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Chapter 4: Problem 31

Show that Kepler's third law, \(\tau \propto a^{3 / 2}\), implies that the force on a planet is proportional to its mass. [This law was originally expressed by Kepler as \(\tau \propto \bar{r}^{3 / 2}\), where \(\bar{r}\) is a 'mean value' of \(r\). For an ellipse, the mean over angle \(\theta\) is in fact \(b\); the mean over time is actually \(a\left(1+\frac{1}{2} e^{2}\right)\); it is the mean over are length - or the median - which is given by \(a !\) Of course, for most planets in our Solar System these values are almost equal.]

Short Answer

Expert verified
Answer: Yes, the force on a planet is proportional to its mass.

Step by step solution

01

Write down the given expression:

Kepler's third law states that the orbital period \(\tau\) of a planet is proportional to the semi-major axis \(a\) raised to the power of 3/2, which is represented mathematically as \(\tau \propto a^{3/2}\).
02

Introduce a constant of proportionality:

Introduce a constant of proportionality \(k\) to transform the proportionality into an equation: \(\tau = k a^{3/2}\).
03

Express the orbital period in terms of velocity and semi-major axis:

The orbital period \(\tau\) can be expressed in terms of the orbital velocity \(v\) and the semi-major axis \(a\) using the formula \(\tau = \frac{2 \pi a}{v}\), where \(2 \pi a\) is the circumference of the elliptical orbit.
04

Substitute the expression for the orbital period in the previous equation:

Replace \(\tau\) in the equation \(\tau = k a^{3/2}\) with the expression obtained in step 3: \(\frac{2 \pi a}{v}=k a^{3/2}\).
05

Rearrange the equation to solve for velocity squared:

Solve the equation for \(v^2\) by rearranging and multiplying both sides by \(\frac{v^2}{2 \pi a}\): \(v^2=\frac{(2 \pi a)^2}{k^2 a^{3}}\).
06

Use Newton's second law of motion and the centripetal force formula:

According to Newton's second law of motion, the force acting on a planet of mass \(m\) is \(F = ma\), and the centripetal force acting on it is \(Fc = \frac{mv^2}{r}\). Since we are considering elliptical orbits, we will use the semi-major axis \(a\) as a generalization of the circular radius \(r\). Thus, the centripetal force becomes \(Fc = \frac{mv^2}{a}\).
07

Substitute the expression for velocity squared in the centripetal force formula:

Replace \(v^2\) in the formula \(Fc = \frac{mv^2}{a}\) with the expression found in step 5: \(Fc = \frac{m(2 \pi a)^2}{k^2 a^{3} \cdot a}\).
08

Simplify the equation and show the proportionality:

Simplify the equation obtained in step 7: \(Fc = \frac{4 \pi^2 a^2 m}{k^2 a^3}\). Further simplification leads to \(Fc = \frac{4 \pi^2 m}{k^2 a}\), which shows that the force on a planet is proportional to its mass, as the other terms are constants. The final expression can be written as \(Fc \propto m\), as required.

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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 ? = ?.

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 \(\Omega \approx\left(c /|k| a^{2}\right) \omega\).

A beam of particles strikes a wall containing \(2 \times 10^{29}\) atoms per \(\mathrm{m}^{3}\). Each atom behaves like a sphere of radius \(3 \times 10^{-15} \mathrm{~m}\). Find the thickness of wall that exactly half the particles will penetrate without scattering. What thickness would be needed to stop all but one particle in \(10^{6} ?\)

*Find the polar equation of the orbit of an isotropic harmonic oscillator by solving the differential equation (4.25), and verify that it is an ellipse with centre at the origin. (Hint: Change to the variable \(v=u^{2}\).) Check also that the period is given correctly by \(\tau=2 m A / J\).
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