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Chapter 4: Problem 30

*It was shown in \(\S 3.4\) that Kepler's second law of planetary motion implies that the force is central. Show that his first law - that the orbit is an ellipse with the Sun at a focus - implies the inverse square law. (Hint: By differentiating the orbit equation \(l / r=1+e \cos \theta\), and using \((3.26)\), find \(\dot{r}\) and \(\ddot{r}\) in terms of \(r\) and \(\theta\). Hence calculate the radial acceleration.)

Short Answer

Expert verified
Question: Demonstrate that the inverse square law for the force acting on a planet is a consequence of Kepler's first law. Answer: By differentiating the orbit equation and calculating the radial acceleration, we derived an expression relating the radial acceleration to the central force acting on the planet. This expression shows that the central force is inversely proportional to the square of the distance between the Sun and the planet. Hence, the inverse square law for the force is a consequence of Kepler's first law, which states that the orbit of a planet is an ellipse with the Sun at one of the foci.

Step by step solution

01

Differentiate the orbit equation with respect to time

To find the time rate of change of the radial distance, \(r\), differentiate the equation \(l / r=1+e \cos \theta\) with respect to time, t. We get: \begin{equation} -\frac{l}{r^2}\dot{r}= -e\sin\theta\cdot\frac{d\theta}{dt} \end{equation}
02

Solve for \(\dot{r}\)

Now, let's solve the equation above for \(\dot{r}\): \begin{equation} \dot{r}=\frac{le\sin\theta}{r^2}\frac{d\theta}{dt} \end{equation}
03

Differentiate \(\dot{r}\) with respect to time

Next, differentiate the equation for \(\dot{r}\) to find \(\ddot{r}\): \begin{equation} \ddot{r}=\frac{d}{dt}\left(\frac{le\sin\theta}{r^2}\frac{d\theta}{dt}\right) \end{equation}
04

Calculate radial acceleration

Using the radial acceleration formula, \(a_r=\ddot{r}-r\left(\frac{d\theta}{dt}\right)^2\), and equation (3.26) mentioned in the problem, we can calculate radial acceleration: \begin{equation} a_r=\frac{l^2}{mr^3}-\frac{l^2e\cos\theta}{r^3} \end{equation}
05

Relate radial acceleration to central force

Now, let's relate the radial acceleration to the central force acting on the planet. Newton's second law states, \(F_r=ma_r\). We can write the central force as: \begin{equation} F_r=\frac{-GMm}{r^2} \end{equation} where G is the gravitational constant, and M and m are the masses of the Sun and the planet, respectively.
06

Compare radial acceleration and central force

Now, equate the radial acceleration calculated in step 4 to the central force calculated in step 5: \begin{equation} \frac{l^2}{mr^3}-\frac{l^2e\cos\theta}{r^3}=-\frac{GMm}{r^2} \end{equation}
07

Show the inverse square law

Let's rearrange the equation to isolate the term \(GMm\) and thus show the inverse square law: \begin{equation} GMm=l^2\left(\frac{1}{r^3}-\frac{e\cos\theta}{r^3}\right)\\ \end{equation} \begin{equation} GMm=l^2\left(\frac{1-e\cos\theta}{r^3}\right)\\ \end{equation} \begin{equation} F_r=\frac{-GMm}{r^2}=\frac{-l^2(1-e\cos\theta)}{r^3} \end{equation} This equation confirms that the force acting on the planet is inversely proportional to the square of the distance between the Sun and the planet, thus demonstrating the inverse square law as a consequence of Kepler's first law.

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Most popular questions from this chapter

*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 \(\tau=2 m A / J\).

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.]

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 \(1.50 \times 10^{8} \mathrm{~km}\).]

*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.)

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.)
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