1. Use Kepler’s relation for circular orbits

For a satellite in circular orbit,

$T = 2\pi \sqrt{\frac{r^3}{GM}}$

So,

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

2. Let the first satellite have

$r_1 = R, \qquad T_1 \propto R^{3/2}$

For the second satellite,

$r_2 = 1.02R, \qquad T_2 \propto (1.02R)^{3/2}$

Thus,

$\frac{T_2}{T_1} = (1.02)^{3/2}$

3. Find the fractional change

Percentage difference in time period:

$\frac{T_2 - T_1}{T_1} \times 100 = \left[(1.02)^{3/2} - 1\right] \times 100$

Now, using binomial approximation for small change:

$(1+x)^{3/2} \approx 1 + \frac{3}{2}x$

with $x=0.02$,

$(1.02)^{3/2} \approx 1 + \frac{3}{2}(0.02) = 1 + 0.03 = 1.03$

Therefore,

$\%\text{ difference} \approx (1.03 - 1)\times 100 = 3\%$

4. Check with direct calculation

$(1.02)^{3/2} = 1.0301\text{(approximately)}$

So actual percentage difference is about

$3.01\%$

which matches $3.0\%$.

5. Evaluate options

• A: $1.5$ ❌
• B: $2.0$ ❌
• C: $0.7$ ❌
• D: $3.0$ ✅

So the correct option is D.