Question

A satellite is launched into 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%

Using Kepler's law of planetary motion:

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

$\Rightarrow T=K{R}^{3/2}$

where K is constant

Now, for radius R

We have $T=K{R}^{3/2}$

Differentiating the time period

$\frac{dT}{dR}=\frac{3}{2}K{R}^{1/2}$

When $R=1.02R:dR=1.02R-R$

$dR=0.02R$

Placing the value,

$\Rightarrow dt=\frac{3}{2}K{R}^{1/2}\times dR\Rightarrow dT=\frac{3}{2}K{R}^{1/2}\times 0.02R$

$\Rightarrow dT=1.5K{R}^{1/2}\times 0.02R=0.03T$

$\%change=\frac{dT}{T}\times 100=\frac{0.03T}{T}\times 100=3$

D is correct