Question

Vectors and Calculus

Bit like this one: http://openstudy.com/study#/updates/57f50118e4b05a233fe64d30

Answer 1

@IrishBoy123 I've not looked at the nitty gritty yet, but that's by far and away the best posting of a question I've seen on this site, possibly "ever". Bravo to you. Oh, how I WISH that others (self incl) could/would follow suit. Bravo encore.

Answer 2

tu @osprey

i think the next part of the process is a drawing.

Answer 3

@IrishBoy123 ARRRRGGHHH .. you mean there's MORE ??? Have another G, if you like it !

Answer 4

@IrishBoy123 "Twickers" sounds vaguely like "London Irish" ? Or am I deaf ?

Answer 5

So "Twickers" isn't Twickenham ?

Answer 6

@IrishBoy123 OK, so Wallabies is Aus rugby union team and Pumas is ? It isn't a question of "being bothered", its a question of "symbols/maths fatigue". That said, your use of the Gaussian surface, and mentioning thereof perhaps lifts the "algebra spirits". Most of what I do is based on guesswork, glimmers and glimpses. I don't know where I'd rank in the "efficiency stakes", and I find it pretty demanding. Problem is that it can also be v. entertaining. As King Lear may not have said, according to W Shakespeare ... "you think me mad ?" Have another G.

Answer 7

lol!!

Pumas = Argentina

Surprisingly good these days as they were invited into the S Hemi tourney....

Answer 8

OUTSIDE the earth

Modelling the earth, with radius R, as a series of concentric shells as per the drawing, for each shell of radius \(r' \) we have \(V' = \dfrac{4}{3} \pi r'^3 \) so \(dV' = 4 \pi r'^2 dr'\) and \(m'= 4 \rho \pi r'^2 dr'\) where \(\rho\) is the uniform density of earth

We are told that the potential function is

\(\phi(r) = \begin{cases}
const. & r \leq R \\
-\dfrac{Cm'}{r} & r > R
\end{cases}
\)

So from \(\vec F = - \nabla \phi\) in spherical, we have

\(\vec F (\vec r) = \begin{cases}
0 & r \leq R \\
-\dfrac{Cm'}{r^2} & r > R
\end{cases}
\)


So for a particle outside the earth's surface at position \(\vec r = r ~ \hat r\) relative to the shell's centre, the force between a single shell and mass m is

\( d \vec F = -\frac{Cm'}{r^2} ~ \hat r\)

\(= -\dfrac{C 4 \rho \pi r'^2 dr'}{r^2} ~ \hat r\) and we can reduce this to a new constant by combining other constant terms so that

\(d \vec F = -\dfrac{C }{r^2} r'^2 dr' ~ \hat r\)

Integrating:

\( \vec F = -\dfrac{C }{r^2} \int\limits_0^R r'^2 dr' ~ \hat r\)

\(= -\dfrac{C }{r^2} R^3 ~ \hat r\) [adjusting the constant C again to include the 1/3]

Now as per the hint, at \(r = R\): \(-mg = -\dfrac{C }{R^2} R^3 \implies C = \dfrac{mg }{R}\)

\(\implies \mathbf { \vec F = -\dfrac{mg R^2 }{r^2} ~ \hat r }\)

For the potential function \(\phi(r)\), with gravitational potential energy \(\phi(r \to \infty) = 0\), the work done **by gravity** to move the mass from \( \infty\) to it's position at \(\vec r = r ~ \hat r\) would be:

\(W_g = \int_{\infty}^{r} \vec F (\vec r) \bullet d \vec r\)

And:

\( \vec F (\vec r) \bullet d \vec r \)

\(= ( -\dfrac{mg R^2 }{r^2} ~ \hat r + 0 ~ \hat \theta + 0 ~ \hat \phi ) \bullet (dr ~ \hat r + r d \theta ~ \hat \theta + r sin \theta d \phi ~ \hat \phi)\)

\(= -\dfrac{mg R^2 }{r^2} ~ dr\)


So:


\( = \int\limits_{\infty}^{R} \left(-\dfrac{mg R^2 }{r^2} \right) dr\)


\( = \left[\dfrac{mg R^2 }{r} \right]_{\infty}^{r} = \dfrac{mg R^2 }{r} \)

This is the loss in gravitational potential energy of the mass from startpoint \(\phi (r \to \infty) = 0\) so its potential energy is now:

\(\mathbf{\phi(r) = - \dfrac{mg R^2 }{r} }\)