$\textstyle \parbox{3.25in}{ We do the usual Newtonian analysis, keeping track o... ...F_\theta = - m_g\, g\, \sin\theta$\ (see diagram for the relevant triangles).}$ $\begin{center}\vbox{\input{pendulum.eepic} }\end{center}$
Setting this applied force equal to times the acceleration $l\ddot{\theta}$ in the $\theta$ directions gives

$\begin{displaymath}m_i l\ddot{\theta} = - m_g\, g\, \sin\theta \ \ .\end{displaymath}$

Making the usual small angle approximation and regrouping gives a simple harmonic oscillator equation

$\begin{displaymath}\ddot{\theta} = -\ \frac{m_g}{m_i}\ \frac{g}{l}\ \theta \end{displaymath}$

with frequency

$\begin{displaymath}\omega = \sqrt{\frac{m_g}{m_i}\ \frac{g}{l}}\end{displaymath}$

and period

$\begin{displaymath}T = \frac{2\pi}{\omega} = 2\pi r\ \sqrt{\frac{m_i\, l}{Gm_g\,M_g}}\ ,\end{displaymath}$

substituting in for the gravitational field .
1. On the surface of the neutron star, we have gravitational acceleration
  
  $\begin{eqnarray*} g &=& \frac{GM}{R^2} \ \ =\ \ \frac{(6.7 \times 10^{-11} \ {\r... ...10^{12}\ {\rm N/kg} \ \ =\ \ 1.9 \times 10^{12}\ {\rm m/s^2}\ \ .\end{eqnarray*}$
  
  Note that this gravitational attraction is $2 \ \times\ 10^{11}$ times that on the earth, since the neutron star is $5\ \times 10^5$ times more massive and the radius of earth.
2. The freefalling observer O sees light follow a horizontal path, reaching at . The external observer sees the light as having an additional vertical displacement, equal to the downward fall of 's frame in the time it takes the photon to reach . Neglecting time dilation means we take
  
  $\begin{displaymath}t' = t = l/c,\ \ \end{displaymath}$
  
  so the vertical displacement is
  
  $\begin{displaymath}\Delta y' = \frac{1}{2}\ g{t'}^2 = \frac{1}{2}\ g{t}^2 = \frac{1}{2}\ \frac{gl^2}{c^2}\ \ .\end{displaymath}$
  
  Thus sees light bend downward through the angle
  
  $\begin{displaymath}\theta \ \ =\ \ \tan^{-1} \ \frac{ \Delta y' }{l} \ \ = \ \ \tan^{-1} \ \frac{ gl }{2c^2}\ \ .\end{displaymath}$
  
  Numerically, this is
  
  $\begin{displaymath}\theta \ \ =\ \ \tan^{-1} \ (10^{-5}\ m^{-1})\ l\ \ .\end{displaymath}$
  
  For the cases given,
  
  $\begin{eqnarray*} l= 1\ {\rm m} &\Rightarrow& \theta \ \ =\ \ \tan^{-1} \ 10^{-5... ...htarrow& \theta \ \ =\ \ \tan^{-1} \ 10^{-3}\ \ =\ \ 10^{-3}\ \ \end{eqnarray*}$
  
  both much greater than the analogous bending on earth.
3. At , is travelling upward with respect to the freefalling frame.
4. Using (b) and (c), the bending angle is given by
  
  $\begin{displaymath}\theta \ \ =\ \ \tan^{-1} \ \frac{ gl }{2c^2}\ \ =\ \ \tan^{-1} \ \frac{ v }{2c} \ \ =\ \ \tan^{-1} \frac{ \beta }{2} \ \ .\end{displaymath}$
  
  Thus the bending angle (a general relativistic effect) becomes nonnegligible only when $\beta$ is nonnegligible -- that is, when special relativistic effects become significant. This means that a full special relativistic treatment is necessarily to accurately calculate the effect, which we haven't attempted here.
We take the lower object to coincide with the origin of the satellite frame, with object 2 having $r_2 = r_1 + \Delta r$ . The main point is that objects 1 and 2, both in circular orbit about the earth, have slightly different angular frequencies, since for each the gravitational acceleration due to the earth balances the centripetal acceleration of the orbit:

$\begin{displaymath}\frac{GM_E}{r^2} = r\omega^2\ \ .\end{displaymath}$

This gives for the frequencies of the two objects

$\begin{displaymath}\omega_1 = \sqrt{\frac{GM_E}{r_1^3}} \hspace{1in} \omega_2 = \sqrt{\frac{GM_E}{r_2^3}} \ \ .\end{displaymath}$

The satellite origin, and object 1, undergo one revolution in time $T = 2\pi/\omega_1$ . In this time, object 2 has orbited with angular frequency $\omega_2$ , so it lags object 1 by the angle

$\begin{displaymath}\Delta \, \theta = 2\pi - \omega_2\, T = 2\pi\ (\ 1 - \frac{ \omega_2}{ \omega_1}\ )\ .\end{displaymath}$

But

$\begin{displaymath}\frac{ \omega_2}{ \omega_1} = \left(\ \frac{r_1}{r_2}\ \right... ...}\ \right)^{3/2} \approx 1 - \frac{3}{2}\ \frac{\Delta r}{r_2} \end{displaymath}$

to first order in $\Delta r$ . Thus object 2 lags by the angle

$\begin{displaymath}\Delta\theta = 3\pi\ \frac{\Delta r}{r_2}\ \ ,\end{displaymath}$

corresponding to the horizontal displacement $3\pi\,\Delta r$ .

About this document ...

This document was generated using the LaTeX2HTML translator Version 2K.1beta (1.47)

Copyright © 1993, 1994, 1995, 1996, Nikos Drakos, Computer Based Learning Unit, University of Leeds.
Copyright © 1997, 1998, 1999, Ross Moore, Mathematics Department, Macquarie University, Sydney.

The command line arguments were:
latex2html -split 0 soln4

The translation was initiated by on 2004-04-06

2004-04-06