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I could not solve the following problem, please help me

Suppose that a satellite is rotating in an elliptical orbit keeping the earth at the center point. Show using Lagrange theory, that the maximum and minimum distances of the satellite from the earth are half of the lengths of the major and minor axes, of the corresponding ellipse, respectively.

avi
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    A satellite in an elliptical orbit around the Earth must have the Earth at one of the focii of the ellipse. The only orbit with the Earth at the center is a circle. – Jeff Snider Feb 02 '14 at 03:33

2 Answers2

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Assume that $b<a$, and let $$ f(x,y) = \frac{x^2}{a^2} + \frac{y^2}{b^2} - 1 $$ So, then, the equation of the elliptical orbit can be expressed as $f(x,y)=0$.

The (squared) distance from earth to a satellite at the point $(x,y)$ is given by $$ d(x,y) = x^2 + y^2 $$ We are interested in finding the minimum and maximum values of $d(x,y)$ subject to the constraint $f(x,y)=0$. Use what you know about Lagrange multipliers to solve this problem.

You should find that the minimum value of $d(x,y)$ is $b^2$ and the maximum value is $a^2$.

bubba
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  • I tried to find the minimum and maximum value after your reply using Lagrange multipliers but could not get the answer, please can you give the solution. – avi Feb 02 '14 at 04:42
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Assume $a>0, b>0$.

If $a=b$ it is straightforward to see that the orbit is a circle with radius $a$, so suppose $a \ne b$.

The constraint for the elliptical orbit is $g(x,y) = ({x \over a})^2 + ({y \over b})^2 =1$. The distance is $f(x,y) = \sqrt{x^2+y^2}$. It is equivalent and more convenience to minimize the square of the distance $s(x,y) = f(x,y)^2$.

Note that $E = \{ (x,y) | ({x \over a})^2 + ({y \over b})^2 =1 \}$ is compact, hence a minimum and maximum exists. Also, $Dg((x,y)) \neq 0$ for all $(x,y) \in E$, so we can use the Lagrange theorem to characterize the extrema.

Then $D s((x,y))+ \lambda Dg ((x,y)) = 0$, which yields $2x(1+ \lambda {1 \over a^2}) = 0$, $2y(1+ \lambda {1 \over b^2}) = 0$.

The solutions to the Lagrange equation are (1) $(x,y) = 0$, (2) $x$ arbitrary, $\lambda = -a^2$, $y=0$ and (3) $y$ arbitrary, $\lambda = -b^2$, $x=0$. Since we must also satisfy the constraint $g(x,y) = 1$, we are left with $x=\pm a$, $y = 0$ and $x=0$, $y = \pm b$. Since the distance is insensitive to sign, we see that one solution (ignoring sign) must be the minimum and the other the maximum, and the corresponding distances are $a,b$ which are half the lengths of the major and minor axes.

copper.hat
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