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The largest and the shortest distance of the earth from the sun are ${r_1}$ and ${r_2}$, its distance from the sun when it is at the perpendicular to the major axis of the orbit drawn from the sun
$\frac{{{r_1} + {r_2}}}{4}$
$\frac{{{r_1}{r_2}}}{{{r_1} + {r_2}}}$
$\frac{{2{r_1}{r_2}}}{{{r_1} + {r_2}}}$
$\frac{{{r_1} + {r_2}}}{3}$
Solution
(c) The earth moves around the sun is elliptical path. so by using the properties of ellipse
${r_1} = (1 + e)\,a$ and ${r_2} = (1 – e)\,a$
$ \Rightarrow \,a = \frac{{{r_1} + {r_2}}}{2}$ and ${r_1}{r_2} = (1 – {e^2})\,{a^2}$
where $a = $ semi major axis
$b =$ semi minor axis
$e =$ eccentricity
Now required distance = semi latusrectum $ = \frac{{{b^2}}}{a}$
$ = \frac{{{a^2}(1 – {e^2})}}{a} = \frac{{({r_1}{r_2})}}{{({r_1} + {r_2})/2}} = \frac{{2{r_1}{r_2}}}{{{r_1} + {r_2}}}$