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Aura Mission. On July 15, 2004 NASA launched the Aura spacecraft to study the earth's climate and atmosphere. This satellite was injected into an orbit \(705 \mathrm{~km}\) 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 \(\mathrm{km} / \mathrm{s})\) is the Aura spacecraft moving?

Short Answer

Expert verified
The orbital period (T) for the Aura spacecraft to complete one orbit is approximately 1.58 hours. The orbital speed (v) of the Aura spacecraft is approximately 7.4 km/s.

Step by step solution

01

Calculate the Total Distance the Satellite is From the Earth's Center

To calculate the time taken by the Aura spacecraft to complete one orbit, we first need to find the total distance of the satellite from the Earth's center. This distance is the sum of the Earth's radius and the height above Earth's surface. The Earth's radius is approximately 6,371 km, so the total distance (r) is \(r = h + R = 705km + 6371km = 7076km = 7.076 x 10^6 m \) (notice the conversion from km to m, it's necessary because the universal gravitational constant (G) uses m^{\^3}kg^{-1}s^{-2}).
02

Calculate the Orbital Period

We can use Kepler's third law to find the orbital period (T). The formula is \(T=2\pi\sqrt{\frac{r^3}{GM}}\) where G is the gravitational constant \(6.67430 \times 10^{-11} m^3kg^{-1}s^{-2}\) and M is the mass of the Earth \(5.972 \times 10^{24} kg\). So, \(T=2\pi\sqrt{\frac{(7.076 \times 10^6m)^3}{(6.67430 \times 10^{-11} m^3kg^{-1}s^{-2} \times 5.972 \times 10^{24} kg)}}\). After calculating this the time obtained is in seconds so convert it into hours.
03

Calculate the Orbital Speed

To calculate the orbital speed (v), we use the formula \(v = \frac{2\pi r}{T}\). We have values for r from step 1 and T from step 2. Substitute those values in the formula and solve for v. After calculating this the speed obtained will be in m/s, so convert it into km/s.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kepler's laws
Kepler's laws are fundamental in understanding the motion of planets and satellites as they orbit larger celestial bodies. These laws describe how objects move in space, making them crucial for calculating orbits. Kepler's First Law states that the orbit of a planet is an ellipse, with the sun at one of the two foci. However, for many satellites, we can approximate their orbits as circular to simplify calculations.

Kepler's Third Law, which is particularly useful in orbital mechanics, relates the time it takes a satellite to complete one orbit to the size of the orbit. Specifically, it tells us that the square of the orbital period (T) is proportional to the cube of the semi-major axis of its orbit (r). This can be mathematically expressed as: \[ T^2 \propto r^3 \].

This law allows us to calculate the time it takes a satellite, like the Aura spacecraft, to orbit the Earth if we know the distance from the Earth's center.
gravitational constant
The gravitational constant, denoted by G, is a key constant in physics that helps us understand the gravitational force between two masses. It appears in Newton's law of universal gravitation, which describes the attractive force between two objects with mass.

The value of G is approximately \(6.67430 \times 10^{-11} \, \text{m}^3\text{kg}^{-1}\text{s}^{-2}\). This constant helps us determine the gravitational force acting on an object in orbit. In the context of the Aura mission, G is used in conjunction with the mass of the Earth and the distance between the satellite and the Earth's center to compute the gravitational pull exerting an influence on the spacecraft's orbit.

Understanding the gravitational constant is essential for solving problems involving orbital motion, as it allows scientists and engineers to predict how satellites behave as they travel around the Earth.
Earth's radius
Earth's radius is a critical factor in many calculations involving the planet, especially its orbiting satellites. The average radius of the Earth is about 6,371 kilometers, and it's important to note this value when solving problems related to satellite orbits.

When placing a satellite in orbit, you must add the altitude of the satellite above the Earth's surface to the Earth's radius. This total distance from the Earth's center is crucial for accurately determining the satellite's orbital period and velocity. In the Aura mission's scenario, the satellite is 705 kilometers above the Earth's surface. Thus, the total distance from the Earth's center must be calculated as the altitude above the Earth's surface plus the Earth's radius.

Accurate measurements of Earth's radius are essential when applying Kepler's laws and understanding the dynamics of orbital motion.
circular orbit
A circular orbit is a type of trajectory that is perfectly circular, with the orbiting object at a constant distance from the central body. While true circular orbits are rare in practice, assuming circular orbits allows for simplifying calculations when examining satellite motion.

In a circular orbit, the speed of the satellite remains constant throughout its path, and the gravitational force provides the necessary centripetal force to maintain this path. For the Aura mission, assuming a circular orbit allows for straightforward calculations of the satellite’s period and speed using formulas that incorporate the gravitational constant and the distance from the Earth's center.

Circular orbits are idealized models, but they are incredibly useful in providing insights into satellite dynamics, helping scientists understand spacecraft behavior more simply.

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Most popular questions from this chapter

A planet orbiting a distant star has radius \(3.24 \times 10^{6} \mathrm{~m}\). The escape speed for an object launched from this planet's surface is \(7.65 \times 10^{3} \mathrm{~m} / \mathrm{s}\). What is the acceleration due to gravity at the surface of the planet?

(a) Calculate how much work is required to launch a spacecraft of mass \(m\) from the surface of the earth (mass \(m_{E}\), radius \(R_{\mathrm{E}}\) ) and place it in a circular low earth orbit - that is, an orbit whose altitude above the earth's surface is much less than \(R_{\mathrm{E}}\). (As an example, the International Space Station is in low earth orbit at an altitude of about \(400 \mathrm{~km},\) much less than \(R_{\mathrm{E}}=6370 \mathrm{~km} .\) ) Ignore the kinetic energy that the spacecraft has on the ground due to the earth's rotation. (b) Calculate the minimum amount of additional work required to move the spacecraft from low earth orbit to a very great distance from the earth. Ignore the gravitational effects of the sun, the moon, and the other planets. (c) Justify the statement "In terms of energy, low earth orbit is halfway to the edge of the universe."

Define the gravitational field \(\vec{g}\) at some point to be equal to the gravitational force \(F_{e}\) on a small object placed at that point divided by the mass \(m\) of the object, so \(\bar{g}=\vec{F}_{e} / m .\) A spherical shell has mass \(M\) and radius \(R .\) What is the magnitude of the gravitational field at the following distances from the center of the shell: (a) \(rR ?\)

Two spherically symmetric planets with no atmosphere have the same average density, but planet \(B\) has twice the radius of planet \(A\). A small satellite of mass \(m_{A}\) has period \(T_{A}\) when it orbits planet \(A\) in a circular orbit that is just above the surface of the planet. A small satellite of mass \(m_{B}\) has period \(T_{B}\) when it orbits planet \(B\) in a circular orbit that is just above the surface of the planet. How does \(T_{B}\) compare to \(T_{A} ?\) with a density of \(2500 \mathrm{~kg} / \mathrm{m}^{3} ?\)

DATA For a spherical planet with mass \(M,\) volume \(V,\) and radius \(R,\) derive an expression for the acceleration due to gravity at the planet's surface, \(g\), in terms of the average density of the planet, \(\rho=M / V,\) and the planet's diameter, \(D=2 R .\) The table gives the values of \(D\) and \(g\) for the eight major planets: $$ \begin{array}{lrc} \text { Planet } & D(\mathrm{~km}) & g\left(\mathrm{~m} / \mathrm{s}^{2}\right) \\ \hline \text { Mercury } & 4879 & 3.7 \\ \text { Venus } & 12,104 & 8.9 \\ \text { Earth } & 12,756 & 9.8 \\ \text { Mars } & 6792 & 3.7 \\ \text { Jupiter } & 142,984 & 23.1 \\ \text { Saturn } & 120,536 & 9.0 \\ \text { Uranus } & 51,118 & 8.7 \\ \text { Neptune } & 49.528 & 11.0 \end{array} $$ (a) Treat the planets as spheres. Your equation for \(g\) as a function of \(\rho\) and \(D\) shows that if the average density of the planets is constant, a graph of \(g\) versus \(D\) will be well represented by a straight line. Graph 8 as a function of \(D\) for the eight major planets. What does the graph tell you about the variation in average density? (b) Calculate the average density for each major planet. List the planets in order of decreasing density, and give the calculated average density of each. (c) The earth is not a uniform sphere and has greater density near its center. It is reasonable to assume this might be true for the other planets. Discuss the effect this has on your analysis. (d) If Saturn had the same average density as the earth, what would be the value of \(g\) at Saturn's surface?

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