# Escape Velocity = Freefall from Height of One Radius?

• fizixfan
In summary: V2 = (2ar)^0.5, where a = surface acceleration of gravity and r = radius of the planetary body. This is derived from the fact that the integral of 1/r^2 is -1/r and that energy per unit mass, corresponding to the latter, differs from acceleration, corresponding to the former, by a factor of r.
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]
http://www.wolframalpha.com/input/?i=escape+velocity+Mars

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mechpeac
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!

fizixfan
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.

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

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.

RUber said:
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.

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)

Khashishi said:
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.

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 (?)

assuming the acceleration due to gravity remained constant during the fall.

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

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

fizixfan
jbriggs444 said:
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.

Thank you jbriggs444! Glad to have this confirmed.

fizixfan said:
Thank you jbriggs444! Glad to have this confirmed.
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.

MrAnchovy said:
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.

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

MrAnchovy said:
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.

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."

fizixfan said:
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."

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.

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.

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.

fizixfan said:
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.
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 ##

fizixfan said:
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."
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.

MrAnchovy said:
I am not sure what circumstances could enable you to know the surface gravitation of a body without knowing ## Gm ##
Crash landing with your spaceship on a random planet. You still need the radius though.

A.T. said:
Crash landing with your spaceship on a random planet. You still need the radius though.
Or, a little less destructive:
1.Put your ship in a circular orbit around the planet.
3.From your altitude and the angular size of the planet from your view port, calculate the radius of the planet.
4.Measure the length of your orbital period.
5. With the information from steps 2,3 and 4, compute your orbital speed.
7. Escape velocity from your orbit altitude will be $\sqrt{2}$ times greater than your orbital speed.
8. Knowing that escape velocity decreases by the squareroot of r, calculate the escape velocity from the surface of the planet.

MrAnchovy said:
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.

The last post is not going to end with a confirmation, obviously. I didn't have room to include the caveat (gravity assumed constant), and I can't go back and change it. Anyone who takes the time to read through all the posts will see that I have been disabused of my erroneous notion. They will also probably lose sight of any possible merit it might have.

Janus said:
Or, a little less destructive:
1.Put your ship in a circular orbit around the planet.
3.From your altitude and the angular size of the planet from your view port, calculate the radius of the planet.
4.Measure the length of your orbital period.
5. With the information from steps 2,3 and 4, compute your orbital speed.
7. Escape velocity from your orbit altitude will be $\sqrt{2}$ times greater than your orbital speed.
8. Knowing that escape velocity decreases by the squareroot of r, calculate the escape velocity from the surface of the planet.

Or, following steps 1 through 7 above, calculate the extra velocity you would need to achieve escape velocity from your orbiting position, and proceed without having to go back to the surface. After all, fuel might be at a premium.

Another possible technique is to use Eratosthenes' method for calculating the radius, r, of the planet (I'm sure someone like Matt Damon would be up to a long walk). Then take a rock and climb up to a height of land with a vertical cliff, drop the rock, and calculate how long it would take to hit the ground using a digital stopwatch. Maybe do this three times and take the average value. Of course, Matt Damon would have accurately estimated the height of the cliff using triangulation or with a laser rangefinder (distance from point A to base of cliff and top of cliff). So then he has an estimate of a, the acceleration due to gravity at the surface. Then, using Vesc ≈ (2ar)^0.5, he would have a very accurate estimate of his escape velocity.

BTW, I also used Vesc vs V2 for Callisto, Titan and Ganymede, and the results were >99.9% similar.

MrAnchovy said:
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 ##

You could measure surface gravitation by dropping a rock off a cliff and seeing how long it takes to hit the ground. As long as you have some way to measure the height of the cliff, you would have your estimate, a = 2d/t^2

fizixfan said:
So do questions like, how long would it take to fall through the earth? Plenty of people still answer this unanswerable question.

That's different. That's a practical impossibility. At the risk of being direct, you made an error in your OP, and this whole thread is trying to find something of value in this error. I don't think there is much.

fizixfan said:
They will also probably lose sight of any possible merit it might have.

I don't see much merit in exploring the consequences of an error. The usual procedure is appropriate - fix the mistakle

russ_watters
fizixfan said:
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.
fizixfan said:
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.

I think the question has been handled not in a physical way- if the numbers do come same if one moves in correct manner then only one gets alternative ways of computing the same physical parameters;
if one wants to calculate escape velocity then one should calculate the work done in carrying a body from the surface of Earth to infinity -that is escaping the gravitational field-
so start with writing the force experienced by a body at r and the work done in moving it away by a displacement dr which will be F. dr -integrate the expression of work done and equate it to the kinetic energy required then one gets the value of that velocity called escape velocity.
at a radial point r the force F equals to GMm/ r^2 where m is the mass of the body escaping and m is the mass of earth.this force is directed towards decresing radius -towards centre. take a scalar product of force and displacement ;integrate it and take the values.
you are bound to get correct value. its wrong to take the average value of g- a bad approximation and that is disturbing your calculation- its value is essential as in nocase your escaping vehicle should fall short of energy required to escape and it will speed towards the Earth and get burned.

That's different. That's a practical impossibility. At the risk of being direct, you made an error in your OP, and this whole thread is trying to find something of value in this error. I don't think there is much.
I don't see much merit in exploring the consequences of an error. The usual procedure is appropriate - fix the mistakle

How do I "fix the mistakle"? And BTW, what is an OP?
Escape Velocity = Freefall from Height of One Radius?
It was a question, not a declaration.

If I were to "drop" an object from a height of one Earth radius (r) what would its velocity be when it hit the earth? The force of gravitation at this altitude would be 1/4 that at sea level, and would increase as the falling object approached the surface. How would I calculate this?

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fizixfan said:
How do I "fix the mistakle"? And BTW, what is an OP?
Escape Velocity = Freefall from Height of One Radius?
It was a question, not a declaration.

If I were to "drop" an object from a height of 4,000 miles what would its velocity be when it hit the earth? The force of gravitation at this altitude would be 1/4 that at sea level, and would increase as the falling object approached the surface. How would I calculate this?

OP means "original post". For your last question, use conservation of energy. The object has ##-\frac{G\ m\ M_{earth}}{r_1}## mechanical energy at ##r_1## distance from the center of the Earth when you "drop" it and ##\frac{1}{2}mv^2-\ \frac{G\ m\ M_{earth}}{R_{earth}}## mechanical energy when it hits the surface. Equate the two and solve for ##v##!

edit: changed "conversation of energy" to "conservation of energy" although this is technically a conversation with energy as a topic.

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fizixfan
fizixfan said:
How do I "fix the mistakle"? And BTW, what is an OP?
Escape Velocity = Freefall from Height of One Radius?
It was a question, not a declaration.

If I were to "drop" an object from a height of one Earth radius (r) what would its velocity be when it hit the earth? The force of gravitation at this altitude would be 1/4 that at sea level, and would increase as the falling object approached the surface. How would I calculate this?
I answered this question in post #12

fizixfan said:
If I were to "drop" an object from a height of one Earth radius (r) what would its velocity be when it hit the earth? The force of gravitation at this altitude would be 1/4 that at sea level, and would increase as the falling object approached the surface. How would I calculate this?
It seems plausible that you are asking about how to calculate the work done by a force applied over a distance when that force is not constant. If the force were constant, we all know that one can simply multiply force by [parallel] distance. When the force is variable, that does not work. If you are familiar with calculus, the answer is that instead of multiplying force by distance, you integrate force over distance.

If you are not familiar with calculus then the idea is that you split up the interval into a bunch of smaller intervals. If these intervals are small enough, the force over each little interval is close to constant and you can multiply the force for each such interval by the length of that interval. Graphically, if you plot distance on the horizontal and force on the vertical axis, you are calculating the area under the curve by dividing it up into a bunch of vertical strips and adding up the area of all of those strips. The total area corresponds to the total work done. The area of each little strip corresponds to the work done crossing that incremental distance.

As you make the strips narrower and narrower, the computed sum gets closer and closer to the correct figure for work done. That is the essence of the integral calculus -- figuring out the limit that the sum tends to as the strips get thinner and thinner.

If you integrate a ##\frac{1}{r^2}## force, the area under the curve turns out to be ##\frac{1}{r_{start}}\ - \ \frac{1}{r_{end}}##. That fact is the underlying source from which a number of formulas already given in this thread have been derived.

Janus said:
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

I tried using your formula V^2 = GM/r (assuming V = (GM/r)^0.5) where
G = 6.67*10^-11 m^3/kg s^2
M = 5.9722*10^24 kg
r = 6,367,444.7 m
and the result was 7,909.47 m/s ?
If I use V = (2GM/r)^0.5, I get Vesc = 11,185.68 m/s = escape velocity

So, is this the correct calculation - that an object dropped from a height of one radius above Earth's surface will reach the surface at a speed of 7,909.47 m/s? Or am I using the wrong units or doing something else wrong?

Please note: I'm no longer trying to "prove" something based on an assumption that violates the laws of physics. I'm here to learn. Thanks.

DocZaius
fizixfan said:
I tried using your formula V^2 = GM/r (assuming V = (GM/r)^0.5) where
G = 6.67*10^-11 m^3/kg s^2
M = 5.9722*10^24 kg
r = 6,367,444.7 m
and the result was 7,909.47 m/s ?
If I use V = (2GM/r)^0.5, I get Vesc = 11,185.68 m/s = escape velocity

So, is this the correct calculation - that an object dropped from a height of one radius above Earth's surface will reach the surface at a speed of 7,909.47 m/s? Or am I using the wrong units or doing something else wrong?
Check it with Google. "escape velocity from Earth's surface" yields https://en.wikipedia.org/wiki/Escape_velocity which yields a figure of 11.2 km/sec and "orbital velocity of earth" yields https://en.wikipedia.org/wiki/Orbital_speed which yields a figure of 7.8 km/sec.

Note that you reported a result with six significant digits even though one of your inputs had only three. It would have been better to round off your results.

fizixfan said:
I tried using your formula V^2 = GM/r (assuming V = (GM/r)^0.5) where
G = 6.67*10^-11 m^3/kg s^2
M = 5.9722*10^24 kg
r = 6,367,444.7 m
and the result was 7,909.47 m/s ?
If I use V = (2GM/r)^0.5, I get Vesc = 11,185.68 m/s = escape velocity

So, is this the correct calculation - that an object dropped from a height of one radius above Earth's surface will reach the surface at a speed of 7,909.47 m/s? Or am I using the wrong units or doing something else wrong?

Please note: I'm no longer trying to "prove" something based on an assumption that violates the laws of physics. I'm here to learn. Thanks.
It's in the right ballpark.
However, I will point out a couple of things about your calculation. One, since you only carry G out to three significant digits, it really makes no sense to carry M and r out to more than that same number of significant digits. Also, if accuracy is important, it is better to use 3.986004418e14 for GM, as we know it to a greater accuracy than we do for either the mass of the Earth, or G. In addition the Earth, being an oblate spheroid has an radius that varies from 6356.8 to 6378.1 km. So the value you give is close to the average between these two, but the actual impact velocity would depend on the actual radius at the point of impact.

fizixfan and jbriggs444
Janus said:
It's in the right ballpark.
However, I will point out a couple of things about your calculation. One, since you only carry G out to three significant digits, it really makes no sense to carry M and r out to more than that same number of significant digits. Also, if accuracy is important, it is better to use 3.986004418e14 for GM, as we know it to a greater accuracy than we do for either the mass of the Earth, or G. In addition the Earth, being an oblate spheroid has an radius that varies from 6356.8 to 6378.1 km. So the value you give is close to the average between these two, but the actual impact velocity would depend on the actual radius at the point of impact.

Yes, you're right about the significant digits. If I use GM = 3.986004418e14 and r = (6356.8+ 6378.1)*1000/2 = 6,367,450 m, I get v = (GM/r)^0.5 = 7,912.00 m/s or 7.912 km/s.

jbriggs444 said:
Check it with Google. "escape velocity from Earth's surface" yields https://en.wikipedia.org/wiki/Escape_velocity which yields a figure of 11.2 km/sec and "orbital velocity of earth" yields https://en.wikipedia.org/wiki/Orbital_speed which yields a figure of 7.8 km/sec.

Note that you reported a result with six significant digits even though one of your inputs had only three. It would have been better to round off your results.

Yes, I was being sloppy with my significant digits. Using Janus's more accurate value of GM (see above), I get an orbital velocity (or the velocity of an object dropped from one Earth radius) of 7.912 km/s.

Similar values are given at http://keisan.casio.com/exec/system/1360310353 (7.907 km/s) and at http://hyperphysics.phy-astr.gsu.edu/hbase/orbv3.html (7.904 km/s where h = 6367.4447 km).

Interestingly (to me at least), Vorbit ≈ Vesc*(2^0.5/2).

fizixfan said:
Interestingly (to me at least), Vorbit ≈ Vesc*(2^0.5/2).
If you look at the formulas for each, that is not just an approximation. It is an equality.

jbriggs444 said:
If you look at the formulas for each, that is not just an approximation. It is an equality.

That's what I thought, but I was a bit hesitant to use the "equals" sign again.

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