Math Problem Statement

The Mangalyaan satellite has a mass of 521 kg. Suppose that it were to orbit Mars at an altitude of 2.26 Mars-radii. (Given: MMars = 6.42x1023 kg, RMars = 3.37x106 m) Calculate the orbital period (in minutes) of the satellite. Consider using the scientific notation entry field

Solution

To calculate the orbital period of the Mangalyaan satellite orbiting Mars, we will use Kepler's Third Law of Planetary Motion:

T=2πr3GMT = 2\pi \sqrt{\frac{r^3}{GM}}

Where:

  • TT is the orbital period,
  • rr is the distance from the center of Mars to the satellite,
  • GG is the gravitational constant, G=6.67430×1011m3kg1s2G = 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2},
  • MM is the mass of Mars.

Step 1: Determine the distance from the center of Mars to the satellite.

The satellite is orbiting at an altitude of 2.26 Mars radii. This means the distance from the center of Mars to the satellite is:

r=2.26×RMars=2.26×3.37×106m=7.6162×106mr = 2.26 \times R_{\text{Mars}} = 2.26 \times 3.37 \times 10^6 \, \text{m} = 7.6162 \times 10^6 \, \text{m}

Step 2: Plug in the values.

We know:

  • MMars=6.42×1023kgM_{\text{Mars}} = 6.42 \times 10^{23} \, \text{kg},
  • G=6.67430×1011m3kg1s2G = 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2},
  • r=7.6162×106mr = 7.6162 \times 10^6 \, \text{m}.

Now, let's calculate the orbital period TT:

T=2π(7.6162×106)36.67430×1011×6.42×1023T = 2\pi \sqrt{\frac{(7.6162 \times 10^6)^3}{6.67430 \times 10^{-11} \times 6.42 \times 10^{23}}}

Step 3: Simplify the equation.

First, calculate the cube of the distance rr:

r3=(7.6162×106)3=4.4094×1020m3r^3 = (7.6162 \times 10^6)^3 = 4.4094 \times 10^{20} \, \text{m}^3

Now calculate the denominator:

GMMars=6.67430×1011×6.42×1023=4.28×1013m3s2GM_{\text{Mars}} = 6.67430 \times 10^{-11} \times 6.42 \times 10^{23} = 4.28 \times 10^{13} \, \text{m}^3 \text{s}^{-2}

Now, calculate the orbital period:

T=2π4.4094×10204.28×1013T = 2\pi \sqrt{\frac{4.4094 \times 10^{20}}{4.28 \times 10^{13}}}

T=2π1.031×107T = 2\pi \sqrt{1.031 \times 10^7}

T=2π×3211.3sT = 2\pi \times 3211.3 \, \text{s}

T=20135.6sT = 20135.6 \, \text{s}

Step 4: Convert the orbital period from seconds to minutes.

T=20135.660335.6minutesT = \frac{20135.6}{60} \approx 335.6 \, \text{minutes}

Final Answer:

The orbital period of the Mangalyaan satellite is approximately 335.6 minutes.

Would you like more details or have any questions about the calculation?

Here are 5 related questions:

  1. How would the orbital period change if the altitude was higher or lower?
  2. What is the significance of Kepler's Third Law in orbital mechanics?
  3. How would the orbital period differ if the satellite were orbiting Earth instead of Mars?
  4. How do you calculate the escape velocity from a planet like Mars?
  5. What are the effects of Mars' gravity on satellite trajectories?

Tip: Always ensure the units are consistent when performing calculations in orbital mechanics, particularly with units of mass, distance, and time.

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Math Problem Analysis

Mathematical Concepts

Orbital Mechanics
Kepler's Third Law
Gravitational Force

Formulas

T = 2π √(r^3 / GM)
r = altitude + planetary radius
GM = gravitational constant × planetary mass

Theorems

Kepler's Third Law of Planetary Motion

Suitable Grade Level

Undergraduate Physics / Advanced High School (Grades 11-12)