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Chapter 13: Problem 33

The dwarf planet Pluto has an elliptical orbit with a semi-major axis of \(5.91 \times 10^{12} \mathrm{~m}\) and eccentricity \(0.249 .\) (a) Calculate Pluto's orbital period. Express your answer in seconds and in earth years. (b) During Pluto's orbit around the sun, what are its closest and farthest distances from the sun?

Short Answer

Expert verified
The orbital period of Pluto in seconds is around \(2.2 脳 10^{8} \mathrm{~seconds}\) and in Earth years is around 7 years. The closest distance (Perihelion) from the Sun is approximately \(4.44 脳 10^{12} \mathrm{~m}\) and the farthest distance (Aphelion) from the Sun is approximately \(7.37 脳 10^{12} \mathrm{~m}\)

Step by step solution

01

Compute the Orbital Period in Seconds

Given the semi-major axis \(a = 5.91 脳 10^{12} \mathrm{~m}\), we can use Kepler's third law which states that the square of the period (T) of a planet鈥檚 orbit is proportional to the cube of the semi-major axis \(a\) of its orbit. The constant of proportionality is the gravitational constant \( G \) times the mass of the sun divided by 4蟺虏. So we can set up the equation: \( T^{2}=\frac{4蟺^{2}}{G M_{sun}}a^{3} \) where \( G = 6.67 脳 10^{-11} \mathrm{~Nm^2/kg^2} \) is the gravitational constant and \( M_{sun} = 1.99 脳 10^{30} \mathrm{~kg} \) is the mass of the Sun. Solve for \( T \) to get \( T = \sqrt{\frac{4蟺^{2}}{G M_{sun}}a^{3}} \)
02

Convert Orbital Period to Years

To express this in Earth years, divide the result by the number of seconds in a year (approx. \( 3.15 脳 10^{7} \mathrm{~seconds} \))
03

Compute Perihelion and Aphelion distances

With the semi-major axis \(a\) and the eccentricity \(e = 0.249\), the perihelion and aphelion (shortest and longest distances from the Sun) can be calculated as \(r_{per} = a(1-e)\) and \(r_{aph} = a(1+e)\) respectively

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Key Concepts

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

Semi-Major Axis
The semi-major axis of an elliptical orbit is the longest radius of the ellipse. It plays a crucial role when calculating the path of celestial bodies like planets. Understanding this concept can help you comprehend how objects travel around a star, like Pluto orbiting the Sun.

In Pluto's case, the semi-major axis is given as \(5.91 \times 10^{12} \mathrm{~m}\). This measurement is significant because it not only determines the size of Pluto's orbit but also influences its orbital period. Using Kepler's Third Law, you may notice a direct relationship between the length of the semi-major axis and how long Pluto takes to complete one orbit around the Sun.
Eccentricity
Eccentricity describes how much an orbit deviates from a perfect circle. It is a dimensionless number that ranges from 0 (a circular orbit) to 1 (a parabolic trajectory). Planetary orbits tend to be ellipses, meaning they have an eccentricity value between 0 and 1.

For Pluto, this is \(0.249\). This means Pluto's orbit is elliptical, but not massively elongated. A larger eccentricity would imply a more stretched out path. Both the perihelion and aphelion, discussed later, rely on this value, showcasing variations in distance within each orbit.
Orbital Period
The orbital period is the time a celestial body takes to complete one entire orbit around its central star. In the exercise, you learned how Kepler's Third Law can help determine Pluto's orbital period using the semi-major axis.

Mathematically, it involves the formula: \[T = \sqrt{\frac{4\pi^{2}a^{3}}{G M_{sun}}}\]where \(G\) is the gravitational constant, \(M_{sun}\) is the Sun's mass, and \(a\) is Pluto's semi-major axis. Calculating this gives you Pluto's orbital period in seconds, and further converting it gives you the period in Earth years, helping you visualize just how long it takes Pluto to orbit the Sun once.
Perihelion
Perihelion is the point in a planet's orbit where it is closest to the Sun. It reflects the influence of the orbit's eccentricity and semi-major axis. Using these measures, you can pinpoint the minimum distance Pluto reaches in its orbit.

For Pluto, the calculation is expressed as:\[r_{per} = a(1-e)\]where \(a\) is the semi-major axis, and \(e\) is the eccentricity. This formula simplifies to finding that closest point, showing how Pluto's journey brings it nearer to the Sun at certain times.
Aphelion
Contrasting with perihelion, the aphelion is where Pluto is furthest from the Sun in its orbit. Much like perihelion, it can be calculated with the semi-major axis and eccentricity.

The formula used is:\[r_{aph} = a(1+e)\]This distance represents one end of Pluto's elliptical path鈥攖he other Big journey Pluto makes around the Sun, reaching this maximum distance before heading back toward the perihelion.
Gravitational Constant
The gravitational constant, usually denoted as \(G\), is a fundamental constant in physics. It appears in the universal law of gravitation and Kepler's laws of planetary motion, which help describe the force and interactions between two bodies like the Sun and Pluto.

\(G = 6.67 \times 10^{-11} \mathrm{~Nm^2/kg^2}\) and, when combined with the mass of the Sun and planetary data such as the semi-major axis, provides insight into natural phenomena like Pluto鈥檚 orbital period. In equations like Kepler's Third Law, \(G\) is a critical factor for resolving astronomical distances and rhythms.

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Most popular questions from this chapter

A thin spherical shell has radius \(r_{A}=4.00 \mathrm{~m}\) and mass \(m_{A}=20.0 \mathrm{~kg} .\) It is concentric with a second thin spherical shell that has radius \(r_{B}=6.00 \mathrm{~m}\) and mass \(m_{B}=40.0 \mathrm{~kg} .\) What is the net gravitational force that the two shells exert on a point mass of \(0.0200 \mathrm{~kg}\) that is a distance \(r\) from the common center of the two shells, for (a) \(r=2.00 \mathrm{~m}\) (inside both shells), (b) \(r=5.00 \mathrm{~m}\) (in the space between the two shells), and (c) \(r=8.00 \mathrm{~m}\) (outside both shells)?

A uniform, solid, \(1000.0 \mathrm{~kg}\) sphere has a radius of \(5.00 \mathrm{~m}\). (a) Find the gravitational force this sphere exerts on a \(2.00 \mathrm{~kg}\) point mass placed at the following distances from the center of the sphere: (i) \(5.01 \mathrm{~m}\), (ii) \(2.50 \mathrm{~m}\). (b) Sketch a qualitative graph of the magnitude of the gravitational force this sphere exerts on a point mass \(m\) as a function of the distance \(r\) of \(m\) from the center of the sphere. Include the region from \(r=0\) to \(r \rightarrow \infty\)

\(\cdot \mathrm{A}\) satellite of mass \(m\) is in a circular orbit around a spherical planet of mass \(m_{\mathrm{p}}\). The kinetic energy of the satellite is \(K_{A}\) when its orbit radius is \(r_{A}\). In terms of \(r_{A}\), what must the orbit radius be in order for the kinetic energy of the satellite to be \(2 K_{A} ?\)

Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed \(2.0 \times 10^{4} \mathrm{~m} / \mathrm{s}\) when at a distance of \(2.5 \times 10^{11} \mathrm{~m}\) from the center of the sun, what is its speed when at a distance of \(5.0 \times 10^{10} \mathrm{~m} ?\)

(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."
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