Not understanding the behaviour of the vis-viva equation

[Philip Robotic]

[Moderator's note: Thread moved from a technical forum hence no formatting template]

Hi everyone!

I came here to seek some explanation.

As I was doing some orbital mechanics I came up with a problem that in short looks like this:

Having the velocity at apoapsis and its distance from the barycentre which "has" 4/5th's of this body's mass (I am doing a two-body problem, I know that the barycentre is just a point and doesn't have a mass ), calculate the speed at periapsis and the distance at that time.

So I used to things to solve it:
1. Conservation of angular momentum
2. Conservation of energy

What these letters mean in my equations:
$v_p$ - velocity at periapsis
$v_a$ - velocity at apoapsis
$d_p$ - distance at periapsis
$d_a$ - distance at apoapsis
$m$ - mass of the orbiting body
$M$ - "mass" of the barycentre

As I mentioned earlier $M=\frac{4}{5}m$

From the conservation of linear momentum I wrote:

$$v_pd_p=v_ad_a$$
Solving for $v_p$ gave me $v_p=\frac{d_a}{d_p}v_a$

Then I wrote down the law of conservation of energy expressed by the sums of gravitational potential energy and kinetic energy (the $m$'s canceled out):

$$\frac{v_a^2}{2}-\frac{GM}{d_a}=\frac{v_p^2}{2}-\frac{GM}{d_p}$$

After plugging in $v_p$ and $M$ from the previous equations and solving for $d_p$ I got:

$$\left(\frac{v_a^2}2-\frac{4Gm}{5d_a}\right)\cdot d_p^2+\frac{4Gm}{5}d_p-\frac{d_a^2v_a^2}{2}=0$$

To verify that this equation works I graphed it. I assigned some random positive values to all variables except $d_p$, as it was my $x$ on the graph.

Here is the link to the graph itself: https://www.geogebra.org/graphing/zbwdtdsv

This is a quadratic equation so it has 2 results. One of them is $-d_a$ and the other one behaves in a way I don't understand. When I increase the velocity at apoapsis the second result starts to approach the value $d_a$, but never gets past it.

As I recall my physics lessons, when the velocity at a certain point in orbit is increased the opposite part of the orbit rises. Shouldn't then the second result at some velocity $v_a$ firstly exceed $d_a$ and go to infinity as it reaches the escape velocity? Why does the function of $d_p$ behave like this?

 

[mfb]

I moved this to our homework section as it is homework-like.

Having the velocity at apoapsis and its distance from the barycentre which "has" 4/5th's of this body's mass (I am doing a two-body problem, I know that the barycentre is just a point and doesn't have a mass ), calculate the speed at periapsis and the distance at that time.

If you know it is wrong why do you write it?
Why don't you just give the two masses of the orbiting objects? The problem you have is unclear.

$v_pd_p=v_ad_a$

That is conservation of angular momentum.

 

[Philip Robotic]

Sorry, I meant angular momentum there. I have corrected it already.

Why don't you just give the two masses of the orbiting objects? The problem you have is unclear.

As I understand the solution would look the same if we had two objects where one has a mass $m$ and the other $\frac{4}{5}m$ and we were looking for $d_p$ where the second body is the point of reference.

But I am not asking for a solution to the problem

I am just asking if somebody could explain why that function

$$\left(\frac{v_a^2}2-\frac{4Gm}{5d_a}\right)\cdot d_p^2+\frac{4Gm}{5}d_p-\frac{d_a^2v_a^2}{2}=0$$​

Never returns $d_a$ greater than $d_p$, as it should (I think).

I am still very new to physics unfortunately so might be (and probably am) missing something basic

 

[gneill]

How is M's kinetic energy being accounted for? This is not a case where M >> m.

 

[Philip Robotic]

How is M's kinetic energy being accounted for? This is not a case where M >> m.

So what I tried doing here is to solve a two-body problem. I attempted at simplifying it by giving the barycentre a "mass" so I can model the body's orbit around the barycentre as if it were a static body in space with a mass of $M$. As it is static from out point of reference it has no kinetic energy.
Is this physically correct actually? I might have made the mistake here...

 

[gneill]

I'm dubious about the validity of the approach, largely because it's borrowing equations that were derived for an (effectively) inertial reference frame and applying them in what must be a non-inertial frame and so ignoring the energy and angular momentum of one of the bodies (unless something clever is happening that I'm not seeing immediately whereby the energy etc. discrepancy somehow vanishes).

I would point out that the total mechanical energy of your mass m should be negative for a bound orbit, so
$$\xi = \frac{v_a^2}{2} - \frac{G M}{d_a} \le 0$$
Using your substitution for M, that would mean
$$v_a^2 \le \frac{8}{5} G \frac{m}{d_a}$$
So forcing the second solution past or even to $d_a$ via an increase in $v_a$ is not right. The root will approach $-d_a$ only as $v_a$ approaches infinity.

 

[mfb]

Clearly describing the problem is the first step.

If the masses are of the same order of magnitude then the distance for the gravitational attraction is not the distance to the barycenter which you use for angular momentum.