<p>This speed is the escape from the solar system speed, from the orbit of the Earth.</p>
<p>If you had a rocket, and it was orbiting the sun at the same distance as the Earth it would be travelling at 29.7 km/s.  If it increased its speed to 41.1 km/s, then it would be on a parabolic trajectory and eventually leave the solar system.</p>
<p>Similarly, and in the context of meteors and comets, a comet that has fallen from very distant parts of the solar system would, by the time it reaches Earth's orbit, be moving at 41.1 km/s</p>

CC BY-SA 4.0
2024-03-10T06:07:03.167Z
2024-03-10T06:07:03.167Z
James K
<p>It is the Solar System &quot;escape speed&quot; for an object in orbit around the Sun, at the distance of the Earth.</p>
<p>The energy of an object in orbit is the sum of its kinetic and potential energy. The kinetic energy and orbital speed is determined by the centripetal acceleration provided by the gravity of the Sun.</p>
<p>The maths turns out that the total energy <em>of a circular orbit</em> is exactly half the potential energy and is negative - since the object is bound. This means the kinetic energy must be equal in magnitude to the binding energy of the orbit and would have to double in order to escape (to make the total energy zero). i.e.
<span class="math-container">$$\frac{1}{2} mv^2 -\frac{GM_\odot m}{r} = -\frac{GM_\odot m}{2r}\ . $$</span></p>
<p>In order to double the kinetic energy, you increase the speed by a factor of <span class="math-container">$\sqrt{2}$</span>. That is the origin of your formula - it is the increase in speed required to escape the Solar System from Earth orbit.</p>

CC BY-SA 4.0
2024-03-11T16:47:24.277Z
Interesting to note that the $\sqrt{2}$ factor applies in any circular orbit around any body. If you are in a stable circular orbit around a body at any distance, increasing your speed by 41% will cause you to escape.
Nuclear Hoagie
CC BY-SA 4.0
2024-03-10T07:55:32.300Z
2024-03-10T07:55:32.300Z
ProfRob
<p>Earth's escape velocity is 11.18 km/s.</p>
<p>If I multiply Earth's orbital speed with the square root of 2,</p>
<p>29.7827 km/s x 1.41 = 42.11 km/s</p>
<p>How is this speed called?</p>
<p>I found this equation in an old German meteor book, but I don't understand.</p>

<solar-system>
Multiply Earth's orbital speed with square root of 2
Robert
# Multiply Earth's orbital speed with square root of 2

(Asked by: username_0 on 2024-03-10)

Earth's escape velocity is 11.18 km/s.

If I multiply Earth's orbital speed with the square root of 2,

29.7827 km/s x 1.41 = 42.11 km/s

How is this speed called?

I found this equation in an old German meteor book, but I don't understand.

**Tags:** solar-system


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## 2 answers

### Answer by username_2 on 2024-03-10. Score: 32 | Accepted answer

It is the Solar System "escape speed" for an object in orbit around the Sun, at the distance of the Earth.

The energy of an object in orbit is the sum of its kinetic and potential energy. The kinetic energy and orbital speed is determined by the centripetal acceleration provided by the gravity of the Sun.

The maths turns out that the total energy *of a circular orbit* is exactly half the potential energy and is negative - since the object is bound. This means the kinetic energy must be equal in magnitude to the binding energy of the orbit and would have to double in order to escape (to make the total energy zero). i.e.
$$\frac{1}{2} mv^2 -\frac{GM\_\odot m}{r} = -\frac{GM\_\odot m}{2r}\ . $$

In order to double the kinetic energy, you increase the speed by a factor of $\sqrt{2}$. That is the origin of your formula - it is the increase in speed required to escape the Solar System from Earth orbit.

> **Comments:**
> **username_3**: Interesting to note that the $\sqrt{2}$ factor applies in any circular orbit around any body. If you are in a stable circular orbit around a body at any distance, increasing your speed by 41% will cause you to escape.  

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### Answer by username_1 on 2024-03-10. Score: 12

This speed is the escape from the solar system speed, from the orbit of the Earth.

If you had a rocket, and it was orbiting the sun at the same distance as the Earth it would be travelling at 29.7 km/s. If it increased its speed to 41.1 km/s, then it would be on a parabolic trajectory and eventually leave the solar system.

Similarly, and in the context of meteors and comets, a comet that has fallen from very distant parts of the solar system would, by the time it reaches Earth's orbit, be moving at 41.1 km/s


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