An electron is moving in an elliptical orbit with........... the nucleus, then find the eccentricity of the orbit.

Question

an electron is moving in ab elliptocal orbit with the nucleus at one of the foci. if the largest distance of the electron from the nucleus is twicw of the smallest distance of the electron from the nucleus, then find the eccentricity of the orbit.

Answer 1

The general equation of an ellipse is x^2/a^2+y^2/b^2 where the origin or centre O is at (0,0); also the eccentricity is given by e, where b^2=a^2(1-e^2) and a and b are the x and y radii of the ellipse respectively (semi-major and semi-minor axes). The foci are at F(f,0) and F'(-f,0).

For any point P(x,y) on the ellipse PF+PF'=k a constant. That is, sqrt((x-f)^2+y^2)+sqrt((x+f)^2+y^2)=k. In particular, at the point Q(a,0) the extent of the x radius, so that P is at Q, sqrt((a-f)^2)+sqrt((a+f)^2)=k; a-f+a+f=k=2a. The shortest distance PF is when P is at Q, so that PF'=2PF (required by the question) and a+f=2(a-f); 3f=a; f=a/3.

At R(0,b) we have RF+RF'=k=2a; that is, 2sqrt(b^2+f^2)=2a, so b^2+f^2=a^2. The hypotenuses of the triangles ROF and ROF' have the same length as the radius a.

But f=a/3 and b^2=a^2(1-e^2), so a^2(1-e^2)+a^2/9=a^2 and 1-e^2+1/9=1, making e=1/3. The eccentricity is 1/3.