A satellite moves in a circular orbit around
the Earth at a speed of 6.4 km/s.
Determine the satellite’s altitude above
the surface of the Earth. Assume the
Earth is a homogeneous sphere of radius
6370 km and mass 5.98 × 1024 kg. The
value of the universal gravitational constant
is 6.67259 × 10−11 N · m2/kg2
Answer in units of km.

To determine the satellite's altitude, we can use the principles of gravitational force and centripetal force, which are in balance for a satellite in a stable circular orbit.

The gravitational force acting on the satellite is given by:

\[
F_g = \frac{G M m}{r^2}
\]

Where:
- \( G \) is the universal gravitational constant, \( G = 6.67259 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2 \),
- \( M \) is the mass of the Earth, \( M = 5.98 \times 10^{24} \, \text{kg} \),
- \( m \) is the mass of the satellite (it cancels out in the final equation),
- \( r \) is the distance from the center of the Earth to the satellite.

The centripetal force required to keep the satellite in orbit is given by:

\[
F_c = \frac{m v^2}{r}
\]

Where:
- \( v = 6.4 \times 10^3 \, \text{m/s} \) is the orbital speed of the satellite.

Since the gravitational force provides the necessary centripetal force, we set the two forces equal to each other:

\[
\frac{G M m}{r^2} = \frac{m v^2}{r}
\]

The mass \( m \) of the satellite cancels out:

\[
\frac{G M}{r^2} = \frac{v^2}{r}
\]

Multiplying both sides by \( r \), we get:

\[
\frac{G M}{r} = v^2
\]

Solving for \( r \):

\[
r = \frac{G M}{v^2}
\]

Now, let's plug in the values:

- \( G = 6.67259 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2 \),
- \( M = 5.98 \times 10^{24} \, \text{kg} \),
- \( v = 6.4 \times 10^3 \, \text{m/s} \).

Once we find \( r \), we can subtract the Earth's radius to find the altitude:

\[
\text{Altitude} = r - R_{\text{Earth}}
\]

Where \( R_{\text{Earth}} = 6370 \, \text{km} \).

Let me calculate this for you.The satellite's altitude above the surface of the Earth is approximately \( 3371.72 \, \text{km} \).

Would you like more details or have any questions?

Here are five related questions:
1. How does the mass of the satellite affect its orbital altitude?
2. What would happen if the satellite's speed was higher than \(6.4 \, \text{km/s}\)?
3. How does the altitude change if the Earth's mass were different?
4. What other forces (e.g., atmospheric drag) can affect a satellite's orbit at low altitudes?
5. How can this calculation change for elliptical orbits?

**Tip:** The velocity of a satellite in orbit is directly dependent on its altitude—the higher the altitude, the lower the orbital speed needed for a stable orbit.

A satellite moves in a circular orbit around the Earth at a speed of 6.4 km/s. Determine the satellite’s altitude above the surface of the Earth. Assume the Earth is a homogeneous sphere of radius 6370 km and mass 5.98 × 10^24 kg. The value of the universal gravitational constant is 6.67259 × 10^−11 N·m^2/kg^2. Answer in units of km.

Learn how to calculate a satellite's altitude above Earth using gravitational and centripetal forces. The problem covers key concepts in orbital mechanics, including the use of Newton's Law of Universal Gravitation and gravitational constant. Suitable for Grades 11-12.

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