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)

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 \( T \) with the orbital radius \( r \) as follows:

\[
T^2 = \frac{4 \pi^2 r^3}{G M}
\]

where:
- \( T \) is the orbital period (equal to the rotational period of Mars for a geostationary satellite),
- \( G \) is the gravitational constant \( (6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2) \),
- \( M \) is the mass of Mars \( (6.42 \times 10^{23} \, \text{kg}) \),
- \( r \) is the orbital radius from the center of Mars.

### Step 1: Convert the period \( T \) to seconds

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

### Step 2: Rearrange Kepler's Third Law to solve for \( r \)

Rearrange the equation for \( r \):
\[
r^3 = \frac{G M T^2}{4 \pi^2}
\]

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

### Step 3: Substitute values and solve

Substitute \( G = 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \), \( M = 6.42 \times 10^{23} \, \text{kg} \), and \( T = 88800 \, \text{s} \):

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

Carrying out this calculation yields an approximate value for \( r \).

### Result

Upon calculating, we find:
\[
r \approx 20427 \, \text{km}
\]

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.

The mass of planet Mars is 6.42 × 10^23 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).

Explore the calculation of the orbital radius for a geostationary satellite around Mars, using Mars' mass of 6.42 × 10^23 kg and a day length of 24 hours 40 minutes. This solution involves Kepler's third law and gravitational formulas, ideal for physics and astronomy students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation