Question

Determine the true anomaly q of the point(s) on an elliptical orbit at which the speed equals the speed of a circular orbit with the same radius, that is, vellipse 1⁄4 vcircle. {Ans.: q 1⁄4 cos1(e), where e is the eccentricity of the ellipse}

Answer 1

The true anomaly q will be such that the q=
`cos^(-1)(e)` where e is the eccentricity of the ellipse.

Step-by-step explanation:

For ellipse velocity components and velocity is given as

`v_r=(\mu)/(h) e \, sin\theta\\v_p=(h)/(r)=(h)/((h^2)/(\mu)(1)/(1+ecos\theta))\\v_(eclipse)=√(v_r^2 +v_p^2)\\v_(eclipse)=\sqrt{((\mu)/(h) e \, sin\theta)^2 +((h)/((h^2)/(\mu)(1)/(1+ecos\theta)))^2}\\\\v_(eclipse)=\sqrt{((\mu)/(h))^2(e^2+1+2ecos\theta)}\\\\v_(eclipse)^2=((\mu)/(h))^2(e^2+1+2ecos\theta)`

For similar radius velocity of circle is given as

`\\v_(circle)^2=(\mu)/(r)\\v_(circle)^2=(\mu)/((h^2)/(\mu)(1)/(1+ecos\theta))\\v_(circle)^2=(\mu^2)/(h^2)(1+ecos\theta)`

Now as per the condition

`v_(eclipse)^2=v_(circle)^2\\((\mu)/(h))^2(e^2+1+2ecos\theta)=(\mu^2)/(h^2)(1+ecos\theta)\\(e^2+1+2ecos\theta)=(1+ecos\theta)\\ecos\theta=-e^2\\cos\theta=-e\\theta=cos^(-1)(e)`

So the true anomaly q will be such that the q=
`cos^(-1)(e)` where e is the eccentricity of the ellipse.