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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

The acceleration due to gravity at the north pole of Neptune is approximately \(11.2 \mathrm{~m} / \mathrm{s}^{2} .\) Neptune has mass \(1.02 \times 10^{26} \mathrm{~kg}\) and radius \(2.46 \times 10^{4} \mathrm{~km}\) and rotates once around its axis in about \(16 \mathrm{~h}\). (a) What is the gravitational force on a \(3.00 \mathrm{~kg}\) object at the north pole of Neptune? (b) What is the apparent weight of this same object at Neptune's equator? (Note that Neptune's "surface" is gaseous, not solid, so it is impossible to stand on it.)

A landing craft with mass \(12,500 \mathrm{~kg}\) is in a circular orbit \(5.75 \times 10^{5} \mathrm{~m}\) above the surface of a planet. The period of the orbit is 5800 s. The astronauts in the lander measure the diameter of the planet to be \(9.60 \times 10^{6} \mathrm{~m}\). The lander sets down at the north pole of the planet. What is the weight of an \(85.6 \mathrm{~kg}\) astronaut as he steps out onto the planet's surface?

An experiment is performed in deep space with two uniform spheres, one with mass \(50.0 \mathrm{~kg}\) and the other with mass \(100.0 \mathrm{~kg}\). They have equal radii, \(r=0.20 \mathrm{~m}\). The spheres are released from rest with their centers \(40.0 \mathrm{~m}\) apart. They accelerate toward each other because of their mutual gravitational attraction. You can ignore all gravitational forces other than that between the two spheres. (a) Explain why linear momentum is conserved. (b) When their centers are \(20.0 \mathrm{~m}\) apart, find (i) the speed of each sphere and (ii) the magnitude of the relative velocity with which one sphere is approaching the other. (c) How far from the initial position of the center of the \(50.0 \mathrm{~kg}\) sphere do the surfaces of the two spheres collide?

There is strong evidence that Europa, a satellite of Jupiter, has a liquid ocean beneath its icy surface. Many scientists think we should land a vehicle there to search for life. Before launching it, we would want to test such a lander under the gravity conditions at the surface of Europa. One way to do this is to put the lander at the end of a rotating arm in an orbiting earth satellite. If the arm is \(4.25 \mathrm{~m}\) long and pivots about one end, at what angular speed (in rpm) should it spin so that the acceleration of the lander is the same as the acceleration due to gravity at the surface of Europa? The mass of Europa is \(4.80 \times 10^{22} \mathrm{~kg}\) and its diameter is \(3120 \mathrm{~km}\)

Two uniform spheres, each with mass \(M\) and radius \(R\), touch each other. What is the magnitude of their gravitational force of attraction?
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