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Comparing the work required for a mass to escape the Earth's gravity to the necessary initial kinetic energy gives us the escape velocity from the surface of the Earth of around $11\;\mathrm{km\cdot s^{-1}}$.

But assuming no air friction and hence no loss in energy on the axis perpendicular to gravity it shouldn't matter what angle the projectile is projected at.

Is this true?

If there were no air and we fired a projectile at $1^\circ$ to the horizontal at a velocity of $11\;\mathrm{km\cdot s^{-1}}$ should it still escape the Earth's gravity despite such a low ( $11000\sin(01^\circ)\;\mathrm{m\cdot s^{-1}}$) vertical velocity?

Could it be that firing a projectile horizontally at the escape velocity in a frictionless medium would result in a perfect orbit?

Also: Taking escape velocity of $11200\;\mathrm{m\cdot s^{-1}}$ and radius af Earth of $6371000\;\mathrm{m}$ and $g=9.80665\;\mathrm{m\cdot s^{-2}}$. On a flat Earth if you fired an object horizontally at $112000\;\mathrm{m\cdot s^{-1}}$ it would fall $0.039\;\mathrm{m}$ after $1000\;\mathrm{m}$. But using trigonometry the Earth would have curved away by $0.078\;\mathrm{m}$. Twice the value necessary for an orbit. Can someone check this?

Przemysław Scherwentke
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Kantura
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  • Thanks Wolphram, but can you verify your statement mathematically? – Kantura Jun 19 '16 at 22:12
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    I believe Wolphram is wrong and Wikipedia nicely comments on this: https://www.wikiwand.com/en/Escape_velocity – Andrii Magalich Jun 19 '16 at 22:17
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    I was wrong, the work to reach infinity obviously does not depend on the path because gravity is a conservative force, I am not sure were my mind was –  Jun 19 '16 at 22:26
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    @Derek, you $g$ is wrong, should be $\approx 9.8$. Also, I think you are missing the point that the direction of the acceleration changes continuously, so it is better to solve the problem explicitly than computing anything in terms of "curving away" – Andrii Magalich Jun 19 '16 at 22:45
  • Thanks Andrii , I fixed my g type. But for small distances I still find that when you fire a particle horizontally at 11200m/s it falls half the distance that the Earth curves away. I imagine they should be the same. There is a factor of 2 hiding somewhere. – Kantura Jun 19 '16 at 22:49
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    @Derek why they should be the same? Take a look at simulation, it has precisely what you want and the horizontally starting ship is not orbiting in a circle. You confuse escape velocity with orbital velocity – Andrii Magalich Jun 19 '16 at 22:51
  • @Andrii : What simulation? – Kantura Jun 19 '16 at 22:55
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    The one I mentioned in the answer: http://bit.ly/1UqRgie – Andrii Magalich Jun 19 '16 at 23:00
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    Oh wow, that's super cool. So there is no "one" orbital velocity. There is a range of velocities which produce orbits of different eccentricities. The minimum of which will result in a crash and the maximum of which is the escape velocity. Yes, I was in fact confusing escape and orbital velocity. If you shoot a projectile horizontally at escape velocity it will orbit at maximum possible eccentricity. Any more and it will escape. Thanks Andrii. – Kantura Jun 19 '16 at 23:13

2 Answers2

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Essentially, yes. The derivation of the escape velocity is based only on the energy balance and energy does not depend on the direction. This follows from the property of the gravitational field to be conservative, so work required to move between any 2 points is independent from the path you take.

Vague intuition: the vertical speed you estimate is about $60 \frac{m}{s} $ which would be nullified by the gravitational acceleration in $\sim 6 \,s$. But in the meantime the ship would move horizontally by almost $65 \, km$ which is far over the horizon. There the acceleration would be more collinear with the velocity of the ship, so it will have much easier time to escape.

In search of inspiration, I found this instructive online simulation

Regarding your second question about the orbit. Actually, no the escape velocity will result in escape. To get to the lowest possible orbit (at a distance $\sim 6400 \, km$ from the center, you need only $7.9 \, \frac{km}{s}$ and increasing speed will result in a higher elliptical orbit.

doppelgreener
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Andrii Magalich
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  • Aren't I already going faster than escape velocity right now? My tangential velocity on the Earth's surface is around 460 m/s. Also, what if I started traveling at escape velocity in the direction of the center of the Earth? Certainly then I would not escape either. – AjaxLeung Dec 19 '19 at 18:23
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    @AjaxLeung 450 m/s is much less, than 11 km/s = 11 000 m/s. Or maybe I don't understand your question. Regarding the center of the Earth — if you would manage not to hit anything, you would go through the center of the planet and escape – Andrii Magalich Dec 19 '19 at 18:33
  • Ha good point, I forgot how metrics worked for a moment. Does this mean that if the Earth rotated quickly enough such that the tangential velocity on the surface was 11km/s, then everything would fly off the surface and escape orbit completely? – AjaxLeung Dec 19 '19 at 20:04
  • @AjaxLeung Just like on a carousel – Andrii Magalich Dec 19 '19 at 20:39
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In newtonian mechanics the angle does not matter, but in relativity it does.

For example:

An object close to the speed of light launched horizontally will orbit circular at a distance of 3GM/c² from the center of mass (the so called photon sphere), but it will escape if launched vertically.

When you launch it at a distance just above 2GM/c² (the so called Schwarzschild radius) from the center of mass it will escape if launched radially, but fall in if launched horizontally.

For more details on the math see https://physics.stackexchange.com/q/262104

Community
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Yukterez
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    Probably true, but definitely won't work out for Earth because of the smallness of Schwarzschild radius – Andrii Magalich Jun 19 '16 at 22:43
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    @Andrii Magalich The effect might be small, but it will be there. Neglecting air resistance won't work on earth either, this is just a Gedankenexperiment [: – Yukterez Jun 19 '16 at 23:30