Math Problem Statement

Consider a satellite moving in an elliptic orbit about a focus under attraction The semi-major axis and the eccentricity of the orbit are 30000 km and 0.6, respectively. The radial velocity of the satellite at the true anomaly of 60° is (where r is the radius, and µ = 398601 km³/s²)

a) 2.37 km/s

b) 7.26 km/s

c) 5.23 km/s

d) 9.49 km/s

Solution

To determine the radial velocity of the satellite at a true anomaly of 6060^\circ, we need to use the given orbital parameters and relevant formulas from orbital mechanics. Here are the steps to solve this problem:

  1. **Given dataTo determine the radial velocity of the satellite at a true anomaly of 6060^\circ, we need to use the given orbital parameters and relevant formulas from orbital mechanics. Here are the steps to solve this problem:

  2. **Given **

    • Semi-major axis a=30000kma = 30000 \, \text{km}
    • Eccentricity e=0.6e = 0.6
    • True anomaly ν=60\nu = 60^\circ
    • Standard gravitational parameter μ=398601km3/s2\mu = 398601 \, \text{km}^3/\text{s}^2
  3. Calculate the radius rr at the true anomaly ν\nu:

    The formula for the radius at any point in an elliptical orbit is: r=a(1e2)1+ecosνr = \frac{a(1 - e^2)}{1 + e \cos \nu}

    Plugging in the values: r=30000×(10.62)1+0.6cos60r = \frac{30000 \times (1 - 0.6^2)}{1 + 0.6 \cos 60^\circ} r=30000×(10.36)1+0.6×0.5r = \frac{30000 \times (1 - 0.36)}{1 + 0.6 \times 0.5} r=30000×0.641+0.3r = \frac{30000 \times 0.64}{1 + 0.3} r=192001.314769.23kmr = \frac{19200}{1.3} \approx 14769.23 \, \text{km}

  4. Calculate the radial velocity vrv_r:

    The radial velocity vrv_r is given by: vr=μpesinνv_r = \sqrt{\frac{\mu}{p}} \cdot e \sin \nu

    where pp is the semi-latus rectum, calculated as: p=a(1e2)=30000×0.64=19200kmp = a (1 - e^2) = 30000 \times 0.64 = 19200 \, \text{km}

    Plugging in the values: vr=398601192000.6sin60v_r = \sqrt{\frac{398601}{19200}} \cdot 0.6 \sin 60^\circ vr=20.760468750.632v_r = \sqrt{20.76046875} \cdot 0.6 \cdot \frac{\sqrt{3}}{2} vr=4.5550.60.866v_r = 4.555 \cdot 0.6 \cdot 0.866 vr2.37km/sv_r \approx 2.37 \, \text{km/s}

So, the radial velocity of the satellite at the true anomaly of 6060^\circ is 2.37km/s2.37 \, \text{km/s}.

Thus, the correct answer is:

a) 2.37 km/s

Would you like further details or have any questions?

Here are 8 related questions to expand on this information:

  1. How do you calculate the tangential velocity of a satellite in an elliptic orbit?
  2. What is the significance of the true anomaly in orbital mechanics?
  3. How does the eccentricity of an orbit affect the shape of the satellite's path?
  4. What is the difference between the semi-major axis and the semi-latus rectum?
  5. How do you find the period of a satellite in an elliptical orbit?
  6. What is the relationship between the radial and tangential velocities in an orbit?
  7. How does changing the semi-major axis affect the orbital period of a satellite?
  8. What is the specific orbital energy of a satellite in an elliptical orbit?

Tip: In orbital mechanics, understanding the geometry of orbits (like semi-major axis, eccentricity, and true anomaly) is crucial for predicting the position and velocity of satellites.

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Math Problem Analysis

Mathematical Concepts

Orbital Mechanics
Elliptical Orbits
Radial Velocity

Formulas

Orbital radius formula
Radial velocity formula

Theorems

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Suitable Grade Level

College