Chapter 8.5, Problem 51E

Solution

a)

To prove:that the perihelion distance of any planet is $a\left(1-e\right)$ and the aphelion distance is $a\left(1+e\right)$ using the fact that $-1\le \cos \theta \le 1$ .

It is proved that the perihelion distance of any planet is $a\left(1-e\right)$ and the aphelion distance is $a\left(1+e\right)$ .

Given information:

The polar equation: $r=\frac{a\left(1-e^2\right)}{\left(1+e\cos \theta \right)}$

Formula used:

  $a^2-b^2=\left(a-b\right)\left(a+b\right)$

Calculation:

The perihelion distance is defined as the closest distance of any planet to the sun.

So, given the fact that $-1\le \cos \theta \le 1$ , substitute $1$ for $\cos \theta$ in the polar equation.

Therefore

  $r=\frac{a\left(1-e^2\right)}{\left(1+e\cdot 1\right)}$

Factorize the numerator using the identity $a^2-b^2=\left(a-b\right)\left(a+b\right)$ and cancel out the common factors.

  $\begin{array}{c}r=\frac{a\left(1-e\right)\left(1+e\right)}{\left(1+e\right)}\\ =a\left(1-e\right)\end{array}$

Hence the perihelion distance is $a\left(1-e\right)$ .

The aphelion distance is defined as the furthest distance of any planet to the sun.

So, given the fact that $-1\le \cos \theta \le 1$ , substitute $-1$ for $\cos \theta$ in the polar equation.

Therefore

  $r=\frac{a\left(1-e^2\right)}{\left(1+e\cdot \left(-1\right)\right)}$

Factorize the numerator using the identity $a^2-b^2=\left(a-b\right)\left(a+b\right)$ and cancel out the common factors.

  $\begin{array}{c}r=\frac{a\left(1-e\right)\left(1+e\right)}{\left(1-e\right)}\\ =a\left(1+e\right)\end{array}$

Hence the aphelion distance is $a\left(1+e\right)$ .

Therefore, it is proved that the perihelion distance of any planet is $a\left(1-e\right)$ and the aphelion distance is $a\left(1+e\right)$ .

b)

To prove: that $a\left(1-e\right)=a-c$ and $a\left(1+e\right)=a+c$ .

Given information:

The polar equation: $r=\frac{a\left(1-e^2\right)}{\left(1+e\cos \theta \right)}$

Formula used:

  $e=\frac{c}{a}$

Calculation:

Substitute $\frac{c}{a}$ for $e$ in $a\left(1-e\right)$ .

  $\begin{array}{l}a\left(1-\frac{c}{a}\right)=a-a\cdot \frac{c}{a}\\ =a-c\end{array}$

Hence it is proved that $a\left(1-e\right)=a-c$ .

Substitute $\frac{c}{a}$ for $e$ in $a\left(1+e\right)$ .

  $\begin{array}{l}a\left(1+\frac{c}{a}\right)=a+a\cdot \frac{c}{a}\\ =a+c\end{array}$

Hence it is proved that $a\left(1+e\right)=a+c$ .

c)

To compute:the perihelion and aphelion distances of each planet listed in the table.

The perihelion and aphelion distances are tabulated as follows:

Planet	Semi major Axis (AU)	Eccentricity	Perihelion	Aphelion
Mercury	0.3871	0.206	$0.307$	$0.467$
Venus	0.7233	0.007	$0.718$	$0.728$
Earth	1.0000	0.017	$0.983$	$1.017$
Mars	1.5237	0.093	$1.382$	$1.665$
Jupiter	5.2026	0.048	$4.953$	$5.452$
Saturn	9.5547	0.056	$9.02$	$10.09$

Given information:

The semimajor axes and eccentricities of the six innermost planets.

Planet	Semimajor Axis (AU)	Eccentricity
Mercury	0.3871	0.206
Venus	0.7233	0.007
Earth	1.0000	0.017
Mars	1.5237	0.093
Jupiter	5.2026	0.048
Saturn	9.5547	0.056

Source: Encrenaz and Bibring The Solar System (2^nd ed.) New York: Springer, p-5.

Formula used:

Formula to compute perihelion distance: $a\left(1-e\right)$

Formula to computer aphelion distance: $a\left(1+e\right)$

Calculation:

To find the perihelion and aphelion distances of Mercury:

Substitute $0.3871$ for $a$ and $0.206$ for $b$ in perihelion formula.

  $\begin{array}{l}0.3871\left(1-0.206\right)=0.3871\left(0.794\right)\\ \approx 0.307\end{array}$

Substitute $0.3871$ for $a$ and $0.206$ for $b$ in aphelion formula.

  $\begin{array}{l}0.3871\left(1+0.206\right)=0.3871\left(1.206\right)\\ \approx 0.467\end{array}$

Therefore, the perihelion and aphelion distances for Mercury are $0.307$ and $0.467$ respectively.

To find the perihelion and aphelion distances of Venus:

Substitute $0.3871$ for $a$ and $0.206$ for $b$ in perihelion formula.

  $\begin{array}{l}0.7233\left(1-0.007\right)=0.7233\left(0.993\right)\\ \approx 0.718\end{array}$

Substitute $0.3871$ for $a$ and $0.206$ for $b$ in aphelion formula.

  $\begin{array}{l}0.7233\left(1+0.007\right)=0.7233\left(1.007\right)\\ \approx 0.728\end{array}$

Therefore, the perihelion and aphelion distances for Venus are $0.718$ and $0.728$ respectively.

To find the perihelion and aphelion distances of Earth:

Substitute $1.0000$ for $a$ and $0.017$ for $b$ in perihelion formula.

  $\begin{array}{l}1.0000\left(1-0.017\right)=1.0000\left(0.983\right)\\ \approx 0.983\end{array}$

Substitute $0.3871$ for $a$ and $0.206$ for $b$ in aphelion formula.

  $\begin{array}{l}1.0000\left(1+0.017\right)=1.0000\left(1.017\right)\\ \approx 1.017\end{array}$

Therefore, the perihelion and aphelion distances for Earth are $0.983$ and $1.017$ respectively.

To find the perihelion and aphelion distances of Mars:

Substitute $1.5237$ for $a$ and $0.093$ for $b$ in perihelion formula.

  $\begin{array}{l}1.5237\left(1-0.093\right)=1.5237\left(0.907\right)\\ \approx 1.382\end{array}$

Substitute $0.3871$ for $a$ and $0.206$ for $b$ in aphelion formula.

  $\begin{array}{l}1.5237\left(1+0.093\right)=1.5237\left(1.093\right)\\ \approx 1.665\end{array}$

Therefore, the perihelion and aphelion distances for Mars are $1.382$ and $1.665$ respectively.

To find the perihelion and aphelion distances of Jupiter:

Substitute $5.2026$ for $a$ and $0.048$ for $b$ in perihelion formula.

  $\begin{array}{l}5.2026\left(1-0.048\right)=5.2026\left(0.952\right)\\ \approx 4.953\end{array}$

Substitute $0.3871$ for $a$ and $0.206$ for $b$ in aphelion formula.

  $\begin{array}{l}5.2026\left(1+0.048\right)=5.2026\left(1.048\right)\\ \approx 5.452\end{array}$

Therefore, the perihelion and aphelion distances for Jupiter are $4.953$ and $5.452$ respectively.

To find the perihelion and aphelion distances of Saturn:

Substitute $9.5547$ for $a$ and $0.056$ for $b$ in perihelion formula.

  $\begin{array}{l}9.5547\left(1-0.056\right)=9.5547\left(0.944\right)\\ \approx 9.020\end{array}$

Substitute $0.3871$ for $a$ and $0.206$ for $b$ in aphelion formula.

  $\begin{array}{l}9.5547\left(1+0.056\right)=9.5547\left(1.056\right)\\ \approx 10.090\end{array}$

Therefore, the perihelion and aphelion distances for Saturn are $9.02$ and $10.09$ respectively.

The perihelion and aphelion distances are tabulated as follows:

Planet	Semimajor Axis (AU)	Eccentricity	Perihelion	Aphelion
Mercury	0.3871	0.206	$0.307$	$0.467$
Venus	0.7233	0.007	$0.718$	$0.728$
Earth	1.0000	0.017	$0.983$	$1.017$
Mars	1.5237	0.093	$1.382$	$1.665$
Jupiter	5.2026	0.048	$4.953$	$5.452$
Saturn	9.5547	0.056	$9.02$	$10.09$

d)

To find:the greatest perihelion and aphelion distances.

The planet Saturn has the greatest difference between the perihelion and aphelion and the difference is $1.07$ .

Given information:

The polar equation: $r=\frac{a\left(1-e^2\right)}{\left(1+e\cos \theta \right)}$

Formula used:

Formula to compute perihelion distance: $a\left(1-e\right)$

Formula to computer aphelion distance: $a\left(1+e\right)$

Calculation:

The difference between perihelion and aphelion distances are tabulated as follows:

Planet	Perihelion	Aphelion	Difference between Perihelion and Aphelion
Mercury	$0.307$	$0.467$	$0.467-0.307=0.16$
Venus	$0.718$	$0.728$	$0.728-0.718=0.01$
Earth	$0.983$	$1.017$	$1.017-0.983=0.034$
Mars	$1.382$	$1.665$	$1.665-1.382=0.283$
Jupiter	$4.953$	$5.452$	$5.452-4.953=0.499$
Saturn	$9.02$	$10.09$	$10.09-9.02=1.07$

The greatest difference between the perihelion and aphelion is the least value in the last column.

Therefore, the planet Saturn has the greatest difference between the perihelion and aphelion and the difference is $1.07$ .