Two satellites $$A$$ and $$B$$ of masses 200 kg and 400 kg are revolving round the earth at height of 600 km and 1600 km respectively. If $$T_A$$ and $$T_B$$ are the time periods of $$A$$ and $$B$$ respectively then the value of $$T_B - T_A$$:

[Given: radius of earth = 6400 km, mass of earth = $$6 \times 10^{24}$$ kg]

Solution & Explanation

1. Establish the Time Period Formula

The time period ($$T$$) of a satellite revolving in a circular orbit around the Earth depends strictly on its orbital radius ($$r = R + h$$) and the mass of the Earth ($$M$$). It is completely independent of the satellite's own mass:

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

Let's pre-calculate the constant value $$2\pi \frac{1}{\sqrt{GM}}$$ using SI units:

• $$G = 6.67 \times 10^{-11} \,\, \text{N m}^2/\text{kg}^2$$
• $$M = 6 \times 10^{24} \,\, \text{kg}$$
• $$\sqrt{GM} = \sqrt{(6.67 \times 10^{-11}) \times (6 \times 10^{24})} = \sqrt{4.002 \times 10^{14}} \approx 2 \times 10^7 \,\, \text{m}^{3/2}\text{s}^{-1}$$

2. Calculate the Time Period for Satellite A ($$T_A$$)

The total orbital radius for satellite $$A$$ is:

$$r_A = R + h_A = 6400 \,\, \text{km} + 600 \,\, \text{km} = 7000 \,\, \text{km} = 7 \times 10^6 \,\, \text{m}$$

Substitute this value into the time period equation:

$$T_A = 2\pi \sqrt{\frac{(7 \times 10^6)^3}{2 \times 10^7 \times 2 \times 10^7}} = \frac{2\pi}{2 \times 10^7} \times (7 \times 10^6)^{1.5}$$

$$T_A \approx 3.14 \times 10^{-7} \times 1.852 \times 10^{10} \approx 5815 \,\, \text{s}$$

3. Calculate the Time Period for Satellite B ($$T_B$$)

The total orbital radius for satellite $$B$$ is:

$$r_B = R + h_B = 6400 \,\, \text{km} + 1600 \,\, \text{km} = 8000 \,\, \text{km} = 8 \times 10^6 \,\, \text{m}$$

Substitute this value into the time period equation:

$$T_B = \frac{2\pi}{2 \times 10^7} \times (8 \times 10^6)^{1.5}$$

$$T_B \approx 3.14 \times 10^{-7} \times 2.263 \times 10^{10} \approx 7106 \,\, \text{s}$$

4. Determine the Difference ($$T_B - T_A$$)

Subtract the two time values to find the final interval:

$$\Delta T = T_B - T_A = 7106 \,\, \text{s} - 5815 \,\, \text{s} = 1291 \,\, \text{s} \approx 1.33 \times 10^3 \,\, \text{s}$$

Correct Option: C ($$1.33 \times 10^3 \,\, \text{s}$$)