# Do transfer orbits toward the central star necessarily result in a higher velocity on arrival due to the star's gravity?

Picture a system with three bodies: a central star and two planets in orbits around the central star. The planets' orbits are perfectly circular (eccentricity 0) and their orbital radii are 1.0 units and 1.5 units, respectively. The two planets have identical mass, and the mass of the planets is much smaller than that of the central star. (We could make the system more complex, but I am pretty sure that this is good enough for now. If you want a more real-life example, take the Sun, Earth and Mars, but make sure to state that you're using that in your answer.)

Now, a spacecraft is launched onto a transfer orbit from one planet to the other. The instant the spacecraft attains escape velocity from the origin planet, its engines are shut down. (The velocity along the transfer orbit is equal to the origin planet's escape velocity.) We can simplify by handwaving: there is no venting or anything else like that which would impact the transfer orbit after the point of engine shutdown.

In this scenario, at the instant when it enters the destination planet's sphere of influence, will a spacecraft launched from the inner planet toward the outer planet have a velocity appreciably different from that which would be the case if the spacecraft was launched from the outer planet toward the inner planet?

You didn't say anything about how you chose the transfer orbit, nor did you specify velocity relative to what. And of course, "appreciably" is not defined.

So we will have to make some assumptions.

First, I will assume that you chose a minimum energy transfer orbit between the two bodies. For bodies with this small of a ratio of radii, that minimum transfer energy is a Hohmann transfer.

Second there are two choices for what the velocity is relative to, one more meaningful than the other. The first choice is relative to the central star. The Hohmann transfer orbit to go from the inner body to the outer body is identical to the Hohmann transfer orbit the other way. Except for the direction, of course (which half of the orbit you are using). By their nature, the orbit will be faster closer to the star than farther.

How much faster? That orbit has an apoapsis just touching the orbit of the outer planet, and a periapsis just touching the orbit of the inner planet. So we have:

$$-{\mu\over a_i+a_o}={v^2\over 2}-{\mu\over r}$$

where $a_i$ is the radius of the inner planet orbit, the perapsis, and $a_o$ is the radius of the outer planet orbit, the apoapsis. The velocity $v$ is determined by the current radius from the central star, $r$. So we have for the velocity of the vehicle at arrival at the two ends of the transfer orbit are:

$$v_o=\sqrt{2\mu{a_i\over a_o\left(a_i+a_o\right)}}$$

$$v_i=\sqrt{2\mu{a_o\over a_i\left(a_i+a_o\right)}}$$

Plugging in your example, assuming that we have picked dimensions for which the radius of the inner planet is one and the $\mu=GM$ of the central star is $30$ (chosen just to make the results a little simpler), we get:

$$v_o=4$$

$$v_i=6$$

The ratio of the velocities is $3\over 2$, equal to one over the ratio of radii. Is that appreciable? I'd say yes. I'll call anything more than about 10% appreciable.

The second choice for what the velocity is relative to has more import. That is the velocity relative to the arrival body. That tells you something about how much you'll have to slow down to stop or how hard you'll hit the atmosphere (though for the full answer you'll also need the mass and size of the planet). The arrival orbit will be faster than the inner planet and slower than the outer planet, so at arrival you will be catching up with the inner planet, or the outer planet will be catching up with you from behind. So we will look at the magnitude of those two velocities. They are (where we will denote the magnitude of the relative velocity as $v_\infty$):

$$v_{\infty o}=\sqrt{\mu\over a_o}-\sqrt{2\mu{a_i\over a_o\left(a_i+a_o\right)}}$$

$$v_{\infty i}=\sqrt{2\mu{a_o\over a_i\left(a_i+a_o\right)}}-\sqrt{\mu\over a_i}$$

In our handy units, they are:

$$v_{\infty o}=\sqrt{20}-4\approx 0.472$$

$$v_{\infty i}=6-\sqrt{30}\approx 0.523$$

The arrival velocity at the inner planet is higher, and the ratio is 1.11, so just a smidge over my definition of 10%. So, yes, it's appreciable, but just barely so. Since the arrival energy ratio is the square of the velocity ratio, and really it's the energy that matters, then it's definitely appreciable at over 20%.

When you've reached escape velocity relative to the planet, the star becomes the dominant gravitational influence: you're in orbit around the star.
And as always when you're in orbit, your orbit's radius and speed are interchangable. So when you travel outward from the star, your speed will decrease, and when you move closer to the star, your speed will increase.

Perhaps this animation will make it more clear, but basically, the craft that leaves a lower/inner orbit planet to arrive at one further fro a star (without further propulsive effort) will arrive there at a slower velocity. Shows the orbits of Mercury through Jupiter, with the 'mid point' of the asteroid belt thrown in for good measure, together with the elliptical orbits that pass between Earth and the other planets. Note how much faster the dots are travelling at the low point of the elliptical orbits, compared to the high point.

N.B. That image is intended to be shown on a dark background. The orbit of Earth is shown in white.

I will use the scenario modeled in my Hohmann Spreadsheet. The spreadsheet calls Mars' orbit 1.52 A.U. and earth's orbit 1 A.U. The sheet assumes circular, coplanar orbits.

Earth to Mars:

Departure Vinf 2.94 km/s.
Arrival Vinf 2.65 km/s. Mars to Earth:

Departure Vinf 2.65 km/s.
Arrival Vinf 2.94 km/s

Quantities are the same but switched. The total Vinfinity is the same regardless if the Hohmann is going out or in. The Vinf at the outer planet is smaller.

The bigger the distance between planets, the more pronounced the difference. For example,

Mercury to Neptune:

Departure Vinf 19.4 km/s.
Arrival Vinf 4.57 km/s.

Vinf is a hyperbolic orbit's speed at infinity. For practical purposes a Hohmann transfer becomes hyperbolic when entering a planet's sphere of influence and Vinf refers to the difference in speed between heliocentric orbits.