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Chapter 5: Problem 1

Find the centrifugal acceleration at the equator of the planet Jupiter and of the Sun. In each case, express your answer also as a fraction of the surface gravity. (The rotation periods are 10 hours and 27 days, respectively, the radii \(7.1 \times 10^{4} \mathrm{~km}\) and \(7.0 \times 10^{5} \mathrm{~km}\), and the masses \(1.9 \times 10^{27} \mathrm{~kg}\) and \(\left.2.0 \times 10^{30} \mathrm{~kg} .\right)\)

Short Answer

Expert verified
Question: Calculate the centrifugal acceleration at the equator of Jupiter and the Sun, and express the answer as a fraction of the surface gravity for both. Answer: The centrifugal acceleration at the equator of Jupiter is approximately \(4.30\mathrm{~m/s^2}\), or \(0.173\) times the surface gravity. The centrifugal acceleration at the equator of the Sun is approximately \(5.82 \times 10^{-3}\mathrm{~m/s^2}\), or \(2.12 \times 10^{-5}\) times the surface gravity.

Step by step solution

01

Calculate linear velocity

First, we will calculate the linear velocity of Jupiter and the Sun using the formula \(v = \frac{2 \pi R}{T}\). We have \(R_J=7.1 \times 10^4 \mathrm{~km}\), \(T_J=10\mathrm{~hours}\) for Jupiter, and \(R_\odot=7.0 \times 10^5 \mathrm{~km}\), \(T_\odot=27\mathrm{~days}\) for the Sun. For Jupiter: \(v_J = \frac{2 \pi (7.1 \times 10^7\mathrm{~m})}{10\mathrm{~hours} \times 3600\mathrm{~s/hour}} \approx 1.75 \times 10^4\mathrm{~m/s}\) For the Sun: \(v_\odot = \frac{2 \pi (7.0 \times 10^8\mathrm{~m})}{27\mathrm{~days} \times 24\mathrm{~hours/day} \times 3600\mathrm{~s/hour}} \approx 2.02 \times 10^3\mathrm{~m/s}\)
02

Calculate centrifugal acceleration

Now, calculate the centrifugal acceleration using the formula \(a_c = \frac{v^2}{R}\). For Jupiter: \(a_{cJ} = \frac{(1.75 \times 10^4\mathrm{~m/s})^2}{7.1 \times 10^7\mathrm{~m}} \approx 4.30 \times 10^0\mathrm{~m/s^2}\) For the Sun: \(a_{c\odot} = \frac{(2.02 \times 10^3\mathrm{~m/s})^2}{7.0 \times 10^8\mathrm{~m}} \approx 5.82 \times 10^{-3}\mathrm{~m/s^2}\)
03

Calculate surface gravity

Calculate the surface gravity using the formula \(g = \frac{GM}{R^2}\). We have \(M_J=1.9 \times 10^{27} \mathrm{~kg}\) for Jupiter, and \(M_\odot=2.0 \times 10^{30}\mathrm{~kg}\) for the Sun. The gravitational constant is \(G = 6.674 \times 10^{-11}\mathrm{~m^3/kg~s^2}\). For Jupiter: \(g_J = \frac{(6.674 \times 10^{-11}\mathrm{~m^3/kg~s^2})(1.9 \times 10^{27}\mathrm{~kg})}{(7.1 \times 10^7\mathrm{~m})^2} \approx 24.79\mathrm{~m/s^2}\) For the Sun: \(g_\odot = \frac{(6.674 \times 10^{-11}\mathrm{~m^3/kg~s^2})(2.0 \times 10^{30}\mathrm{~kg})}{(7.0 \times 10^8\mathrm{~m})^2} \approx 274.07\mathrm{~m/s^2}\)
04

Express centrifugal acceleration as a fraction of surface gravity

Divide the calculated centrifugal acceleration by the surface gravity to express the centrifugal acceleration as a fraction of the surface gravity. For Jupiter: \(\frac{a_{cJ}}{g_J} = \frac{4.30 \times 10^0\mathrm{~m/s^2}}{24.79\mathrm{~m/s^2}} \approx 0.173\) For the Sun: \(\frac{a_{c\odot}}{g_\odot} = \frac{5.82 \times 10^{-3}\mathrm{~m/s^2}}{274.07\mathrm{~m/s^2}} \approx 2.12 \times 10^{-5}\) The centrifugal acceleration at the equator of Jupiter is approximately \(4.30\mathrm{~m/s^2}\), or \(0.173\) times the surface gravity. The centrifugal acceleration at the equator of the Sun is approximately \(5.82 \times 10^{-3}\mathrm{~m/s^2}\), or \(2.12 \times 10^{-5}\) times the surface gravity.

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Key Concepts

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

Linear Velocity
Understanding linear velocity is essential when examining the movement of celestial bodies. Let's start with its definition. Linear velocity, symbolized as 'v', is the rate at which an object covers distance in a particular direction, and it is a vector quantity having both magnitude and direction. In the context of a rotating planet or the Sun, the linear velocity at the equator can be determined by the formula:
\( v = \frac{2 \times \pi \times R}{T} \)
Here, \( R \) is the radius of the celestial body and \( T \) is the rotation period. The rotation period is the time it takes for the celestial body to complete one full rotation on its axis. It's expressed in seconds, minutes, or hours.
  • For Jupiter with a radius of 7.1 x 104 km and rotation period of 10 hours, the linear velocity is high due to the short rotation period.
  • For the Sun with a much larger radius of 7.0 x 105 km but a longer rotation period of 27 days, the linear velocity is considerably lower.
Linear velocity is crucial, as it directly influences the centrifugal force experienced by objects at the equator, impacting the distribution of matter and atmospheric phenomena on the celestial body.
Surface Gravity
Surface gravity, often represented by 'g', is the gravitational acceleration experienced by an object on the surface of a celestial body. It is determined by the mass of the body and the distance from its center, calculated using the formula:
\( g = \frac{GM}{R^2} \)
where \( G \) is the gravitational constant, \( M \) is the mass of the celestial body, and \( R \) is its radius. Surface gravity gives us an understanding of how strong a body’s gravitational pull is at its surface.
  • For example, Jupiter has a much lower surface gravity than the Sun due to its smaller mass relative to its radius.
  • Surface gravity influences an object's weight and the escape velocity needed for it to leave the surface of the celestial body.
Understanding surface gravity is essential when examining why objects, including atmosphere and liquids, behave differently on different celestial bodies. It also affects the body's overall shape and the height of its mountains and depressions.
Rotation Periods of Celestial Bodies
The rotation period of a celestial body is the time it takes to complete one full spin around its axis. This spin determines the length of a day on the body, affecting the life cycles of potential inhabitants and influencing the planet's climate and weather patterns.
  • Shorter rotation periods, like Jupiter's 10 hours, can lead to a more pronounced bulging at the equator, increased centrifugal force, and potentially stronger atmospheric dynamics.
  • Longer rotation periods, such as the Sun's 27-day cycle, result in less bulging and relatively weaker centrifugal forces.
The rotation period is also a factor in calculating both the linear velocity at the equator and the centrifugal acceleration, which contributes to the feeling of apparent weightlessness for objects on the equator. Knowledge of a celestial body's rotation period helps scientists understand the dynamics of its atmosphere, the weather patterns, and how the body interacts with moons and rings in its orbit.

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Most popular questions from this chapter

The water in a circular lake of radius \(1 \mathrm{~km}\) in latitude \(60^{\circ}\) is at rest relative to the Earth. Find the depth by which the centre is depressed relative to the shore by the centrifugal force. For comparison, find the height by which the centre is raised by the curvature of the Earth's surface. (Earth radius \(=6400 \mathrm{~km}\).)

The co-ordinates \((x, y, z)\) of a particle with respect to a uniformly rotating frame may be related to those with respect to a fixed inertial frame, \(\left(x^{*}, y^{*}, z^{*}\right)\), by the transformation $$ \left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{ccc} \cos \omega t & \sin \omega t & 0 \\ -\sin \omega t & \cos \omega t & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{c} x^{*} \\ y^{*} \\ z^{*} \end{array}\right] $$ (Here, we use matrix notation: this stands for three separate equations, \(x=\cos \omega t \cdot x^{*}+\sin \omega t \cdot y^{*}\) etc.) Write down the inverse relation giving \(\left(x^{*}, y^{*}, z^{*}\right)\) in terms of \((x, y, z)\). By differentiating with respect to \(t\), rederive the relation (5.15) between \(\mathrm{d}^{2} \boldsymbol{r} / \mathrm{d} t^{2}\) and \(\ddot{\boldsymbol{r}}\). [Hint: Note that \(\ddot{\boldsymbol{r}}=(\ddot{x}, \ddot{y}, \ddot{z})\), while \(\mathrm{d}^{2} \boldsymbol{r} / \mathrm{d} t^{2}\) is the vector obtained by applying the above transformation \(\left.\operatorname{to}\left(\ddot{x}^{*}, \ddot{y}^{*}, \ddot{z}^{*}\right) .\right]\)

The orbit of an electron (charge \(-e)\) around a nucleus (charge \(Z e\) ) is a circle of radius \(a\) in a plane perpendicular to a uniform magnetic field \(\boldsymbol{B}\). By writing the equation of motion in a frame rotating with the electron, show that the angular velocity \(\omega\) is given by one of the roots of the equation \(m \omega^{2}-e B \omega-Z e^{2} / 4 \pi \epsilon_{0} a^{3}=0\) Verify that for small values of \(B\), this agrees with \(\S 5.5\). Evaluate the two roots if \(B=10^{5} \mathrm{~T}, Z=1\) and \(a=5.3 \times 10^{-11} \mathrm{~m}\). (Note, however, that in reality \(a\) would be changed by the field.)

Write down the equation of motion for a charged particle in uniform, parallel electric and magnetic fields, both in the \(z\)-direction, and solve it, given that the particle starts from the origin with velocity \((v, 0,0)\). A screen is placed at \(x=a\), where \(a \ll m v / q B .\) Show that the locus of points of arrival of particles with given \(m\) and \(q\), but different speeds \(v\), is approximately a parabola. How does this locus depend on \(m\) and \(q ?\)

Find the equations of motion for a particle in a frame rotating with variable angular velocity \(\boldsymbol{\omega}\), and show that there is another apparent force of the form \(-m \dot{\boldsymbol{\omega}} \wedge \boldsymbol{r}\). Discuss the physical origin of this force.
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