Question

A space probe is to be launched from a space station 200 miles above Earth. Determine its escape velocity in miles/s. Take Earth's radius to be 3960 miles.

Answer 1

@SithsAndGiggles

Answer 2

If you consider the free body diagram of the probe, the only force acting on it (presumably) is the gravitational force of the Earth, so you have the DE
\[m\frac{d^2x}{dt^2}=-\frac{mgR^2}{(x+R)^2}\]
(where \(x(t)\) is the position of the probe, \(m\) is the mass of the probe, \(g\) is the acceleration due to gravity, and \(R\) is the radius of the Earth)

To solve this you'll need two initial conditions (initial velocity and height), but you're only given one. According to this link, you can write the initial velocity as \(\sqrt{2gR}\). The derivation is all here: http://www.unc.edu/math/Faculty/mccombs/232/appdiffeqkey.pdf

Answer 3

So after plugging the values, I got...
\[\frac{ d ^{2}x }{dt ^{2}}=-\frac{ 9801}{338}\]

Answer 4

Now what do I do?

Answer 5

Your RHS is missing the dependent variable...

Answer 6

So what should it be? What dependent variable did I miss?

Answer 7

Plugging in all relevant values, the ODE is
\[\frac{d^2x}{dt^2}=-\frac{3960^2g}{(x+3960)^2}\]
(Since \(R\) is in units of miles and \(d^2x/dt^2\) is in miles per second per second, you need to convert \(g\) appropriately.)

Answer 8

And then how do I solve for x(t) from there?

Answer 9

According to this link ( http://www.unc.edu/math/Faculty/mccombs/232/appdiffeqkey.pdf ), the ODE is
\[\frac{dv}{dt}=-\frac{3960^2g}{(x+3960)^2}\]
I'm not sure if I interpreted the equivalence properly, i.e. if \(\dfrac{dv}{dt}=\dfrac{d^2x}{dt^2}\)...

Answer 10

I think it holds, but it complicates the problem. Instead, you need to make a substitution so that the independent variable is \(x\), and not \(t\).

To do this, notice that \(\dfrac{dv}{dt}=\dfrac{dv}{dx}\times\dfrac{dx}{dt}\). Since \(v=\dfrac{dx}{dt}\), you then have \(\dfrac{dv}{dt}=v\dfrac{dv}{dx}\). Now the equation is
\[v\frac{dv}{dx}=-\frac{3960^2g}{(x+3960)^2}\]
This equation is separable.

Answer 11

But shouldn't we have a value for g?

Answer 12

Yes, it's the constant \(9.81\dfrac{\text{m}}{\text{s}^2}\), Like I said earlier, you need to convert this to \(\dfrac{\text{mi}}{\text{s}^2}\) before you plug it in. If it helps, go ahead and solve
\[v\frac{dv}{dx}=-\frac{gR^2}{(x+R)^2}\]
first, then plug in the constants later.

Answer 13

So I solved \[v \frac{ dv }{ dx }=-\frac{ 3960^{2} g}{ (x+3960)^{2} }\]

Answer 14

and got\[\frac{ v^2 }{ 2 }=\frac{ 3960^2g }{ x+3960 }+C\]

Answer 15

\[v^2=\frac{ 31363200g }{ x+3960 }+C\]

Answer 16

Right. We don't have an explicit initial condition, so for the time being, we let the initial launch velocity be \(v_0\) when \(t=0\). At this time, though, we also have that \(x(0)=200\text{ mi}\). In other words, for \(v(x)\), you have \(v(200)=v_0\).

Plugging this into the general solution, you get
\[\frac{{v_0}^2}{2}=\frac{3960^2g}{200+3960}+C\]
Solve for \(C\) in terms of \(v_0\).

Answer 17

Once you find \(C\), you find the escape velocity by examining what happens to \(v_0\) as the height grows arbitrarily large; that is, compute \(\lim\limits_{h\to\infty}v_0\)

Answer 18

(where \(h\) denotes the height; replace it with \(x\) for this context.)

Answer 19

So I got\[C=v _{0}^2-\frac{ 31363200g }{ 4160 }\]

Answer 20

Note that when solving for \(C\), it's no longer a general constant, so \(2C\not\to C\). Instead, the particular solution should be
\[\frac{ v^2 }{ 2 }=\frac{R^2g }{ x+R}+\underbrace{\frac{{v_0}^2}{2}-\frac{R^2g}{200+R}}_{C}\]
(I've replaced 3960 with \(R\) for a general radius).

According to that link, when the height is maximized, the velocity is 0 (I'm still trying to figure out the physical interpretation of that, but I'll take their word for it at the moment), i.e. for max height \(x_{\text{max}}\), you have \(v=0\). Plugging this in gives
\[0=\frac{R^2g}{x_{\text{max}}+R}+\frac{{v_0}^2}{2}-\frac{R^2g}{200+R}\]
and solving for \(v_0\), you have
\[v_0=\sqrt{2\left(\frac{gR^2}{200+R}-\frac{gR^2}{x_{\text{max}}+R}\right)}\]
As \(x_{\text{max}}\to\infty\), you thus get
\[v_0=\sqrt{\frac{2gR^2}{R+200}}\]
where \(g=9.81\dfrac{\text{m}}{\text{s}^2}=0.006096\dfrac{\text{mi}}{\text{s}^2}\) and \(R=3960\text{ mi}\). In other words, the escape velocity should be about \(6.77\dfrac{\text{mi}}{\text{s}}\).

Answer 21

Thank you!