Orbiting body around a star

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?

• does horizontal mean tangential?
– JEB
Commented Mar 26 at 12:56
• @JEB yup tangential Commented Mar 26 at 13:02
• Voting to reopen. Clearly a conceptual question about the correct interpretation of the vis-viva equation and not a "do my homework for me" question. Commented Mar 26 at 15:42
• @StarGazer The equation you gave is the vis-viva equation for one body orbiting another when $m << M$ - see en.wikipedia.org/wiki/Vis-viva_equation. Note that $v$ in this equation is the orbital speed and not just its tangential component. The fundamental issue here is that knowing $v$ and $r$ at a single point does not uniquely determine the shape of the orbit. Commented Mar 26 at 15:46
• @gandalf61 Thank you for being the reasonable one here and voting to reopen the question. Well i completely agree to your answer but i just have a small visualization which contradicts this. Lets say a small body is kept near a star(as in the question). Now initially it would fall straight into the star but if we gave it a slight push gaining an orbital/tangential velocity, there would be a particular orbit(elliptical) and thats the orbit i want the values of Commented Mar 26 at 15:55

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,

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

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.

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.)