Problem 22 - University Physics with Modern Physics

Question

On July 15, 2004, NASA launched the \(Aura\) spacecraft to study the earth's climate and atmosphere. This satellite was injected into an orbit 705 km above the earth's surface. Assume a circular orbit. (a) How many hours does it take this satellite to make one orbit? (b) How fast (in km/s) is the \(Aura\) spacecraft moving?

Step-by-Step Solution

Answer

(a) Approximately 1.7 hours; (b) Approximately 7.6 km/s.

Step 1: Identify Known Values

First, determine the known values for our problem. The radius of the Earth is approximately 6,371 km. The satellite is orbiting at an altitude of 705 km above the Earth's surface. Therefore, the total radius of the orbit is the sum of Earth's radius and the altitude: \(R = 6371 + 705 = 7076\) km.

Step 2: Calculate the Orbital Period

Use Kepler's Third Law and the formula for orbital period \(T\), given by \(T = 2\pi \sqrt{\frac{R^3}{GM}}\), where \(G\) is the gravitational constant \(6.674 \times 10^{-11} \mathrm{m^3 \cdot kg^{-1} \cdot s^{-2}}\), and \(M\) is Earth's mass, \(5.972 \times 10^{24} \mathrm{kg}\). Convert \(R\) to meters: \(R = 7076 \times 10^3\) meters. Substitute the values to find \(T\).

Step 3: Convert the Orbital Period to Hours

Once we have \(T\) in seconds, convert it to hours by dividing by 3600 (the number of seconds in an hour). This gives us the satellite's orbital period in hours.

Step 4: Calculate Orbital Speed

Use the formula for orbital speed \(v = \frac{2\pi R}{T}\). With \(R\) in kilometers and \(T\) in seconds, calculate the speed \(v\) in km/s by plugging the values into the equation.

Key Concepts

Kepler's Third Law

Kepler’s Third Law is a fundamental principle for understanding satellite orbits. It is used to relate the time a satellite takes to orbit a planet (its "orbital period") to its orbital radius (the distance from the center of the planet). Essentially, it states that the square of the orbital period (\(T^2\)) of a satellite is proportional to the cube of the semi-major axis (radius, for circular orbits) of its orbit (\(R^3\)). The formula is:

• \(T^2 \propto R^3\)

For planets orbiting the sun, Kepler's Third Law helps predict the time it takes for a planet to complete an orbit. In our exercise, this law is critical as it provides the basis for calculating the orbital period of the satellite from its orbit radius, which we will detail next.

Orbital Period

The orbital period is the time it takes for a satellite to complete one full orbit around a celestial body, in this case, Earth. To find this period, we can use Kepler’s Third Law in its quantitative form:

• \(T = 2\pi \sqrt{\frac{R^3}{GM}}\)

In this formula, \(T\) is the orbital period, \(R\) is the total radius of the orbit (including Earth's radius and the satellite’s altitude), \(G\) is the gravitational constant, and \(M\) is the mass of the Earth. For practical calculations, remember to convert units, such as the radius from kilometers to meters. The resultant \(T\) we obtain is in seconds, which we can convert into hours by dividing by 3600.

Orbital Speed

Orbital speed refers to how fast a satellite travels along its orbit. It can be derived from the orbital period and radius using the formula:

• \(v = \frac{2\pi R}{T}\)

Here, \(v\) represents the orbital speed, \(R\) is the radius of the orbit, and \(T\) is the orbital period. Converting \(R\) into kilometers and using the period in seconds, this formula will yield the speed in kilometers per second (km/s). Understanding this concept is crucial, as it tells us how rapidly the satellite is moving through space. A correct calculation can help with navigation and maintain satellite stability.

Gravitational Constant

The gravitational constant, symbolized as \(G\), is a key factor in the calculations of celestial mechanics. It is a measure of the strength of gravity and is universally applied in gravitational equations. Its value is:

• \(G = 6.674 \times 10^{-11} \text{ m}^3\cdot\text{kg}^{-1}\cdot\text{s}^{-2}\)

In our satellite orbit calculations, \(G\) helps determine the gravitational pull between the Earth and the satellite. It is combined with the mass of Earth in the formula for determining the orbital period. This constant allows us to apply the law of universal gravitation in varying astronomical contexts, including satellite operations around Earth.

Earth's Radius

Earth’s radius is a critical component in calculating satellite orbits. It is measured from the center of the Earth to the surface and is approximately 6,371 kilometers. However, when calculating the total orbital radius \(R\) of a satellite, you must add the altitude of the satellite to this radius:

• \(R = \text{Earth's radius} + \text{satellite altitude}\)

In the exercise, the altitude of the satellite was given as 705 km. So, the total orbital radius used in calculations becomes \(6371 + 705 = 7076\) km. This total radius is critical to accurately determining both the orbital period and speed, showcasing how essential precise measurements are in space calculations.