A satellite is launched into a circular orbit of radius R around the earth. A second satellite is launched into an orbit of radius (1.01) R. The period of the second satellite is larger than that of the first one by approximately

The correct option is 1.5%.

To find the fractional change in the orbital period of the satellite when its orbital radius increases, we can use Kepler's Third Law of Planetary Motion. Kepler's Third Law states that the square of the orbital period of a satellite ($T$) is directly proportional to the cube of the radius of its circular orbit ($r$). This can be mathematically expressed as:

$T^2\propto r^3$

Taking the square root on both sides gives:

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

We can write this as an equation with a constant of proportionality $k$:

$T=k{r}^{3/2}$

Taking the natural logarithm ($\ln$) on both sides of the equation, we get:

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

To find the relationship between small fractional changes in $T$ and $r$, we differentiate both sides of the equation:

$\frac{dT}{T}=\frac{3}{2}\frac{dr}{r}$

For small changes, we can approximate the differentials with finite changes ($\Delta$):

$\frac{\Delta T}{T}\approx \frac{3}{2}\frac{\Delta r}{r}$

According to the problem, the first satellite has a radius of $r_1=R$, and the second satellite has a radius of $r_2=1.01R$.
The change in the orbital radius is:

$\Delta r=r_2-r_1=1.01R-R=0.01R$

The fractional change in the radius is:

$\frac{\Delta r}{r}=\frac{0.01R}{R}=0.01$

Substituting this fractional change into our approximation equation:

$\frac{\Delta T}{T}\approx \frac{3}{2}\times 0.01=0.015$

To convert this fractional change to a percentage, we multiply by 100%:

$\text{Percentage increase in period}=0.015\times 100\%=1.5\%$

Thus, the period of the second satellite is larger than that of the first one by approximately 1.5%.