Math Problem Statement

A spy satellite is in circular orbit around Earth. It makes one revolution in 7.50 h. Mass of Earth is 5.974 × 1024 kg, radius of Earth is 6371 km and Gravitational constant G is = 6.674 × 10−11 N·m2/kg2.

How high above Earth’s surface is the satellite?

Solution

To determine the height of the satellite above Earth's surface, we need to use the following concepts and equations from orbital mechanics:

  1. Gravitational force is providing the necessary centripetal force for the satellite to stay in orbit.

    The gravitational force equation is: F=GMmr2F = \frac{G M m}{r^2} where:

    • G=6.674×1011Nm2/kg2G = 6.674 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2 is the gravitational constant,
    • M=5.974×1024kgM = 5.974 \times 10^{24} \, \text{kg} is the mass of the Earth,
    • mm is the mass of the satellite,
    • rr is the distance from the center of the Earth to the satellite.
  2. Centripetal force required to maintain the satellite in a circular orbit is given by: F=mv2rF = m \frac{v^2}{r} where:

    • vv is the orbital velocity of the satellite,
    • rr is the distance from the center of the Earth to the satellite.
  3. From these two expressions, we equate the gravitational force to the centripetal force: GMmr2=mv2r\frac{G M m}{r^2} = m \frac{v^2}{r}

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

  4. The satellite’s velocity is also related to the orbital period TT (the time it takes for one complete revolution) by: v=2πrTv = \frac{2 \pi r}{T}

  5. Substituting this expression for vv into the previous equation: GMr=(2πrT)2\frac{G M}{r} = \left( \frac{2 \pi r}{T} \right)^2

  6. Rearranging for rr, we get: r3=GMT24π2r^3 = \frac{G M T^2}{4 \pi^2}

  7. Now, we can solve for rr (the orbital radius from the center of the Earth) and then subtract the Earth's radius to get the satellite's height above the surface.

Let's calculate this step-by-step.The height of the satellite above Earth's surface is approximately 13,083km13,083 \, \text{km}.

Would you like more details on the calculation, or have any further questions?

Here are 5 related questions you might explore:

  1. How does changing the satellite's orbital period affect its altitude?
  2. What is the satellite's orbital velocity at this altitude?
  3. How would the height change if the satellite orbited a different planet?
  4. What is the gravitational force acting on the satellite at this altitude?
  5. How does Earth's mass influence satellite orbits?

Tip: Orbital period and altitude are linked via Kepler's third law, which states that a satellite farther from Earth takes longer to complete an orbit.

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Math Problem Analysis

Mathematical Concepts

Orbital Mechanics
Gravitational Force
Centripetal Force
Kepler's Third Law

Formulas

Gravitational force: F = G * M * m / r^2
Centripetal force: F = m * v^2 / r
Orbital velocity: v = 2 * π * r / T
Orbital radius: r^3 = G * M * T^2 / 4 * π^2

Theorems

Kepler's Third Law

Suitable Grade Level

Undergraduate