Question

A satellite is launched into a circular orbit of radius R around earth while a second satellite is launched into an orbit of radius 1.02R. The percentage difference in the time period is:

• A. 0.7%
• B. 1.0%
• C. 1.5%
• D. 3.0%

Answer 1

Answer: (D)

3.0%

According to Kepler's law,

$⟹T^2\propto R^3$

$⟹T\propto R^{\left(3/2\right)}$ ......................equation $\left(1\right)$

where,

$T=$ time period

$R=$ radius

Now, differentiating both sides we get

$dT=(\frac{3}{2})R^{(\frac{1}{2})}dR$

Now, Dividing both sides by equation $\left(1\right)$, we get

$⟹\frac{dT}{T}=\frac{(\frac{3}{2})R^{(\frac{1}{2})}}{R^{(\frac{3}{2})}}dR$

$⟹\frac{dT}{T}=\frac{3}{2}\frac{dR}{R}$ .............................equation $\left(2\right)$

Now,

Percentage difference in radius, $=\frac{dR}{R}\times 100\%=\frac{1.02R-R}{R}\times 100\%=2\%$

So, Percentage difference in time period , $=\frac{dT}{T}\times 100=\frac{3}{2}\times 2=3\%$