# Calculate Initial Velocity For Orbital (Gravity) Slingshot

I am trying to find the initial velocity to slingshot a planet around the sun and through a gap.

The green ball is the planet, and the yellow ball is the sun. In this trial I need to get the planet to go around the sun and through the gap at 278Gm. I have tried different approaches, but nothing seems to be even remotely correct. Anything under 20k m/s will land you in the sun and anything over 50k will slingshot you out of the system.

I want to know what formula to use so that I can solve this type of problem.

• Well, depending on how long your attempts take, you could just use bisection until you find the right velocity, though it's not a formula. – Kyle Kanos Dec 8 '14 at 18:45
• As far as I remember, a slingshot manoeuver can only be done against a moving planet, and a gravity assist can only be done by losing mass near the bottom of a gravity well. Your case of a planet (which does not lose mass) plus a sun (which can be assumed as stationary, at least so I guess) seems to me to allow neither. – LSerni Dec 8 '14 at 19:06

## 2 Answers

If you are in a circular orbit what you need is a Hohmann transfer, from Wikipedia:

In orbital mechanics, the Hohmann transfer orbit /ˈhoʊ.mʌn/ is an elliptical orbit used to transfer between two circular orbits of different radii in the same plane.

It works like this assuming the planet is in a circular orbit.

Then the amount of delta v needed to go from the green orbit to the yellow orbit is.

where units are

• $v \,\!$ is the speed of an orbiting body
• $\mu = GM\,\!$ is the standard gravitational parameter of the primary body, assuming $M+m$ is not significantly bigger than $M$ (which makes $v_M \ll v$)
• $r \,\!$ is the distance of the orbiting body from the primary focus
• $a \,\!$ is the semi-major axis of the body's orbit.

Using an online calculator I deduce that the delta v you need is 25.07 km/s

This is independent of the mass of the planet.

Ok, let's start over with a different approach, what is the velocity exactly. Lets just use our trusted elliptical orbits.

Then using equations from this link you can calculate the speed at any point of a eclipse with,

$$v^2 = \mu\left(\frac{2}{r} - \frac{1}{a}\right)$$

which leads to 44.31 km/s at perihelion.

• The OP only wants to know the initial velocity (range), so the planet would not already be in a circular orbit. – fibonatic Dec 8 '14 at 19:54
• A oke my mistake then that just needs to be added. – Mellester Dec 8 '14 at 19:55

There seem to be three kinds of slingshot manoeuver. You can bleed some kinetic energy off a moving body, sort of like an ancient slingshot; that allows a spacecraft to either increase or decrease its own kinetic energy. Or you can get more "bang for the buck" with the assistance of a gravity well, be it moving or fixed, by expelling mass after having increased your kinetic energy at the expense of gravitational potential energy.

Your case of a planet (which does not lose mass) plus a sun (which can be assumed as stationary, at least so I guess) seems to me to allow neither.

The third kind is to change direction (in 3-D), and only needs a gravity well, but it seems to me you're not interested in that.

But if it's not under propulsion, your planet must be moving on an elliptical orbit, where the sun is one of its foci (sorry; don't know the actual English word). If your values of 100 Gm and 278 Gm are relative to the apsides of its orbit, then you can use the conservation of energy to work out the exact values. You will need to know the masses involved, though.

Your solution will have a slight error due to relativistic considerations.

In the case of a out-of-system planet, it will be moving in a hyperbolic orbit and I don't think there's a way to get to point B from point A (but I may well be mistaken).

• If points A and B lie on opposite sides of the sun, so with a 180° angle, and if either point A or B would be the periapsis, then a hyperbolic trajectory would not be able to pass through both, since the maximum angle relative periapsis would be $\cos^{-1}{-e^{-1}}$ ($e$ is the eccentricity of the orbit which would be bigger than one for and hyperbolic trajectory). However if neither of the points would have to be the periapsis, then it would be possible, since $\lim_{e\to\infty}{2\cos^{-1}{-e^{-1}}}=\pi$. – fibonatic Dec 8 '14 at 19:51
• Your confusing terms. A slingshot and a gravity assist are the same thing. There does indeed need to be a third mass in the system to want to do slingshot. – Mellester Dec 8 '14 at 19:53
• Thanks for the clarification. I've checked the nomenclature and it's as you say, even if, this way, I feel it becomes quite ambiguous. Amending answer anyway. – LSerni Dec 8 '14 at 19:59
• I believe the bang for the buck is called the oberth effect. I have also edited my answer to include calculations using a eclipse – Mellester Dec 8 '14 at 20:50
• Yes, I had embedded that in the wikipedia link. On an unrelated note, I think the OP was a bit unclear about the original trajectory of the planet. – LSerni Dec 9 '14 at 13:27