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The orbit of the planet mercury around the sun is in elliptical shape with sun at a focus. The semi-major axis is of length $36$ million miles and the eccentricity of the orbit is $0.206$ The greatest possible distance between mercury and sun.

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Toolbox:
  • Standard forms of equation of the ellipse with major axes 2a, minor axis 2b $(a >b)$ eccentricity e and $b^2=a^2(1-e^2)$ or $e^2=1-\large\frac{b^2}{a^2}$
  • $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$
  • http://clay6.com/mpaimg/20_2_Toolbox.png
  • Foci$(\pm ae,o),$ center $(0,0)$,vertices $(\pm a,0)$
  • End points of Latus Rectum $\; (ae,\pm \large\frac{b^2}{a})$ and $(-ae,\pm \large\frac{b^2}{a})$
  • Directrices $x=\pm \large\frac{a}{e}$.
  • The major axis is $y=0$ (x- axis) and the minor axes is $x=0$ (y- axis)
  • $\large\frac{x^2}{b^2}+\frac{y^2}{a^2}$$=1$
  • http://clay6.com/mpaimg/20_2_Toolbox1.png
  • Foci$(0,\pm ae),$ center $(0,0)$,vertices $(0,\pm a)$
  • End points of Latus Rectum $(\pm \large\frac{b^2}{a}$$,ae)$ and $(\pm \large\frac{b^2}{a}$$,-ae)$
  • Directrices $y=\pm \large\frac{a}{e}$.
  • The major axis is $x=0$ (y- axis) and the minor axes is $y=0$ (x- axis)
  • If a point moves so that the sum of its distances from two fixed points is a constant,then the path traced as an ellipse with major axis of length equal to the constant sum and foci at the two fixed points.
Step 1:
In the orbit of mercury,$e=0.206$ and semi-major axis $a=36$(in millions of miles).
Let $F(ae,0),F'(-ae,0)$ be the foci and $A(a,0),A'(-a,0)$ be the vertices of the elliptical path with the sun at $F$.
Step 2:
The planet is at the greatest distance when it is at $A'$.
The distance $FA'=a(1+e)$
$\qquad\qquad\qquad=36(1+0.206)$
$\qquad\qquad\qquad=36\times 1.206$
$\qquad\qquad\qquad=43.416$ million miles.
answered Jun 18, 2013 by sreemathi.v
edited Jun 18, 2013 by sreemathi.v
 

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