Orbiting body around a star

Question

Let us assume there's a body with mass $m$ and velocity $v$, at a distance $r$ from another body of mass $M$. The velocity vector is perpendicular to the radial vector. With these values, how do we find the apogee and perigee of the elliptical orbit of the body?

The problem I'm facing here is that we can assume $r$ to be any distance for the ellipse, i.e., apogee $r_a = r$, or perigee $r_p = r$. So: $$v=\sqrt{2GM\left(\frac{1}{r}-\frac{1}{r_p + r_a}\right)}$$

Assuming $r=r_a$ we can solve for $r_p$ and vice versa.

Therefore, in each case, we find different values. But in reality, there is only one set of values for a body with tangential velocity $v$. So whats the concept here?

Answer 1

That equation relating the mass, orbital speed, radial distance, and semi-major axis is known as the vis-viva equation. Its standard form is

$$v^2 = \mu\left(\frac2r - \frac1a\right)$$

where $\mu$ is the standard gravitational parameter $GM$, where $M$ is the sum of the masses of the two bodies. It's common to neglect the mass of the smaller body when it's much smaller than the mass of the larger body.

I have a derivation of the vis-viva equation here

In your scenario, the initial velocity vector is perpendicular to the radial vector. Now in a Kepler ellipse, that can only happen at periapsis or apoapsis. That is, your initial $r$ is either $r_a$ or $r_p$. So we can rearrange vis-viva to calculate $a$, and then use $2a = r_p + r_a$ to calculate the other $r$. Then we just compare the two $r$ values to figure out which one's which.

Rearranging, we get

$$\frac1a = \frac2r - \frac{v^2}\mu$$ $$a = \frac{\mu r}{2\mu - rv^2}$$

With a little more algebra, we get the other radius,

$$r_2 = \frac{r^2v^2}{2\mu - rv^2}$$

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Note that $a$ becomes infinite when $2\mu - rv^2 = 0$. That's the escape velocity, which gives a parabolic trajectory. And when $2\mu - rv^2 < 0$, $a$ becomes negative, and we have a hyperbolic trajectory.

Answer 2

A conceptual answer:

If the initial velocity is solely tangential, then the object is already at either apogee or perigee, since these are the only two points in an elliptical orbit where this is the case.

In those circumstances, to find an expression for the other extremum is a simple application of conservation of angular momentum, since $$r_p v_p = r_a v_a = rv \tag*{(1)}$$

However, we do not know the tangential velocity at the other extremum. To get this we can use conservation of energy (the sum of kinetic and potential energy) $$ \frac{v_a^2}{2} - \frac{GM}{r_a} = \frac{v_p^2}{2} - \frac{GM}{r_p} = \frac{v^2}{2} - \frac{GM}{r}\ . \tag*{(2)} $$

Thus you have two equations and two unknowns (either $v_a$, $r_a$ or $v_p$, $r_p$) and you can get an expression for these in terms of $r$ and $v$. You will find a quadratic in $r_a$ (or $r_p$) which has one solution which is $r$ and another that is either bigger than $r$ (in which case it is the apogee and $r$ is the perigee) or smaller than $r$ (in which case it is the perigee and $r$ is the apogee), depending on whether $GM/rv^2$ is smaller or bigger than 1 respectively. (And if $2GM -rv^2 < 0$ there is no bound solution at all.)