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Chapter 4: Problem 2

Find the radii of synchronous orbits about Jupiter and about the Sun. [Their mean rotation periods are 10 hours and 27 days, respectively. The mass of Jupiter is 318 times that of the Earth. The semi-major axis of the Earth's orbit, or astronomical unit (AU) is \(1.50 \times 10^{8} \mathrm{~km}\).]

Short Answer

Expert verified
Answer: The radii of the synchronous orbits around Jupiter and the Sun are approximately \(2.89 \times 10^5 \mathrm{~km}\) and \(1.97 \times 10^8 \mathrm{~km}\), respectively.

Step by step solution

01

Kepler's Third Law for orbital period

Kepler's Third Law states that the cube of the semi-major axis of an orbit is directly proportional to the square of the orbital period, i.e., \(\frac{T^2}{r^3} = \frac{4\pi^2}{GM}\), where \(T\) is the orbital period, \(r\) is the radius of the orbit, \(G\) is the gravitational constant, and \(M\) is the mass of the central object (Jupiter or the Sun in this case). In the case of a synchronous orbit, the orbital period \(T\) is equal to the rotation period of the central object. Step 2: Finding the synchronous orbit radius for Jupiter
02

Calculate the mass and rotation period of Jupiter

Given that the mass of Jupiter is 318 times that of Earth, we can find its mass by multiplying the mass of Earth by 318: \(M_J = 318 M_\oplus = 318 \times 5.972 \times 10^{24} \mathrm{~kg}\). The rotation period of Jupiter is given as 10 hours. Convert this to seconds: \(T_J = 10 \times 3600 \mathrm{~s}\). Step 3: Calculate the radius of the synchronous orbit around Jupiter
03

Applying Kepler's Third Law for Jupiter

Now we can use Kepler's Third Law to find the radius of the synchronous orbit around Jupiter: \(\frac{T_J^2}{r_J^3} = \frac{4\pi^2}{GM_J}\). Solving for \(r_J\), we get: \(r_J = \sqrt[3]{\frac{T_J^2GM_J}{4\pi^2}}\). Step 4: Finding the synchronous orbit radius for the Sun
04

Calculate the mass and rotation period of the Sun

The mass of the Sun is approximately \(M_\odot = 1.989 \times 10^{30} \mathrm{~kg}\). The rotation period of the Sun is given as 27 days. Convert this to seconds: \(T_\odot = 27 \times 24 \times 3600 \mathrm{~s}\). Step 5: Calculate the radius of the synchronous orbit around the Sun
05

Applying Kepler's Third Law for the Sun

Now we can use Kepler's Third Law to find the radius of the synchronous orbit around the Sun: \(\frac{T_\odot^2}{r_\odot^3} = \frac{4\pi^2}{GM_\odot}\). Solving for \(r_\odot\), we get: \(r_\odot = \sqrt[3]{\frac{T_\odot^2GM_\odot}{4\pi^2}}\). Step 6: Compute the values
06

Calculate the radii of the synchronous orbits

Plug the values for the gravitational constant, the mass, and rotation periods of Jupiter and the Sun into the expressions from Steps 3 and 5 to obtain the radii of their synchronous orbits. For Jupiter: \(r_J \approx \sqrt[3]{\frac{(10\times3600)^2(6.674\times10^{-11})(318\times(5.972\times10^{24}))}{4\pi^2}} \approx 2.89 \times 10^5 \mathrm{~km}\). For the Sun: \(r_\odot \approx \sqrt[3]{\frac{(27\times24\times3600)^2(6.674\times10^{-11})(1.989\times10^{30})}{4\pi^2}} \approx 1.97 \times 10^8 \mathrm{~km}\). The radii of the synchronous orbits around Jupiter and the Sun are approximately \(2.89 \times 10^5 \mathrm{~km}\) and \(1.97 \times 10^8 \mathrm{~km}\), respectively.

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Most popular questions from this chapter

. A particle of mass m moves under the action of a harmonic oscillator force with potential energy 1 2 kr2. Initially, it is moving in a circle of radius a. Find the orbital speed v. It is then given a blow of impulse mv in a direction making an angle ? with its original velocity. Use the conservation laws to determine the minimum and maximum distances from the origin during the subsequent motion. Explain your results physically for the two limiting cases ? = 0 and ? = ?.

Show that the comet discussed at the end of \(\$ 4.4\) crosses the Earth's orbit at opposite ends of a diameter. Find the time it spends inside the Earth's orbit. (To evaluate the area required, write the equation of the orbit in Cartesian co-ordinates. See Appendix B.)

*If the Earth's orbit is divided in two by the latus rectum, show that the difference in time spent in the two halves, in years, is \frac{2}{\pi}\left(e \sqrt{1-e^{2}}+\arcsin e\right) and hence for small e about twice as large as the difference computed in the example in \(\S 4.4\). (Hint: Use Cartesian co-ordinates to evaluate the required area. The identity \(\pi / 2-\arcsin \sqrt{1-e^{2}}=\arcsin e\) may be useful.)

A star of mass M and radius R is moving with velocity v through a cloud of particles of density ?. If all the particles that collide with the star are trapped by it, show that the mass of the star will increase at a rate $$ \frac{\mathrm{d} M}{\mathrm{~d} t}=\pi \rho v\left(R^{2}+\frac{2 G M R}{v^{2}}\right) $$ Given that \(M=10^{31} \mathrm{~kg}\) and \(R=10^{8} \mathrm{~km}\), find how the effective crosssectional area compares with the geometric cross-section \(\pi R^{2}\) for velocities of \(1000 \mathrm{~km} \mathrm{~s}^{-1}, 100 \mathrm{~km} \mathrm{~s}^{-1}\) and \(10 \mathrm{~km} \mathrm{~s}^{-1}\).

*The potential energy of a particle of mass \(m\) is \(V(r)=k / r+c / 3 r^{3}\), where \(k<0\) and \(c\) is a small constant. (The gravitational potential energy in the equatorial plane of the Earth has approximately this form, because of its flattened shape - see Chapter 6.) Find the angular velocity \(\omega\) in a circular orbit of radius \(a\), and the angular frequency \(\omega^{\prime}\) of small radial oscillations about this circular orbit. Hence show that a nearly circular orbit is approximately an ellipse whose axes precess at an angular velocity \(\Omega \approx\left(c /|k| a^{2}\right) \omega\).
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