Pytels Dynamics 12.8: Missile dynamics, acceleration and escape velocity

[Alexanddros81]

Homework Statement

A missile is launched from the surface of a planet with the speed v₀ at t=0.
According to the theory of universal gravitation, the speed v of the missile after launch
is given by
v² = 2gr₀(r₀ / r -1) + v₀²

where g is the gravitational acceleration on the surface of the planet and r₀ is the mean
radius of the planet.
(a) Determine the acceleration of the missile in terms of r.
(b) Find the escape velocity that is the minimum value of v₀ for which the missile will not return to the planet.
c)Using the result of part (b) calculate the escape velocity for earth, where g = 9.81m/s² and
r₀ = 6370km

Homework Equations

The Attempt at a Solution

I have tried part (a) . My attempt is found in the attached file.
Another way to go is to use the v² = 2a(x-x₀) + v₀²
and compare with given equation. I have tried that and couldn't get -g(r₀/r)²
How do i solve part (b)? any hints

 

[haruspex]

I have tried part (a)

$\sqrt{a-b}$ is not the same as $\sqrt a- \sqrt b$.
Try differentiating without taking the square root.

For b, if it does not return, how large does r get?

 

[Alexanddros81]

Check for part a and part b in attached files
If it doesn't return then r→∞
Is part b correct?

 

[haruspex]

Check for part a and part b in attached files
If it doesn't return then r→∞
Is part b correct?

Remarkably, you got the right answers to both a and b via flawed algebra.

In a, you differentiated the left hand side with respect to t and the right with respect to r. You must always do the same thing to both sides of an equation. If you differentiate both sides wrt t then on the right you get a factor $\frac{dr}{dt}$ (chain rule).
But in the next line, the derivative of v² with respect to t is $2v\frac{dv}{dt}=2va$, not just 2a. These two errors canceled out to give the right result.

In b, you again have made the blunder of distributing √ across a sum of terms. $\sqrt{v^2+2gr_0}$ is not $\sqrt{v^2}+\sqrt{2gr_0}$. Is √1+√4=√(1+4)? You got away with that because in the case you are interested in v=0.

 

[Alexanddros81]

Ok got it!
On the left side we let z=v² and take the dz/dt = (dz/dr)(dr/dt) = 2va
On the right side we do the same : 2gr₀² dz/dt where z=1/r => 2gr₀²(dz/dr)(dr/dt) =
= 2gr₀²(-1)r^-2v
the 2 and v cancel out
This is implicit differentiation isn't it?

 

[haruspex]

Ok got it!
On the left side we let z=v² and take the dz/dt = (dz/dr)(dr/dt) = 2va
On the right side we do the same : 2gr₀² dz/dt where z=1/r => 2gr₀²(dz/dr)(dr/dt) =
= 2gr₀²(-1)r^-2v
the 2 and v cancel out
This is implicit differentiation isn't it?

Yes. But I think you mean dz/dt=dz/dv dv/dt on the left.

 

[Alexanddros81]

yes.