Chapter 8.3, Problem 56E

Solution

The value of $a$ , $b$ , $c$ , $e$ and the coordinates of the center of the Sun if the center of the hyperbolic orbit is $\left(0,0\right)$ and the Sun lies on the positive $x$ axis.

It has been determined that $a=156.8$ , $b=252$ , $c=296.8$ and $e\approx 1.89$ .

It has also been determined that the coordinates of the center of the Sun are $\left(296.8,0\right)$ .

Given:

A comet following a hyperbolic path about the Sun has a perihelion of $140$ Gm. When the line from the comet to the Sun is perpendicular to the focal axis of the orbit, the comet is $405$ Gm from the Sun.

Concept used:

The perihelion distance of a hyperbola is $c-a$ .

Calculation:

It is given that a comet follows a hyperbolic path about the Sun such that the center of the hyperbolic orbit is $\left(0,0\right)$ and the Sun lies on the positive $x$ axis.

Then, the equation of the hyperbola is given as $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ .

As discussed, the perihelion distance of a hyperbola is $c-a$ .

It is given that the comet following a hyperbolic path about the Sun has a perihelion of $140$ Gm.

Then, $c-a=140$ .

Now, for a hyperbola, $c^2=a^2+b^2$ .

Put $x=c$ in $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ to get,

  $\begin{array}{c}\frac{c^2}{a^2}-\frac{y^2}{b^2}=1\\ \frac{y^2}{b^2}=\frac{c^2}{a^2}-1\\ y^2=b^2\left(\frac{c^2}{a^2}-1\right)\end{array}$

Put $c^2=a^2+b^2$ in $y^2=b^2\left(\frac{c^2}{a^2}-1\right)$ to get,

  $\begin{array}{c}{y}^2=b^2\left(\frac{a^2+b^2}{a^2}-1\right)\\ =b^2\left(1+\frac{b^2}{a^2}-1\right)\\ =b^2\left(\frac{b^2}{a^2}\right)\\ =\frac{b^4}{a^2}\end{array}$

So, $y=\pm \frac{b^2}{a}$ .

It is given that when the line from the comet to the Sun is perpendicular to the focal axis of the orbit, the comet is $405$ Gm from the Sun.

This implies that $\frac{b^2}{a}=405$ .

Simplifying,

  $b^2=405a$

Put $b^2=405a$ in $c^2=a^2+b^2$ to get,

  $c^2=a^2+405a$

Simplifying,

  $c^2-a^2=405a$

On further simplification,

  $\left(c-a\right)\left(c+a\right)=405a$

Put $c-a=140$ in the above equation to get,

  $140\left(c+a\right)=405a$

Simplifying,

  $140c+140a=405a$

On further simplification,

  $140c=265a$

Now, $c-a=140$ . Then, $c=140+a$ .

Put $c=140+a$ in $140c=265a$ to get,

  $\begin{array}{c}140\left(140+a\right)=265a\\ 19600+140a=265a\\ 125a=19600\\ a=156.8\end{array}$

So, $a=156.8$ .

Put $a=156.8$ in $b^2=405a$ to get,

  $b^2=63504$

Simplifying,

  $b=\pm 252$

Taking the positive value, $b=252$ .

Put $a=156.8$ in $c=140+a$ to get,

  $c=296.8$

Put $c=296.8$ and $a=156.8$ in $e=\frac{c}{a}$ to get,

  $e=\frac{296.8}{156.8}$

Simplifying,

  $e\approx 1.89$

Now, the Sun is at the focus, which is the point $\left(c,0\right)$ .

Then, the coordinates of the center of the Sun are $\left(296.8,0\right)$ .

Conclusion:

It has been determined that $a=156.8$ , $b=252$ , $c=296.8$ and $e\approx 1.89$ .

It has also been determined that the coordinates of the center of the Sun are $\left(296.8,0\right)$ .