A satellite is launched into a circular orbit of radius ‘R’ around earth while a second satellite is launched into an orbit of radius 1.02 R. The percentage difference in the time periods of the two satellites is

The correct option is 3.

Step-by-step Derivation:

According to Kepler's Third Law of planetary motion, the square of the orbital time period of a satellite is directly proportional to the cube of the radius of its circular orbit. This relation can be written as:

$T^2\propto r^3$

Taking the square root on both sides, we get:

$T\propto r^{3/2}$

For small changes, we can find the fractional error or relative change by taking the natural logarithm on both sides and differentiating:

$\ln \left(T\right)=\frac{3}{2}\ln \left(r\right)+\text{constant}$

Differentiating both sides, we get the relationship between the fractional changes:

$\frac{\Delta T}{T}=\frac{3}{2}\times \frac{\Delta r}{r}$

To express this as a percentage difference, we multiply both sides by 100:

$(\frac{\Delta T}{T}\times 100)=\frac{3}{2}\times (\frac{\Delta r}{r}\times 100)$

Now, let us calculate the percentage change in the orbit's radius. The initial radius is $r_1=R$ and the final radius is $r_2=1.02R$. The change in radius $\Delta r$ is:

$\Delta r=1.02R-R=0.02R$

The percentage difference in the radius is:

$\frac{\Delta r}{r}\times 100=\frac{0.02R}{R}\times 100=2\%$

Substituting the percentage change of radius back into the relation for the time period:

$\text{Percentage difference in time period}=\frac{3}{2}\times 2\%=3\%$

Thus, the percentage difference in the time periods of the two satellites is 3.