One of the frequent questions asked in general physics, mechanics and astrodynamics as well as general astronomy is to do with

__escape velocity__, i.e.

*'What is escape velocity*?' The answers given by various profs, lecturers, physics teachers can range from the purely tautological, i.e. "

*It is the velocity needed to escape from the Earth's gravity, or 11.2 kilometers per second*", to the convoluted - involving the potential for a small, very slow moving object to gradually thrust its way to the critical speed. (With no thought to the fact that something like a "toy helicopter" would disintegrate long before achieving such speed, even it experienced prolonged acceleration over time, e.g. such as 1 m/s/s).

Man of the convoluted answers come in response to apparently innocent (but untutored questions) such as:

**1) Couldn't an object of any mass move "free" of Earth's gravity ( at least free of falling back to Earth) whether it is moving away from Earth at 1 mph or 20,000 mph?**

**2) Does escape velocity refer to a velocity that must be reached in order to "coast" out of gravity's effective reach?**

**3)Or is the escape velocity specific orbiting velocity?**

In the case of questions (1) and (2) the person has no clue that the force of gravity, say arising from the Earth, doesn't just suddenly cease at a specific boundary. To appreciate this one can make use of the Newtonian force equation for (gravitational attraction) and set it equal to the weight (product of mass m by acceleration of gravity) at Earth's (mass M

_{E}) surface, whence.

mg
= G M

_{E}m/ r^{2}
g = GM

_{E}/r^{2}The important aspect of this equation is the inverse square relation of g to the distance, r. One can thereby see that even at an enormous distance (say r = 10 million km) there will still be a minuscule g. More important, this concept leads the way to an even more fundamental one that bears directly on the escape velocity. We call this the gravitational potential V.

We can pursue this manner of definition, and at the same time disallow any definitions that make use of objects which may start out at say, 1 m/s and experience continual thrust (acceleration) to finally reach the escape velocity. While this is certainly possible it is not regarded as a practical modus operandi for attaining

**v**economically and hence doesn't meet the criteria for acceptable definition. (For example, starting out with such a low velocity and only allowing a = 1 m/s

_{esc}^{2 }would take 11,200 secs or just over three hours to attain escape velocity - which is absurd.)

Indeed, if anyone proposed this in an advanced celestial mechanics or astrodynamics class he'd likely be laughed out of the lecture room.

This elicits the question: 'What does?'

Perhaps the best definition I have found which meets the criteria is the one from the

__Oxford Dictionary of Physics__- which I reproduce below:

"The minimum speed needed by a space vehicle, rocket etc. to escape from the gravitational field of the moon, Earth or other celestial body. The gravitational force between a rocket of mass m and a celestial body of mass M is:

F = GMm/ r

^{2 }

Therefore, the gravitational potential energy of the rocket with respect to its possible position very far from the celestial body (on which it is resting) can be shown to be:

V = - GMm/ r

assuming by convention the potential energy is zero at an infinite distance from the celestial body.

If the rocket is to escape the gravitational field it must have a kinetic energy (K) that exceeds this potential energy, i.e.

m

**v**

_{esc}^{2}/ r must be greater than MmG/r

or

**v**> [2MG/ r]

_{esc}^{1/2}

This is the value of the escape velocity"

---

In the case of

**v**with respect to Earth, we can work it out. The Earth's radius is 6396 km or call it 6.4 x 10

_{esc}^{6}m.

G, the Newtonian gravitational constant is G = 6.7 x 10

^{-11}Nm

^{2}/ kg

^{2}

The mass of the Earth is 6 x 10

^{24}kg

Then:

[2MG/ r]

^{1/2}= [2 (6 x 10

^{24}kg) (6.7 x 10

^{-11}Nm

^{2}/ kg

^{2})/ 6.4 x 10

^{6}m ]

^{1/2}

This comes out to

**v**= 11,200 m/s or 11.2 km/s

_{esc}We can frame this in terms of

**conservation of energy**also, such that the kinetic (K) plus potential energy (V) for the gravitating system must equal zero.

Or:

0 = K + V = mv

^{2 }/ r + (-GMm/r) = mv

^{2 }/ r - GMm / r

With a glance at the attached image one can see a plot with V, the gravitational potential energy (GPE) vs. r.

As shown here, the zero of GPE is taken at infinity, i.e. V = 0 at r = oo. The dotted line applies to an object - say satellite- in

**, so that we see in any such case E < 0 which means the object is in a gravitational potential "well". All objects. i.e. planets or moons, bound to a more massive object in some orbit are said to be in a gravitational potential "well" and their value of V is negative.**

*a bound orbit*By contrast, if the dotted line were situated above the zero line we would have an unbound or 'hyperbolic' orbit. This is only an orbit in the loosest sense, in that it fulfills one of the Keplerian conic sections most students learn in freshman calculus, i.e.

.

Also, the escape velocity is not the

*specific orbiting velocity*since the energy needed to attain the former is greater than the latter. In general, for a body-object much less massive than say the Earth (m << M) we have the mean orbital speed v(0)

= v / [2]

^{1/2 }

I.e. lower than escape velocity by a factor of root 2.

Most interested readers should be able to figure out from the Oxford definition and also the GPE curve provided why the escape velocity

*cannot be less*than what it is, i.e. for the Earth! (Hint: Apply the conservation of energy)

__Suggested insight problems__:

1) Use the principle of

*conservation of energy*to find the escape velocity for the planet Mars. Take Mars' mass M = 6.4 x 10

^{23}kg and its radius R = 3390 km.

2) Mars' moon Deimos takes 30.3 hours to make one orbit. Using Kepler's 3rd law of planetary motion work out its mean distance (a) and thereby estimate the value of the gravitational potential V, of Deimos relative to Mars.

3) The condition for a gas molecule of mass m to escape from a planet of mass M and radius r is given as:

6 [3kT / m]

^{1/2}> [2MG/ r]

^{1/2}

Where k is

**, and T is the surface temperature in degrees Kelvin. Use the above condition to ascertain the extent to which primitive oxygen on Mars could have escaped. Take T = 300 K and use the Martian planetary properties in (1).**

*the Boltzmann constant*Note: calculations show that if the mean molecular speed is as much as 1/3 the velocity of escape then the planet will lose up to half the gas from its atmosphere (of that composition) in a few weeks. If the molecular speed is even one -fifth of

**v**the gas will disperse into space in a few hundred million years. The ability of fractions of gas to escape is directly a result of the

_{esc}*Maxwell-Boltzmann distribution*of gas velocities. Can you explain?

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