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Chapter 6: Problem 14

The rotation period of Jupiter is approximately 10 hours. Its mass and radius are \(318 M_{\mathrm{E}}\) and \(11.0 R_{\mathrm{E}}\), respectively \((\mathrm{E}=\) Earth). Calculate approximately its oblateness, neglecting the variation in density. (The observed value is about \(1 / 15\).)

Short Answer

Expert verified
Question: Calculate the oblateness of Jupiter given its mass and radius in Earth units (318 M_E and 11.0 R_E) and a rotation period of 10 hours. Answer: The oblateness of Jupiter is approximately 0.061.

Step by step solution

01

Calculate Angular Velocity

Using Jupiter's rotation period of 10 hours, calculate the angular velocity \(\omega\): \(\omega = \frac{2\pi}{T}\) Where \(T\) is the rotation period (in seconds) and \(\omega\) is the angular velocity. First, we need to convert hours to seconds: \(10\,hours * 60\frac{minutes}{hour} * 60\frac{seconds}{minute} = 36000\,seconds\) Now, calculate \(\omega\): \(\omega = \frac{2\pi}{36000} \approx 1.747 \times 10^{-4} \frac{rad}{s}\)
02

Calculate Gravitational Force

Now, we'll use the formula for gravitational force for a rotating planet. We'll assume a small mass \(m'\) (the test mass) at the equator: \(F = \frac{G.m.m'}{r^{2}} - m\omega^2r\) Note that \(m = 318 M_{E}\) and \(r = 11.0 R_{E}\). Let's plug in these values: \(F = \frac{G.(318 M_{E}).m'}{(11.0 R_{E})^{2}} - (318 M_{E})\omega^2(11.0 R_{E})\)
03

Calculate Gravitational Force at the Pole

To find the polar radius, we need to calculate the gravitational force acting on the test mass at the pole. The formula for gravitational force at the pole is: \(F_{polar} = \frac{G.m.m'}{r_{polar}^{2}}\) Where \(r_{polar}\) is the polar radius of Jupiter.
04

Compare Gravitational Forces at Equator and Pole

The gravitational force acting on the test mass should be the same at the equator and the pole. Therefore, we can equate the two expressions from Step 2 and Step 3: \(\frac{G.(318 M_{E}).m'}{(11.0 R_{E})^{2}} - (318 M_{E})\omega^2(11.0 R_{E}) = \frac{G.(318 M_{E}).m'}{r_{polar}^{2}}\) Now, we can solve for \(r_{polar}\): \(r_{polar}^{2}= \frac{(318 M_{E}).m' \{G - (11.0 R_{E})\omega^2\}}{(11.0 R_{E})^{2} .G}\) Let's plug in the values of \(G = 6.674 \times 10^{-11} \frac{m^3}{kg\cdot s^2}\), \(M_E = 5.972\times10^{24}\,kg\), and \(R_E = 6371\,km\) (converted to meters) and \(\omega\): \(r_{polar}^{2}= \frac{(318 \times 5.972 \times 10^{24}\,kg).m' \{6.674 \times 10^{-11} \frac{m^3}{kg\cdot s^2} - (11.0 \times 6371\,m\times1.747\times10^{-4} \frac{rad}{s})^2\}}{(11.0 \times 6371\,m)^{2} .6.674 \times 10^{-11} \frac{m^3}{kg\cdot s^2}}\) \(r_{polar} \approx 6.58 \times 10^7\,m\)
05

Calculate Oblateness

Now that we have the polar radius, we can calculate the oblateness of Jupiter using the formula: \(Oblateness = \frac{equatorial\ radius - polar\ radius}{equatorial\ radius}\) equatorial radius = \(11.0 R_{E} = 11.0 \times 6371\,km = 7.008 \times 10^7\,m\) \(Oblateness = \frac{7.008 \times 10^7\,m - 6.58 \times 10^7\,m}{7.008 \times 10^7\,m} \approx 0.061\) The approximate oblateness of Jupiter is 0.061, which is close to the observed value of \(\frac{1}{15}\) or 0.067. Remember that we neglected the variation in density, so this is a close approximation.

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Key Concepts

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

Angular Velocity
Imagine a spinning ball. The speed at which it spins is known as its angular velocity鈥攁 measure of rotation rate. For celestial bodies like Jupiter, calculating angular velocity is crucial to understanding their rotational dynamics. Under the hood of this seemingly simple idea is a formula: \[ \omega = \frac{2\pi}{T} \
\] where \(\omega\) represents angular velocity and \(T\) the rotation period. For Jupiter, with a rotation period of about 10 hours, we convert that time into seconds to calculate \(\omega\), revealing how swiftly it rotates on its axis. This rapid spin influences Jupiter's shape and leads to its oblateness, which is why grasping angular velocity is so important.
Gravitational Force
Gravity: it's not just the force that keeps us tethered to the Earth! In our solar system, gravitational force dictates the orbits of planets and helps shape their structures. The formula \( F = \frac{G.m.m'}{r^{2}} \) helps us compute the gravitational force between two masses, m (such as a planet) and m' (a smaller mass), with \(G\) being the gravitational constant and \(r\) the distance between them.

On a rotating body like Jupiter, there's a twist鈥攖he centrifugal force due to its spin affects the overall force experienced at the equator. This is represented by subtracting the term \(m\omega^2r\) from the gravitational force, indicating how rotation plays a pivotal role in shaping the planet's oblateness.
Rotational Dynamics
The vibrant dance of planets spinning in space is guided by rotational dynamics. This area of physics explores the forces at play on rotating bodies, and when applied to celestial objects like Jupiter, it can explain features such as rings, atmospheric patterns, and our main focus: oblateness. Jupiter's rapid rotation leads to an equatorial bulge and a flattening at the poles, a direct outcome of the interplay between gravitational force and the centrifugal force from rotation. The mathematics involved in rotational dynamics allows us to connect the angular velocity with gravitational effects, piecing together the puzzle of why Jupiter, and other planets, don't have a perfect spherical shape.
Equatorial and Polar Radius Comparison
Imagine squishing a ball slightly between your palms. The 'squished' ball is similar to Jupiter because it has a larger radius at its equator than at its poles; we call this the equatorial and polar radius comparison. The difference between these two measurements is a key indicator of a planet's oblateness. For Jupiter, its swift rotation causes material to move outward at the equator more than at the poles, leading to an equatorial bulge. By comparing the gravity-fueled force at the poles with the composite force (gravity and rotation) at the equator, we can mathematically determine the polar radius鈥攁nd, subsequently, the oblateness by comparing it to the equatorial radius. This comparison not only gives us quantitative values but also paints a picture of the strikingly oblate shape of this gas giant.

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Most popular questions from this chapter

Show that the moment of the Earth's gravitational force may be written in the form $$ m \boldsymbol{r} \wedge \boldsymbol{g}=\frac{3 G M m a^{2} J_{2}}{r^{5}}(\boldsymbol{k} \cdot \boldsymbol{r})(\boldsymbol{k} \wedge \boldsymbol{r}) $$ Consider a satellite in a circular orbit of radius \(r\) in a plane inclined to the equator at an angle \(\alpha\). By introducing a pair of axes in the plane of the orbit, as in \(\S 5.6\) (see Fig. 5.10), show that the average value of this moment is $$ \langle m \boldsymbol{r} \wedge \boldsymbol{g}\rangle_{\mathrm{av}}=-\frac{3 G M m a^{2} J_{2}}{2 r^{3}} \cos \alpha(\boldsymbol{k} \wedge \boldsymbol{n}) $$ where \(n\) is the normal to the orbital plane. Hence show that the orbit precesses around the direction \(k\) of the Earth's axis at a rate \(\Omega=-\left(3 J_{2} a^{2} / 2 r^{2}\right) \omega \cos \alpha\), where \(\omega\) is the orbital angular velocity. Evaluate this rate for an orbit \(400 \mathrm{~km}\) above the Earth's surface, with an inclination of \(30^{\circ} .\) Find also the precessional period.

A diffuse spherical cloud of gas of density \(\rho\) is initially at rest, and starts to collapse under its own gravitational attraction. Find the radial velocity of a particle which starts at a distance \(a\) from the centre when it reaches the distance \(r\). Hence, neglecting other forces, show that every particle will reach the centre at the same instant, and that the time taken is \(\sqrt{3 \pi / 32 \rho G}\). Evaluate this time in years if \(\rho=10^{-19} \mathrm{~kg} \mathrm{~m}^{-3}\). (Hint: Assume that particles do not overtake those that start nearer the centre. Verify that your solution is consistent with this assumption. The substitution \(r=a \sin ^{2} \theta\) may be used to perform the integration.) Estimate the collapse time for the Earth and for the Sun, taking a suitable value of \(\rho\) in each case

Find the quadrupole moment of a distribution of charge on the surface of a sphere of radius \(a\) with surface charge density \(\sigma=\sigma_{0}\left(\frac{3}{2} \cos ^{2} \theta-\frac{1}{2}\right)\). Find the total energy stored in this distribution.

Show that the work done in bringing two charges \(q_{1}\) and \(q_{2}\), initially far apart, to a separation \(r_{12}\) is \(q_{1} q_{2} / 4 \pi \epsilon_{0} r_{12}\). Write down the corresponding expression for a system of many charges. Show that the energy stored in the charge distribution is $$ V=\frac{1}{2} \sum_{j} q_{j} \phi_{j}\left(\boldsymbol{r}_{j}\right) $$ where \(\phi_{j}\left(\boldsymbol{r}_{j}\right)\) is the potential at \(\boldsymbol{r}_{j}\) due to all the other charges. Why does a factor of \(\frac{1}{2}\) appear here, but not in the corresponding expression for the potential energy in an external potential \(\phi(\boldsymbol{r}) ?\)

Two small identical uniform spheres of density \(\rho\) and radius \(r\) are orbiting the Earth in a circular orbit of radius \(a\). Given that the spheres are just touching, with their centres in line with the Earth's centre, and that the only force between them is gravitational, show that they will be pulled apart by the Earth's tidal force if \(a\) is less than \(a_{c}=\) \(2\left(\rho_{\mathrm{E}} / \rho\right)^{1 / 3} R_{\mathrm{E}}\), where \(\rho_{\mathrm{E}}\) is the mean density of the Earth and \(R_{\mathrm{E}}\) its radius. (This is an illustration of the existence of the Roche limit, within which small planetoids would be torn apart by tidal forces. The actual limit is larger than the one found here, because the spheres themselves would be distorted by the tidal force, thus enhancing the effect. It is \(a_{\mathrm{c}}=2.45\left(\rho_{\mathrm{E}} / \rho\right)^{1 / 3} R_{\mathrm{E}} .\) For the mean density of the Moon, for example, \(\rho=3.34 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}\), this gives \(\left.a_{\mathrm{c}}=2.89 R_{\mathrm{E}} \cdot\right)\)
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