To solve these problems, we use the concepts of circular motion and the law of gravitation.

Given:

1. Orbital speed ($v$) = $4000 \, \text{m/s}$
2. Mass of the Earth ($M_e$) = $5.972 \times 10^{24} \, \text{kg}$
3. Gravitational constant ($G$) = $6.674 \times 10^{-11} \, \text{N·m}^2/\text{kg}^2$
4. Radius of Earth ($R_e$) = $6.371 \times 10^6 \, \text{m}$

Formulae:

1. Centripetal force = Gravitational force: $\frac{v^2}{r} = \frac{G M_e}{r^2}$ Solving for $r$ (orbital radius): $r = \frac{G M_e}{v^2}$

2. Altitude above Earth’s surface: $h = r - R_e$

3. Orbital period ($T$): $T = \frac{2\pi r}{v}$

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(a) Altitude of the satellite:

Substitute the known values into the equation for $r$: $r = \frac{(6.674 \times 10^{-11})(5.972 \times 10^{24})}{(4000)^2}$ $r = 1.668 \times 10^7 \, \text{m}$

Now, find the altitude $h$: $h = r - R_e = (1.668 \times 10^7) - (6.371 \times 10^6)$ $h = 1.031 \times 10^7 \, \text{m} \, \text{(or } 10,310 \, \text{km)}.$

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(b) Orbital period:

Using the formula for $T$: $T = \frac{2\pi r}{v} = \frac{2\pi (1.668 \times 10^7)}{4000}$ $T = 2.62 \times 10^4 \, \text{s}$

Convert to hours: $T = \frac{2.62 \times 10^4}{3600} = 7.28 \, \text{hours}.$

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Final Answers:

(a) Altitude: $10,310 \, \text{km}$
(b) Period: $7.28 \, \text{hours}$

Would you like more details or have any questions?

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Follow-up Questions:

1. What assumptions are made about the satellite's orbit?
2. How does altitude impact the orbital speed of a satellite?
3. What is the relationship between orbital radius and period?
4. How would the orbital parameters change for a geostationary satellite?
5. How does the mass of the satellite affect its orbit?

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Tip:

For a satellite in low-Earth orbit, the altitude and orbital period are closely tied to the gravitational force of Earth, which weakens with distance squared. Always use precise values for constants to reduce errors.