Math Problem Statement

The mass of planet Mars is 6.42 × 1023 kg and mean radius of 3390 km. The duration of a day on Mars is 24 hours 40 minutes. Perhaps in the future, a geostationary satellite may be planned for Mars. Determine the orbital radius (in km) of such a satellite (measured from center of Mars)

Solution

To determine the orbital radius for a geostationary satellite around Mars, we need to find the radius at which the satellite's orbital period matches the rotational period of Mars. We can apply Kepler's third law here, which connects the period of orbit TT with the orbital radius rr as follows:

T2=4π2r3GMT^2 = \frac{4 \pi^2 r^3}{G M}

where:

  • TT is the orbital period (equal to the rotational period of Mars for a geostationary satellite),
  • GG is the gravitational constant (6.674×1011Nm2/kg2)(6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2),
  • MM is the mass of Mars (6.42×1023kg)(6.42 \times 10^{23} \, \text{kg}),
  • rr is the orbital radius from the center of Mars.

Step 1: Convert the period TT to seconds

The day on Mars is 24 hours 40 minutes, so: T=24×3600+40×60=88800secondsT = 24 \times 3600 + 40 \times 60 = 88800 \, \text{seconds}

Step 2: Rearrange Kepler's Third Law to solve for rr

Rearrange the equation for rr: r3=GMT24π2r^3 = \frac{G M T^2}{4 \pi^2}

and then take the cube root: r=GMT24π23r = \sqrt[3]{\frac{G M T^2}{4 \pi^2}}

Step 3: Substitute values and solve

Substitute G=6.674×1011Nm2/kg2G = 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2, M=6.42×1023kgM = 6.42 \times 10^{23} \, \text{kg}, and T=88800sT = 88800 \, \text{s}:

r=(6.674×1011)(6.42×1023)(88800)24π23r = \sqrt[3]{\frac{(6.674 \times 10^{-11})(6.42 \times 10^{23})(88800)^2}{4 \pi^2}}

Carrying out this calculation yields an approximate value for rr.

Result

Upon calculating, we find: r20427kmr \approx 20427 \, \text{km}

Thus, the orbital radius required for a geostationary satellite around Mars is approximately 20427 km from the center of Mars.

Would you like further explanation or a breakdown of any part?

Related Questions

  1. How would the orbital radius change if the rotational period of Mars were different?
  2. What other factors affect a satellite’s stability in geostationary orbit?
  3. How does Mars’ lower gravity impact the satellite’s orbital characteristics compared to Earth?
  4. Could the geostationary satellite experience tidal effects from Mars' moons?
  5. What would be the energy requirements to place a satellite in this orbit?

Tip:

Remember that Kepler’s laws apply universally to any orbiting body, making them useful for analyzing celestial mechanics across various planets.

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Math Problem Analysis

Mathematical Concepts

Orbital Mechanics
Kepler's Third Law

Formulas

T^2 = (4 * π^2 * r^3) / (G * M)
r = (G * M * T^2 / (4 * π^2))^(1/3)

Theorems

Kepler's Third Law of Planetary Motion

Suitable Grade Level

Undergraduate Physics/Astronomy