Escape Velocity = Freefall from Height of One Radius?

  • #26
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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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  • #27
Janus
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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
 
  • #28
jbriggs444
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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.
 
  • #29
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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.
 
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  • #30
jbriggs444
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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.
 
  • #31
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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.
 
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  • #32
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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.
 
  • #33
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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).
 
  • #34
jbriggs444
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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.
 
  • #35
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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.
 
  • #36
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Isn't this whole discussion just a little silly? All it amounts to is that if object B (e.g., a rocket) leaves object A (e.g., earth) with enough velocity to reach a height equal to the radius of the object A, it has almost escaped. In fact, if it's initial velocity had been just a bit higher, it would have escaped! So of course the initial velocity needed to reach height h=r is quite close to Vesc (h=∞).

Repeat your calculations to see what initial velocity is needed to reach h=2r and you'll magically find that it is an even closer approximation to Vesc! :smile:
 
  • #37
jbriggs444
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Isn't this whole discussion just a little silly? All it amounts to is that if object B (e.g., a rocket) leaves object A (e.g., earth) with enough velocity to reach a height equal to the radius of the object A, it has almost escaped. In fact, if it's initial velocity had been just a bit higher, it would have escaped! So of course the initial velocity needed to reach height h=r is quite close to Vesc (h=∞).
Welcome to the Physics Forums!

But by what measure is 70% of escape velocity "quite close"?
 
  • #38
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But by what measure is 70% of escape velocity "quite close"?
I also do not understand why people wish to change the definition of escape velocity and agree with you -perhaps the weak field decreasing as inverse square of distance leads them to such calculations of approximation.
however it is best to calculate the potential at the surface of any planet and equate it to the kinetic energy required to escape .
 
  • #39
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Isn't this whole discussion just a little silly? All it amounts to is that if object B (e.g., a rocket) leaves object A (e.g., earth) with enough velocity to reach a height equal to the radius of the object A, it has almost escaped. In fact, if it's initial velocity had been just a bit higher, it would have escaped! So of course the initial velocity needed to reach height h=r is quite close to Vesc (h=∞).
Actually, once that rocket reaches an altitude of 1r, it will at that moment be at rest with respect to the Earth and find itself ~7.9 km/sec short of the velocity it would need to be traveling in order to continue away from the Earth without falling back.
Repeat your calculations to see what initial velocity is needed to reach h=2r and you'll magically find that it is an even closer approximation to Vesc! :smile:
Closer, but still ~6.5 km/sec short at the top of the trajectory.
 
  • #40
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My comment was based on the original premise (post #1) that the acceleration is constant, in which case, quoting from post #7:

"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 (?)"

That's what I was referring to. The difference in velocity between a drop from height h and from height ∞ obviously approaches 0 as h approaches ∞, so I think my comment stands, given the original assumptions. I was trying to point out that it wasn't just luck; it would have been even better and more mysterious luck if he had arbitrarily started with h=πr. :smile:

However, the same principle holds when more accurate, realistic calculations are employed, except that the distance at which V2 and Vesc begin to agree closely moves further out. I didn't bother to calculate the correct h at which V2/Vesc = 0.99.
 
  • #41
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As the escape velocity is given by
v2=2gR
where R is the radius of the planet and g is the acceleration of gravity on the surface, you can associate the formula with any uniform accelerated motion with acceleration g and over distance R.
And is an exact match, not an approximation. There is nothing special in imagining that R is height above ground. You can go towards the center of the Earth with constant acceleration g or on a tangent or whatever.

If my car reaches 100 km/h over a distance of 100 m, it will have an average acceleration of 3.8 m/s2.
It will reach the same 100 km/h if I drop it from a height of 100 m on Mercury, assuming g has the constant surface value of about 3.8 m/s2. Does this mean that there is a deep connection between Nissan and the Mercurians? :)

Is this such a big subject to discus for 41 posts (including mine :) )
 
  • #42
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"Is this such a big subject to discus for 41 posts (including mine :) )"

That's precisely the point I was trying to make in post #36 when I said how silly it was to have so many posts going on and on about details that had nothing to do with the original post that used a simple constant g model.
 
  • #43
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"Is this such a big subject to discus for 41 posts (including mine :) )"

That's precisely the point I was trying to make in post #36 when I said how silly it was to have so many posts going on and on about details that had nothing to do with the original post that used a simple constant g model.
The thread had been dead for 48 hours until you woke it up again saying how silly it was that a discussion that had finished was still going on...
 
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  • #44
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Thread is done.
 

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