratio of the time periods of satellite A and satellite B:
$\frac{T_A}{T_B}={\left(\frac{r_A}{r_B}\right)}^{3/2}$

Substitute the given values into the equation:
$\frac{T_A}{T_B}={\left(\frac{r}{2r}\right)}^{3/2}={\left(\frac{1}{2}\right)}^{3/2}$

We can simplify the power of 3/2 as follows:
${\left(\frac{1}{2}\right)}^{3/2}=\frac{1}{2^{3/2}}=\frac{1}{2\sqrt{2}}$

Thus, the ratio of their time periods is:
$T_A:T_B=1:2\sqrt{2}$