Expert: Philip A. Stahl - 5/25/2009

Question
QUESTION: I've always been told Earth's escape velocity is 25,000 mph, but never understood WHY? If a space ship lifted off, hypothetically, with loads of fuel and attained...say...3,000 mph and was able to maintain that speed, wouldn't it eventually escape Earth's gravity? After all, the farther it gets from Earth, the weaker gravity's pull. Why does it have to go 25,000 mph?

ANSWER: Hello,

You probably have never understood escape velocity - the concept, because it was always presented to you in a non-mathematical way that was superficial. Hence, there was no way you could corroborate it for any given velocity and it seemed ill-conceived.

It is a tragedy that too many in the U.S. flee and fear math and the use of mathematical arguments, points ...because that is often the optimum way to press on and really grasp a concept. If then a math basis were available to you, you'd easily see why a velocity of 3,000 mph simply wouldn't make the cut no matter how long it was sustained.

So let's start from basics and not be fearful of employing the math (AND physics!) needed! Consider the start point at the surface of Earth (say sea level as at Cape Kennedy or close to it!) and r(I) = R(E) = 6400 km, the radius of the Earth. We designate the initial velocity v(I) = v, and call the final velocity v(F) = 0 since it will always be this value at maximum altitude.

Now, we need to use or frame the TOTAL energy of the system which is:

E(TOTAL) =  KE   + PE

or kinetic energy plus potential energy (in the latter case, GRAVITATIONAL potential energy)

Then:

E(TOTAL) =  mv(I)^2/ 2 - GM(E)m/r(I) =  mv(F)^2/2 - GM(E)m/r(F)


where G is the Newtonian gravitational constant: 6.7 x 10^-11 N-m^2/kg^2

M(E) is the mass of Earth =  6 x 10^24 kg

and m is the mass of the object to be launched

Now, since we know for the purposes of escape velocity:

r(F) = R(max)  = final alt. relative to R(E)

v(F) = 0

and r(I) = R(E)= initial alt. then:

mv(I)^2/2  - GM(E)m/ R(E) =  -GM(E)m/ R(max)

since v(F) = 0 and that term (mv(F)^2/2) = 0 so drops out

Using basic algebra to solve for the initial velocity (e.g. velocity needed to attain R(max):

v(I)^2 =  2GM(E) (1/ R(E)  -  1 / R(max)

Thus, if the initial velocity or v(I) is known, it can be used to compute the maximu altitude or h =  R(max) - R(E)

This then permits us to compute the *minimum velocity* an object needs to escape the Earth's gravity well or gravitational field.

Conceptually, this corresponds to a position such that the object can reach infinity with final speed 0 or v(F) = 0 when R(max) = oo.

Now, if R(max) = oo

the term:  1 / R(max) =   1/ oo = 0

so:

v(I)^2 =  2GM(E) [1/ R(E)]

Take the square root of both sides:

v(I) =   {2GM(E)/ R(E)}^1/2

Since all values on the RHS are known, this can be computed:

v(I) = [{2 x 6.7 x 10^-11 N-m^2/kg^2 x 6 x 10^24 kg)/ 6.4 x 10^6 m]^1/2


v(I) =  1.12 x 10^4 m/s

Now, multiply this by 3600 (seconds in one minute) and you get:

v(I) = 4.035 x 10^7 m/hour

and divide this by 1000 m/ km:

v(I) = 4.035 x 10^4  km/hour

and convert to miles per hour - realizing that 1 km = 0.625 miles

v(I) = 4.035 x 10^4 km/h (0.625 mi/km) =  2.522 x 10^4 miles/h

or, just over 25,000 mph

Hope this helps, but if not, please ask further questions!


---------- FOLLOW-UP ----------

QUESTION: Philip,

I appreciate your time and quick reply, but....WOW! That went right over my head. It was simply a long math equation. My degree's in psychology! :)  So with no common basis of understanding, it didn't help much. Do you have a non-math explanation by any chance? Again, I'm thankful for your time.

Answer
Hello,

Well, there is a non-math explanation but: a) it won't eliminate fully your 3,000 mph belief (that an object can attain escape velocity) and b) won't establish why a velocity of 25,200 mph does fulfill the requirement.

Basically, one can think of the Earth and its gravitational field like a gravitational "well". Visualize a lead ball resting atop a rubber sheet held taut. The ball produces a deformation in the sheet, a "hole" or "well" depression about 2-3" deep.

Visualize an object coming from this lead ball and trying to make it to the edge of the hole, and thence to escape it and maybe roll away separately. This is the analog of the escape velocity concept.

If the object just barely makes it to the edge and then "loop the loops" around the circumference of the "well" (without leaving it) it's the analog of an artificial satellite which as attained Earth orbit at ~ 18,000 mph. If the object actually can *leave* the well (moving out onto the flatt rubber sheet)  then it has attained 25,000 mph as I had worked out for you earlier.

Again, there is no way -minus the math-  to show either of those velocities are the respective ones needed - for orbit and escape. Once one adopts the descriptive approach, quantifying the speeds is simply not feasible.

Btw, no insult intended, but even with a psych degree don't you have to know algebra and statistics? I would have thought the presentation was well within that scope since it mostly relied on basic algebra. Ok, "intermediate" algebra, the type one needs to be au fait with on SAT or GRE exams.