1. Use the satellite time-period formula

For a satellite in circular orbit,

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

where:

• $r$ = orbital radius from the center of Earth
• $G$ = gravitational constant
• $M$ = mass of Earth

The mass of the satellite does not affect the time period.

2. Find orbital radii

Radius of Earth:

$R = 6400\ \text{km} = 6.4 \times 10^6\ \text{m}$

For satellite A:

$h_A = 600\ \text{km} = 0.6 \times 10^6\ \text{m}$ $r_A = R + h_A = 6.4 \times 10^6 + 0.6 \times 10^6 = 7.0 \times 10^6\ \text{m}$

For satellite B:

$h_B = 1600\ \text{km} = 1.6 \times 10^6\ \text{m}$ $r_B = R + h_B = 6.4 \times 10^6 + 1.6 \times 10^6 = 8.0 \times 10^6\ \text{m}$

3. Compute $GM$

Given:

$G = 6.67 \times 10^{-11}\ \text{SI}$ $M = 6 \times 10^{24}\ \text{kg}$

So,

$GM = 6.67 \times 10^{-11} \times 6 \times 10^{24}$ $GM = 40.02 \times 10^{13} = 4.002 \times 10^{14}$

4. Calculate $T_A$

$T_A = 2\pi \sqrt{\frac{(7 \times 10^6)^3}{4.002 \times 10^{14}}}$

Now,

$(7 \times 10^6)^3 = 343 \times 10^{18} = 3.43 \times 10^{20}$

Thus,

$\frac{r_A^3}{GM} = \frac{3.43 \times 10^{20}}{4.002 \times 10^{14}} \approx 8.57 \times 10^5$

$\sqrt{8.57 \times 10^5} \approx 925.7$

Therefore,

$T_A = 2\pi \times 925.7 \approx 5816\ \text{s}$

5. Calculate $T_B$

$T_B = 2\pi \sqrt{\frac{(8 \times 10^6)^3}{4.002 \times 10^{14}}}$

Now,

$(8 \times 10^6)^3 = 512 \times 10^{18} = 5.12 \times 10^{20}$

Thus,

$\frac{r_B^3}{GM} = \frac{5.12 \times 10^{20}}{4.002 \times 10^{14}} \approx 1.279 \times 10^6$

$\sqrt{1.279 \times 10^6} \approx 1131.0$

Therefore,

$T_B = 2\pi \times 1131.0 \approx 7106\ \text{s}$

6. Find the difference

$T_B - T_A = 7106 - 5816 = 1290\ \text{s}$

$T_B - T_A \approx 1.29 \times 10^3\ \text{s}$

This is closest to:

$1.33 \times 10^3\ \text{s}$

7. Option check

• A: $1.33 \times 10^3\,\text{s}$ ✅
• B: $4.24 \times 10^2\,\text{s}$ ❌
• C: $3.33 \times 10^2\,\text{s}$ ❌
• D: $4.24 \times 10^3\,\text{s}$ ❌

So the correct option is A.