# I Escape Velocity = Freefall from Height of One Radius?

Tags:
1. Feb 9, 2016

### fizixfan

I was doing some calculations using the escape velocities from Earth, Moon and Mars. Then by chance I calculated the velocities attained when an object was "dropped" from a height of the radius on each of these bodies, assuming the acceleration due to gravity remained constant during the fall. Escape velocity is the minimum velocity needed to escape a gravitational field.

For example, escape velocity on earth, Vesc = 40,270 km/h (given).
Using the simple formula V2 = (2ar)^0.5, where
V2 = velocity after falling distance r; a = acceleration due to gravity; r = radius of earth (initial velocity = 0), and

a = 9.80665 m/s^2
r = 6,378,000 m

V2 = (2*9.80665*6378000)^0.5 = 11,184.53 m/s = 40,264.29 km/h

This value of 40,264 km/h compares very closely to the escape velocity of 40,270 km/h

I thought perhaps this was a coincidence, so I calculated V2 for the Moon and Mars and compared them to the known escape velocities for each body.

Moon:
Escape velocity = Vesc = 8,533.6 km/h
a = 1.62 m/s^2
r = 1,737,150 m
V2 = (2*1.62*1737150)^0.5 = 2,372.418 m/s = 8,540.7 km/h
Again, a very close fit.

Mars:
Escape velocity = Vesc = 18,108 km/h
a = 3.72761 m/s^2
r = 3,389,500 m
V2 = (2*3.72761*3389500)^0.5 = 5026.875 m/s = 18,097 km/h
Another very close approximation.

I haven't done the calculations for the other planets, moons, or the sun, but I expect I would get similar results. From these results I speculate that Vesc = V2. This seems to be a quick and dirty formula for calculating escape velocity, and might be handy for a stranded astronaut. But am I really just reiterating Vesc = (2Gm/r)^0.5? (G = gravitational constant; m = mass, e.g., of earth; r = radius, e.g., of earth).

I any case, using this method avoids a lot of the very intricate calculations necessary for determining Vesc the standard way, as long as the acceleration due to gravity and radius are known.

Sources:
http://www.livescience.com/50312-how-long-to-fall-through-earth.html
[PLAIN]http://keisan.casio.com/exec/system/1360310353[/PLAIN] [Broken]
http://www.wolframalpha.com/input/?i=escape+velocity+Mars

Last edited by a moderator: May 7, 2017
2. Feb 9, 2016

### mgkii

Not sure you'd get the same results for the Sun or the gas giants; the average density of the body will impact the mass/radius ratio and therefore the escape velocity/radius ratio. Could be that you've fallen on a happy coincidence, but it could equally be something deeper - the numbers do seem to match within a small range!

3. Feb 9, 2016

### RUber

Given that $a = G\frac{m}{r^2},$ you are essentially saying that you found:
$\sqrt{2ar} = \sqrt{2\frac{Gm}{r}}$
Expanding a,
$\sqrt{2G\frac{m}{r^2}r} = \sqrt{2G\frac{m}{r}} .$
It looks like these principles might all be based on the same physics.

4. Feb 9, 2016

### mgkii

Ooops! I'd missed the OP had used "a" in the equation

5. Feb 9, 2016

### Khashishi

Escape velocity is the velocity you get from dropping from infinite distance away, not just one radius away. You got the right answer but for the wrong reasons. You basically made two errors which happened to luckily cancel out.
Surface gravity g is just $g=GM/R$. If you plug in g into the equation
$v_{esc} = \sqrt{\frac{2GM}{r}}$
you get
$v_{esc} = \sqrt{2gr}$
which is what you have (re-labeling g as a). But you have to think logically through it and not invent the wrong reasons.

6. Feb 9, 2016

### fizixfan

I got the same results using V2 = (2ar)^0.5 for the Sun as well: V2 = 617.2 km/s vs Vesc = 617.6 km/s
Where a = 273.7 m/s^2 and r = 696,000,000 m.

Using the Vesc formula for the Earth in Excel-friendly format: Vesc = (2*G*m/r)^0.05
where
G =
a =
r =
I get Vesc =(((2*(6.67*10^-11)*(5.9721986*10^24))/6367.4447*1000)^0.5)/10^6 = 11.19 km/s
whereas using V2 = (2ar)^0.5 = 11.18 km/s (calculation shown above in my first post)

7. Feb 9, 2016

### fizixfan

Isn't escape velocity the velocity required to escape the gravitational field of a body? Wouldn't dropping from an infinite distance away be the cosmic escape velocity, or at least the velocity required to escape the solar system?

The V2/Vesc values I calculated for the Sun, Earth, Moon and Mars are in 99.9% agreement. I'm not a rocket scientist, but I think this is more that just luck (?)

8. Feb 9, 2016

### mathman

Above is incorrect - inverse square law makes it g/4 at the start.

9. Feb 10, 2016

### jbriggs444

Yes. The claim being made is that escape velocity from the planetary surface is equal to the velocity that a falling object would have if it fell from a height equal to the planetary radius and did so at an acceleration equal to the surface acceleration of gravity the whole way. That claim is correct.

Ultimately, this derives from the fact that the integral of $\frac{1}{r^2}$ is $-\frac{1}{r}$. The latter corresponds to energy per unit mass. The former corresponds to acceleration. Hand-waving past the details, they differ by a factor of r. Escape energy per unit mass = r times surface acceleration.

10. Feb 10, 2016

### fizixfan

Thank you jbriggs444! Glad to have this confirmed.

11. Feb 10, 2016

### MrAnchovy

But note that the acceleration due to gravity at a height of r above the Earth's surface is NOT g, it is given by $\frac {GM}{(2r)^2} = \frac g4$ as mathman pointed out, and a body falling from a height r will not accelerate at a constant rate.

This means you CANNOT say (as the tile of this thread asserts) that Escape Velocity = Freefall from Height of One Radius. What you can say is that escape velocity is equal to the velocity attained after acceleration at a constant rate equal to surface gravity over a distance equal to the planetary radius.

12. Feb 10, 2016

### Janus

Staff Emeritus
However, you can say that an object dropped from a height of one radius above the surface will reach the surface(r) at a speed equal to the orbital speed at the surface.

Gravitational potential energy at r: -GMm/r
Gravitational potential energy at 2r: -GMm/(2r)
Difference in potential energy: GMm/r-GMm/(2r)
Kinetic energy; mv^2/2
Kinetic energy at r after being dropped from 2r is equal to potential energy difference between r and r2:
mv^2/2 = GMm/r-GMm/(2/r)
v^2/2=GM/r-GM/(2r)
v^2 = 2GM/r-GM/r
v^2 = GM/r

13. Feb 11, 2016

### fizixfan

I did say in my original post, "I calculated the velocities attained when an object was 'dropped' from a height of the radius on each of these bodies, assuming the acceleration due to gravity remained constant during the fall."

14. Feb 11, 2016

Staff Emeritus
Yes you did. But since this assumption is incorrect, the solutions will be as well. "What would the laws of physics say would happen if we violate the laws of physics" is asked a lot here, but it is unanswerable.

15. Feb 11, 2016

### fizixfan

Yes, it does violate the law of physics. So do questions like, how long would it take to fall through the earth? Plenty of people still answer this unanswerable question.

But it does give a very close estimate of escape velocity if the gravitational constant (G) and mass of the planet (m) are not known. So it has some use in that respect.

16. Feb 11, 2016

### MrAnchovy

I think you are missing the point that the two are identically equal due to the fact that $- \frac 1g$ differentiates to $\frac 1{g^2}$.

I am not sure what circumstances could enable you to know the surface gravitation of a body without knowing $Gm$

17. Feb 11, 2016

### MrAnchovy

Yes you did, but not everyone who reads the title and clicks through to the thread will read through all the posts, and if the thread ends with a post that appears to confirm the assertion in the title they will be mislead.

18. Feb 11, 2016

### A.T.

Crash landing with your space ship on a random planet. You still need the radius though.

19. Feb 11, 2016

### Janus

Staff Emeritus
Or, a little less destructive:
1.Put your ship in a circular orbit around the planet.
7. Escape velocity from your orbit altitude will be $\sqrt{2}$ times greater than your orbital speed.