Math Problem Statement

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.

Solution

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:

Fg=GMmr2F_g = \frac{G M m}{r^2}

Where:

  • GG is the universal gravitational constant, G=6.67259×1011Nm2/kg2G = 6.67259 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2,
  • MM is the mass of the Earth, M=5.98×1024kgM = 5.98 \times 10^{24} \, \text{kg},
  • mm is the mass of the satellite (it cancels out in the final equation),
  • rr 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:

Fc=mv2rF_c = \frac{m v^2}{r}

Where:

  • v=6.4×103m/sv = 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:

GMmr2=mv2r\frac{G M m}{r^2} = \frac{m v^2}{r}

The mass mm of the satellite cancels out:

GMr2=v2r\frac{G M}{r^2} = \frac{v^2}{r}

Multiplying both sides by rr, we get:

GMr=v2\frac{G M}{r} = v^2

Solving for rr:

r=GMv2r = \frac{G M}{v^2}

Now, let's plug in the values:

  • G=6.67259×1011Nm2/kg2G = 6.67259 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2,
  • M=5.98×1024kgM = 5.98 \times 10^{24} \, \text{kg},
  • v=6.4×103m/sv = 6.4 \times 10^3 \, \text{m/s}.

Once we find rr, we can subtract the Earth's radius to find the altitude:

Altitude=rREarth\text{Altitude} = r - R_{\text{Earth}}

Where REarth=6370kmR_{\text{Earth}} = 6370 \, \text{km}.

Let me calculate this for you.The satellite's altitude above the surface of the Earth is approximately 3371.72km3371.72 \, \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 6.4km/s6.4 \, \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.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Gravitational Force
Centripetal Force
Orbital Mechanics

Formulas

F_g = (G * M * m) / r^2
F_c = (m * v^2) / r
Gravitational force equals centripetal force: G * M / r^2 = v^2 / r
r = G * M / v^2
Altitude = r - Earth's radius

Theorems

Newton's Law of Universal Gravitation

Suitable Grade Level

Grades 11-12