# What if Earth gained 1 km/s orbital velocity?

How would Earth's orbit be affected if we (hypothetically) added $$1 {\rm km/s}$$ to its orbital velocity? Would Earth reach close to Mars' orbit? Could Earth get gravitational assist from Mars and go to outer solar system?

• Welcome to astronomy SE! I edited your question slightly, trying to make it a bit clearer. I also added two relevant tags. Jul 14 at 7:40
• Earth, ~30km/s. Venus ~35km/s, Mars ~24km/s. If you added 1km/sec instantaneously then Earth's orbit would be eccentric, rising above our normal distance from the sun towards Mars at apoapsis and returning to our nominal orbital distance at periapsis. If you wanted to keep a roughly circular orbit you would need to slow down first, allowing the Earth to drop closer to the sun. You could then slow down again to circularize and we would end up in a slightly tighter orbit, closer to Venus and 1km/s faster. So which of these is your question?
– J...
Jul 14 at 14:32
• For comparison, Earth gains about 1km/s of orbital velocity every year as it goes from aphelion to perihelion, and loses it on the other half of the year. Jul 14 at 15:37
• This could happen if the Earth was hit by a lone planet entering the solar system. The Earth could even be "marbled" out of the system. It could also be sent to the Sun. Both scenarios are not too beneficient for people and Nature. If Nature provided such a hitting loner it would be "bye-bye" for mankind. Jul 14 at 16:01
• You need anywhere from 4.23 to 4.89 km/s of delta V to get to Mars from Earth. So 1 km/s is a decent change, but not enough to get near Mars. May I suggest using this tool to select an efficient date for this hypothetical velocity gain?
– Wyck
Jul 14 at 18:03

## 2 Answers

I am assuming that by adding 1 km/s, you mean increasing the tangential speed of the Earth by 1 km/s. This would increase both the kinetic energy and the angular momentum.

This is a relatively small increase in both quantities and it would send the Earth into a slightly more eccentric orbit than it has now. It would not come close to Mars.

Using the vis-viva equation, we can say $$a^{-1} = \frac{2}{r} - \frac{v^2}{GM_\odot}\ ,$$ where $$a$$ is the new semi-major axis, $$r$$ is the radial separation of the Earth and Sun when the impulse is given and $$v$$ is the new speed. From there we know that the aphelion radius is $$r_a = a(1+e)$$, where $$e$$ is the new eccentricity.

Exactly what happens depends where in the Earth's orbit, the impulse is given. The biggest effect will be when Earth is at perihelion, increasing $$v$$ to 31.3 km/s at $$r_p= 0.983$$ au. The vis-viva equation gives $$a=1.075$$ au, so that $$e=0.085$$ and $$r_a = 1.167$$ au. Still some way short of the 1.38 au perihelion of Mars.

A better question might be - how much speed do we need to add to the Earth's orbit to get it to Mars. If you can assume the impulse is given at perihelion and it intercepts Mars at its perhelion (not quite right because they aren't in the same plane), then the equations above suggest $$> 2.2$$ km/s is needed.

• Is there a crazy billionaire considering attaching rocker boosters to Earth as an alternative to going to Mars by rocket ship? Or maybe as a silly solution to global warming? Jul 14 at 14:55
• @Barmar Oddly, I was back-of-the-enveloping just that idea (boost orbit to cool the planet) in the shower this morning. Turns out SciAm actually did the same thing. It would take about ten trillion times as much energy as we currently produce on the world's electrical grids annually.
– J...
Jul 14 at 15:11
• "but it is significantly more than 1 km/s" [citation needed] Jul 14 at 20:21
• @PhilFrost It looks like it could be as small as 3 km/s if you timed it right. This can be calculated using the equations above. Jul 14 at 21:59
• @user45266 I'm still trying to hire my henchpeople (I'm a woke supervillain). Jul 14 at 22:28

Using calculations from here: https://www.vanderbilt.edu/AnS/physics/astrocourses/ast201/orbitalvelocity.html

New semi-major axis

$$a = \frac{150000000000 \cdot 0.0000000000667 \cdot 2\cdot 10^{30}}{2 \cdot 0.0000000000667 \cdot 2 \cdot 10^{30} - 150000000000 \cdot 30000 \cdot 30000)}\\ = 151.821\,\mathrm{million\,km}$$

New orbital period $$p = \frac{\sqrt{4\cdot\pi^2\cdot a^3/(0.0000000000667\cdot (2\cdot 10^{30} + 5.972 \cdot 10^{24}))}}{60\cdot 60\cdot 24}\\ = 372.465\,\mathrm{days}$$

• With the input precision you use, you can only give the results at most as accurate as 150 million km and 370 days. Everything else is pretending an accuracy your input values don't warrant. 88.59394756% of all statistics give an over-precise accuracy. Jul 14 at 15:26
• @justhalf Cross sign is generally use in scientific notation (see above). Also dots may sometimes be confusing with the decimal points. I usually prefer cross than dots. Jul 15 at 6:29
• @NilayGhosh I tend to disagree with using the cross sign for scientific notation, that's a a very cultural thing (depending on country, scientific community etc.). For instance, some fields prefer to write $m E n = m \cdot 10^n$, e.g. $1.23{\rm e}45$ for $1.23 \cdot 10^{45}$. Jul 15 at 8:37
• Where I grew up, dots were used for regular multiplication in school and uni (maths courses); the cross was reserved for the vector cross product. That's also a country where we use "," to separate decimals, not "." as in the English world.
– AnoE
Jul 15 at 8:58
• According to ISO: If the point is used as the decimal sign, the cross and not the half-high dot should be used as the multiplication sign between numbers expressed with digits. If the comma is used as the decimal sign, both the cross and the half-high dot may be used as the multiplication sign between numbers expressed with digits.
– user41187
Jul 15 at 14:11