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Suppose that we decide to send a spacecraft to Neptune, using the simple transfer described in Example 8.6 (page 318). The craft starts in a circular orbit close to the earth (radius 1 AU or astronomical unit) and is to end up in a circular orbit near Neptune (radius about \(30 \mathrm{AU}\) ). Use Kepler's third law to show that the transfer will take about 31 years. (In practice we can do a lot better than this by arranging that the craft gets a gravitational boost as it passes Jupiter.)

Short Answer

Expert verified
The transfer takes about 31 years using Kepler's Third Law.

Step by step solution

01

Understand Kepler's Third Law

Kepler's Third Law states that the square of the period of orbit (T) of a planet around the Sun is proportional to the cube of the semi-major axis (a) of its orbit. Mathematically, this is expressed as \( T^2 \propto a^3 \). For transfer orbits, a similar setup can be used.
02

Define the Semi-Major Axis for the Transfer Orbit

The semi-major axis \( a \) for the spacecraft's transfer orbit is the average of the Earth's orbit radius and Neptune's orbit radius. Thus, \( a = \frac{1 + 30}{2} \mathrm{AU} = 15.5 \mathrm{AU} \).
03

Apply Kepler's Third Law

For solar system objects, we use the proportional relation \( T^2 = a^3 \), with \( T \) in years when \( a \) is in astronomical units. Substitute into this relation: \( T^2 = (15.5)^3 \).
04

Calculate Duration of Transfer Orbit

Solve \( T^2 = (15.5)^3 \), so \( T = \sqrt{15.5^3} \). Compute \( 15.5^3 = 3723.875 \), then \( T = \sqrt{3723.875} \approx 61 \) years (for the full round trip).
05

Determine One-Way Transfer Time

The calculated 61 years is for the full orbit around the Sun. Therefore, one-way to Neptune will take \( \frac{61}{2} = 30.5 \) years, approximately 31 years.

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Key Concepts

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

transfer orbit
A transfer orbit is a trajectory followed by a spacecraft that allows it to move between two celestial bodies in space, such as planets in the solar system. This is essential when sending missions to distant planets or moons. In our exercise, the spacecraft begins its journey in Earth's orbit and aims to reach Neptune's orbit. To achieve this, the spacecraft will leverage what is known as a Hohmann Transfer Orbit. This type of orbit is an efficient elliptical path that minimizes the fuel needed for the journey by aligning with the gravitational pull of the Sun.

Specifically, for the transfer from Earth to Neptune, the spacecraft initially travels along an elliptical orbit that has its perihelion (closest point to the Sun) at Earth's orbit and its aphelion (farthest point from the Sun) at Neptune's orbit. This makes the path energy-efficient, as it uses the least amount of propulsion.

There are a few key aspects involved in designing a transfer orbit:
  • The timing of the launch has to be precise to ensure that when the spacecraft arrives at its aphelion, Neptune is positioned where the spacecraft can enter its orbit.
  • Adjustments can be made for the spacecraft to receive gravitational boosts from other planets, like Jupiter, to reduce travel time.
semi-major axis
The semi-major axis is a crucial parameter in understanding orbits, particularly elliptical ones. It represents the longest diameter of an ellipse, extending from the center through the longest part of the ellipse to the perimeter, essentially half the longest diameter. In the context of celestial mechanics and Kepler's laws, it provides a measure of the size of an orbit.

In our exercise, we determined the semi-major axis of the transfer orbit by averaging the distances of Earth's and Neptune's orbits from the Sun. Since Earth's orbit is 1 astronomical unit (AU) from the Sun and Neptune's orbit is roughly 30 AU from the Sun, their average gives us a semi-major axis of 15.5 AU for the transfer orbit.

Why is this important? According to Kepler's Third Law, the time it takes for a body to orbit the Sun (orbital period) is closely related to the semi-major axis of its orbit. By knowing the semi-major axis, we can predict the time period necessary for the spacecraft to travel from Earth to Neptune. Essentially, a longer semi-major axis results in a longer orbital period.

Knowing about the semi-major axis helps us use Kepler's Third Law to forecast travel duration and optimize mission planning by adjusting trajectory before launch.
astronomical unit
An astronomical unit (AU) is a unit of length used in astronomy to describe distances within our solar system. It is approximately equal to the average distance from the Earth to the Sun, which is about 149.6 million kilometers or roughly 93 million miles.

The concept of an astronomical unit makes it easier to express and compare vast distances between planets and other celestial bodies in the solar system without using cumbersome large numbers.

In this exercise, we used astronomical units to measure the spacecraft's starting and target orbits, specifically 1 AU for Earth and around 30 AU for Neptune. This measurement simplifies the application of Kepler's Third Law in computing the semi-major axis and orbital period since AU offers a consistent and relatable scale.

When it comes to understanding space travel and telescope observations, astronomical units provide a helpful reference for interpreting distances. For students and astronomers alike, it offers a convenient way to visualize, calculate, and communicate the vast expanses of space with greater clarity and context.

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Most popular questions from this chapter

Consider a particle of reduced mass \(\mu\) orbiting in a central force with \(U=k r^{n}\) where \(k n>0\). (a) Explain what the condition \(k n>0\) tells us about the force. Sketch the effective potential energy \(U_{\mathrm{eff}}\) for the cases that \(n=2,-1,\) and \(-3 .\) (b) Find the radius at which the particle (with given angular momentum \(\ell\) ) can orbit at a fixed radius. For what values of \(n\) is this circular orbit stable? Do your sketches confirm this conclusion? (c) For the stable case, show that the period of small oscillations about the circular orbit is \(\tau_{\mathrm{osc}}=\tau_{\mathrm{orb}} / \sqrt{n+2} .\) Argue that if \(\sqrt{n+2}\) is a rational number, these orbits are closed. Sketch them for the cases that \(n=2,-1,\) and 7.

(a) Using elementary Newtonian mechanics find the period of a mass \(m_{1}\) in a circular orbit of radius \(r\) around a fixed mass \(m_{2} .\) (b) Using the separation into \(\mathrm{CM}\) and relative motions, find the corresponding period for the case that \(m_{2}\) is not fixed and the masses circle each other a constant distance \(r\) apart. Discuss the limit of this result if \(m_{2} \rightarrow \infty\). (c) What would be the orbital period if the earth were replaced by a star of mass equal to the solar mass, in a circular orbit, with the distance between the sun and star equal to the present earth-sun distance? (The mass of the sun is more than 300,000 times that of the earth.)

What would become of the earth's orbit (which you may consider to be a circle) if half of the sun's mass were suddenly to disappear? Would the earth remain bound to the sun? [Hints: Consider what happens to the earth's KE and PE at the moment of the great disappearance. The virial theorem for the circular orbit (Problem 4.41) helps with this one.] Treat the sun (or what remains of it) as fixed.

In deriving Kepler's third law (8.55) we made an approximation based on the fact that the sun's mass \(M_{\mathrm{s}}\) is much greater than that of the planet \(m .\) Show that the law should actually read \(\tau^{2}=\left[4 \pi^{2} / G\left(M_{\mathrm{s}}+m\right)\right] a^{3},\) and hence that the "constant" of proportionality is actually a little different for different planets. Given that the mass of the heaviest planet (Jupiter) is about \(2 \times 10^{27} \mathrm{kg}\), while \(M_{\mathrm{s}}\) is about \(2 \times 10^{30} \mathrm{kg}\) (and some planets have masses several orders of magnitude less than Jupiter), by what percent would you expect the "constant" in Kepler's third law to vary among the planets?

Prove that for circular orbits around a given gravitational force center (such as the sun) the speed of the orbiting body is inversely proportional to the square root of the orbital radius.

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