2
$\begingroup$

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!!

$\endgroup$
4
  • $\begingroup$ I'm sure the $3-4-5$ right triangle is there to make the numbers work out nicely ... $\endgroup$ Jul 30, 2021 at 18:45
  • $\begingroup$ @EthanBolker yes! Its answer is given 9 so i worked out reversely and found out that it follows the relation $e=\sqrt{1-cos^2\theta}$ but i dont know how to arrive at it $\endgroup$ Jul 31, 2021 at 6:03
  • $\begingroup$ Probably $e-1$ should be corrected to $1-e$. $\endgroup$ Jul 31, 2021 at 21:30
  • $\begingroup$ @Intelligentipauca Thanks for the suggestion! Edited $\endgroup$ Aug 1, 2021 at 11:37

2 Answers 2

2
$\begingroup$

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.

enter image description here

$\endgroup$
1
$\begingroup$

The angle between the radius vector $ \ \overrightarrow{r} \ $ (from the orbit's focus) and the velocity vector $ \ \overrightarrow{v} \ $ is the "zenith angle" $ \ \gamma \ \ . $ (I should mention that many of these terms and symbols vary among sources and presentations on orbital mechanics.) The "specific orbital angular momentum" $ \ \overrightarrow{h} \ = \ \frac{ \ \overrightarrow{L}}{m} \ $ (some call this $ \ \overrightarrow{L} \ $) is then $ \ \overrightarrow{h} \ = \ \overrightarrow{r} \times \overrightarrow{v} \ \ , $ which has magnitude $ \ r · v · \sin \gamma \ \ . $ (It is, of course, a conserved quantity, or, as mechanicians say, a "constant of the motion".) An angle that is often used instead is the "flight-path angle" $ \ \phi \ $ between the velocity vector and the perpendicular to the radius vector (or "local horizontal"). This is the angle given in the problem as $ \ \theta \ \ , $ so $ \ \cos \theta \ = \ \frac35 \ = \ \cos \phi \ = \ \sin \gamma \ \ . $

For the orbital properties, the initially circular orbit of radius $ \ r_0 \ $ has the "circular speed" $ \ v_0 \ = \ v_{circ} \ = \ \sqrt{\frac{GM}{r_0}} \ \ ; $ this is unchanged by the alteration of the orbit, so the specific angular momentum becomes $ \ h \ = \ r · v · \cos \phi \ = \ r_0 · \sqrt{\frac{GM}{r_0}} · \frac35 \ = \ \frac35 \sqrt{GMr_0} \ \ , $ whereas it was formerly $ \ r · v · \cos 0 \ = \ 1 · \sqrt{GMr_0} \ \ . $ (Since the speed at $ \ r_0 \ $ remains the same, the orbital semi-major axis $ \ a \ = \ r_0 \ $ [from the circular orbit] is also unaffected, as also mentioned by Intelligenti pauca ).

The derivation of the important relation we need is carried out in many texts on orbital mechanics (I mostly went to Danby's Fundamentals of Celestial Mechanics, 2nd ed., Section 6.2), so I won't run through it here, other than to say that starting with $$ \ \frac{d^2}{dt^2} \ \overrightarrow{r} \ = \ -\frac{GM}{r^3} \ \overrightarrow{r} $$ (the gravitational acceleration vector), the vector product $$ \ \overrightarrow{h} \times \frac{d^2}{dt^2} \ \overrightarrow{r} \ = \ -\frac{GM}{r^3} \ \overrightarrow{h} \times \overrightarrow{r} \ = \ -GM \ \frac{d}{dt} \left(\frac{\overrightarrow{r}}{r} \right) $$ is integrated, ultimately to produce the polar equation for the orbit. A relation that emerges from this is $ h^2 \ = \ GMa · (1 - e^2) \ \ , $ with $ \ e \ $ being the orbital eccentricity. For the orbit under discussion, this becomes $$ \left( \ \frac35 \sqrt{GMr_0} \ \right)^2 \ = \ GM· r_0 · (1 - e^2) \ \ \Rightarrow \ \ \frac{9}{25} \ = \ (1 - e^2) \ \ \Rightarrow \ \ e \ = \ \frac45 \ \ . $$

Other sources give various derived expressions. Roy ( Orbital Motion, 3rd ed., Section 4.5) gives, for example, $$ e \ \ = \ \ \left[ \ 1 \ - \ \frac{r}{a^2} · (2a - r) · \sin^2 \gamma \ \right]^{1/2} $$ $$ \rightarrow \ \ e \ \ = \ \ \left[ \ 1 \ - \ \frac{r_0}{r_0^2} · (2r_0 - r_0) · \left(\frac35 \right)^2 \ \right]^{1/2} \ \ = \ \ \left[ \ 1 \ - \ 1 · \frac{9}{25} \ \right]^{1/2} \ \ = \ \ \frac45 \ \ . $$

(And, yes, the differences in your derived expressions should read $ \ " 1 - e " \ $ ; otherwise, your results won't even make sense for a circular orbit $ \ [e = 0] \ \ . $ )

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .