A satellite A of mass m is revolving round the earth at a height ‘r’ from the centre. Another satellite B of mass 2m is revolving at a height 2r. The ratio of their time periods will be

To find the ratio of the time periods of the two satellites, we can use Kepler's Third Law of Planetary Motion.

Kepler's Third Law states that the square of the time period ($T^2$) of a satellite revolving around a central body is directly proportional to the cube of the radius of its orbit ($R^3$):
$T^2\propto R^3$
This can also be written as:
$T\propto R^{3/2}$
Note that the time period of a satellite is independent of its mass ($m$). Therefore, the masses $m$ and $2m$ of satellites A and B will not affect their orbital periods.

Let $T_A$ and $T_B$ be the time periods of satellites A and B, respectively.
Let $R_A=r$ be the orbital radius of satellite A, and $R_B=2r$ be the orbital radius of satellite B.

Taking the ratio of their time periods:
$\frac{T_A}{T_B}={\left(\frac{R_A}{R_B}\right)}^{3/2}$

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

Now, evaluate the exponent:
${\left(\frac{1}{2}\right)}^{3/2}=\frac{1}{2^{3/2}}=\frac{1}{2^1·2^{1/2}}=\frac{1}{2\sqrt{2}}$

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

The correct option is 1: 2√2.