Question

• a satellite orbits earth at an altitude of 300 km. find its orbital speed and period.
• a satellite has an orbital radius of 7.0×10⁶ m. calculate its orbital speed.
• which satellite moves faster: one at 300 km altitude or one at 36,000 km altitude? explain briefly.
• if the orbital radius of a satellite is doubled, how does the orbital speed change? (no calculation required.)

Explanation:

Step1: Recall Orbital Speed Formula

The orbital speed \( v \) of a satellite in circular orbit around Earth is given by \( v = \sqrt{\frac{GM}{r}} \), where \( G \) is the gravitational constant (\( 6.67 \times 10^{-11} \, \text{N·m}^2/\text{kg}^2 \)), \( M \) is Earth's mass (\( 5.97 \times 10^{24} \, \text{kg} \)), and \( r \) is the orbital radius (Earth's radius \( R_E = 6.37 \times 10^6 \, \text{m} \) plus altitude \( h \)).

Step2: Solve First Satellite's Orbital Speed

For the first satellite, \( h = 300 \, \text{km} = 3 \times 10^5 \, \text{m} \), so \( r_1 = R_E + h = 6.37 \times 10^6 + 3 \times 10^5 = 6.67 \times 10^6 \, \text{m} \).
Substitute into \( v \):
\( v_1 = \sqrt{\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.67 \times 10^6}} \).
Simplify: \( \frac{GM}{r_1} = \frac{6.67 \times 5.97 \times 10^{13}}{6.67 \times 10^6} = 5.97 \times 10^7 \), so \( v_1 = \sqrt{5.97 \times 10^7} \approx 7727 \, \text{m/s} \approx 7.7 \, \text{km/s} \).

Step3: Solve First Satellite's Period

Orbital period \( T \) is \( T = \frac{2\pi r}{v} \). Using \( r_1 = 6.67 \times 10^6 \, \text{m} \) and \( v_1 \approx 7727 \, \text{m/s} \):
\( T_1 = \frac{2\pi \times 6.67 \times 10^6}{7727} \approx 5430 \, \text{s} \approx 1.51 \, \text{hours} \).

Step4: Solve Second Satellite's Orbital Speed

Given \( r_2 = 7.0 \times 10^6 \, \text{m} \), use \( v_2 = \sqrt{\frac{GM}{r_2}} \).
\( \frac{GM}{r_2} = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{7.0 \times 10^6} \approx 5.71 \times 10^7 \), so \( v_2 = \sqrt{5.71 \times 10^7} \approx 7556 \, \text{m/s} \approx 7.6 \, \text{km/s} \).

Step5: Compare Speeds (300 km vs 36,000 km Altitude)

Orbital speed \( v \propto \frac{1}{\sqrt{r}} \) (from \( v = \sqrt{\frac{GM}{r}} \)). A 36,000 km altitude satellite has \( r_3 = 6.37 \times 10^6 + 3.6 \times 10^7 = 4.237 \times 10^7 \, \text{m} \), which is larger than \( r_1 \). Since \( v \) decreases with increasing \( r \), the 300 km altitude satellite moves faster.

Step6: Effect of Doubling Orbital Radius on Speed

From \( v = \sqrt{\frac{GM}{r}} \), if \( r' = 2r \), then \( v' = \sqrt{\frac{GM}{2r}} = \frac{1}{\sqrt{2}} \sqrt{\frac{GM}{r}} = \frac{v}{\sqrt{2}} \approx 0.707v \). So speed decreases by a factor of \( \sqrt{2} \).

Answer:

• Orbital speed (300 km altitude): \( \approx 7.7 \, \text{km/s} \), Period: \( \approx 1.51 \, \text{hours} \)
• Orbital speed ( \( 7.0 \times 10^6 \, \text{m} \) radius): \( \approx 7.6 \, \text{km/s} \)
• Faster satellite: 300 km altitude (because \( v \propto 1/\sqrt{r} \), smaller \( r \) means higher \( v \))
• Orbital speed change: Decreases to \( \frac{1}{\sqrt{2}} \) of original (or ~70.7% of original)