This question came in my physics mock test
A small satellite of mass 'm' is revolving around earth in a circular orbit of radius $r_0$ with speed $v_0$. At certain point of its orbit, the direction of motion of satellite is suddenly changed by angle $θ = cos^{–1}(\frac35)$ by turning its velocity vector in the same plane of motion, such that speed remains constant. The satellite, consequently goes to elliptical orbit around earth. The ratio of speed at perigee to speed at apogee is :-
I did solved this question by conservation of angular momentum and energy .But I thought of another approach using the formula of speed in elliptical orbit at perigee and apogee given by
$$V_{max}=\sqrt{{\frac{GM}{a}} \frac{1+e}{1-e}}$$
$$V_{min}=\sqrt{{\frac{GM}{a}} \frac{1-e}{1+e}}$$
By taking the ratio the constants cancel up and i am left with
$$\frac{V_{max}}{V_{min}} =\frac{1+e}{1-e}$$ so i need to find the Ecentricity using the angle $cos^{-1}(3/5)$
Is there any method to calculate the eccentricity using the angle??
Thanks in Advance!
REGARDS!!
Best Answer
We know that the energy of a satellite depends only on the semi-major axis $a$ of its elliptical orbit. As the energy of the satellite does not change when it suddenly switches from a circular to an elliptic orbit, it follows that the semi-major axis of the ellipse is the same as the radius of the circle.
Hence, when the switch occurs, the satellite is at a distance $a$ from a the Earth (first focus) and also from the second focus (because the sum of its distances from the foci is $2a$): foci and satellite form an isosceles triangle, with sides $a$, $a$ and $2ea$ (by definition of eccentricity). Angle $\theta$ is one half of the angle at vertex of that triangle, and we thus get:
$${4\over5}=\sin\theta={ea\over a}=e,$$
which is exactly your result.