The Johns Hopkins University Applied Physics Laboratory website depicts the New Horizons flight path as almost a straight line after the Jupiter flyby. Could a spacecraft travel from the outer solar system to the Earth in a similar straight line with enough propulsion. Is that amount of propulsion feasible with current technologies, even for a small spacecraft?

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    $\begingroup$ It's not straight. An incoming object would travel in a similar trajectory, but played in reverse. $\endgroup$ – JCRM Jul 10 at 19:40
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    $\begingroup$ It's an asymptote, not straight $\endgroup$ – Antzi Jul 11 at 1:23

Could a spacecraft fly in a direct line from the outer solar system to Earth?

Ignoring tiny deviations due to long-range attraction to the larger planets, sure! absolutely! no problem! you betcha!

And you will need zero propulsion to do it. Sit back and enjoy the ride!

At the right time of year you can position yourself at say 100 AU from the Sun in the plane of the Earth's orbit and just let go. With no tangential velocity, the Sun's gravity will tug on you and accelerate you along a straight-line trajectory, directly towards the Sun.

Since you have positioned yourself in the Earth's orbital plane, and timed your start at the right time of the year, the Earth will move directly in front of you as you pass 1 AU and you'll slam into it producing a truly spectacular fireball!

How fast will you be moving then?

Specific energy is the kinetic plus potential energy per unit mass. So it's

$$u = \frac{v^2}{2} - \frac{GM_S}{r}$$

where $GM_S$ is $G$ times the mass of the Sun, otherwise known as the standard gravitational parameter of the Sun or about 1.327E+20 m^3/s^2.

At 100 AU and not moving, your total specific energy $u$ is -8.85E+06 Joules/kg, which is a shallow potential well, you are almost free. Your total energy won't change, so you can calculate your specific kinetic energy as you are starting to approach the Earth using:

$$-\frac{GM_S}{\text{100 AU}} = \frac{v^2}{2} - \frac{GM_S}{\text{1 AU}}$$

$$\frac{v^2}{2} = GM_S \left(\frac{1}{\text{1 AU}} - \frac{1}{\text{100 AU}}\right)$$

$$v = \sqrt{2 GM_S \left(\frac{1}{\text{1 AU}} - \frac{1}{\text{100 AU}}\right)}.$$

That gives 42,850 meters/sec, which by no small coincidence is 0.5% lower than $\sqrt{2}$ times the Earth's orbital velocity.

What about Earth's gravity? That's why I said "as you are starting to approach the Earth." Once the Earth starts getting close you will accelerate even fast.

The Earth's escape velocity is

$$v_{esc} = \sqrt{\frac{2 GM_E}{\text{6378137 meters}}}$$

is another 11,180 meters/sec ($GM_E$ is about 3.986E+14 m^3/s^2), so your final impact velocity, should you survive it, will be roughly 53,000 meters/sec.

Welcome home!

  • $\begingroup$ How would no tangential velocity be established? $\endgroup$ – Bob516 Jul 11 at 14:16
  • $\begingroup$ @Bob516 You haven't mentioned any specific incoming trajectory in your question, so I chose a simple one. Full stop at 100 AU. $\endgroup$ – uhoh Jul 11 at 14:24
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    $\begingroup$ Related question Free Fall to Earth from Solar North Pole? $\endgroup$ – James Jenkins Jul 12 at 17:14

Trajectories are reversible. So if you had a spacecraft heading into the solar system at the same velocity that NH is currently heading out, it would follow the same trajectory but in reverse. If you time things correctly, it could narrowly miss Jupiter and be deflected into a solar orbit which could then hit the Earth and it would reenter at the same velocity that NH was launched at (about 16 km/s). This would need no more fuel than NH used after it had left Earth (ie very little).

On the other hand if you start at rest relative to the Sun, or maybe in a distant orbit around Neptune or something of the kind, you would need a lot of fuel to get onto that trajectory -- enough for a delta-V of about 13 km/s. That's an awful lot for deep space, but fairly routine for a launch beyond Earth orbit. So if you could somehow obtain a fully fueled Falcon 9 and a suitable upper stage (for instance) you could actually put a modest payload onto that trajectory. As far as we know very large cryogenic fuelled rockets are not currently available at that location, but keep checking your supplier's web page for delivery options...


Theoretically: Yes

If the net force on an object is zero, then Newton's 1st Law says that it will continue in a straight line. By continually adjusting the amount and direction of your thrust, so that the thrust cancels out the force of the Sun's gravity, you can do just that.

Practically: No

Doing the above takes ridiculous amounts of energy -- which in turn means spacecraft mass, volume, and cost -- for even a tiny payload. No spacecraft does this in practice. Real spacecraft instead use orbital mechanics to travel.


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