# Galaxies Lecture 22

### Tidal Forces

In addition to the normal tidal forces which act to stretch an orbitting body along the radial action of gravity, tidal shear results when an object (like a globular cluster) differentially orbits around a massive body (that is, the near edge is moving in orbit more quickly than the far edge). The degree of shear depends on the relation between the differential rotation and the self-gravity of the system. In practice, we are in a rotating frame (our sun) which is also rotating, so we either need to find an inertial frame of reference or account for the Coriolis forces.

In a rotating frame, the force per unit mass is:

${\displaystyle {\ddot {R}}=-{\theta ^{2} \over R}+R\Omega ^{2}\,\!}$

where R is the distance from the galactic center.

{\displaystyle {\begin{aligned}{\frac {d}{dR}}{F_{R}}r&=\left({\theta \over R^{2}}-{2\theta \over R}{\frac {d}{dR}}{\theta }+\Omega ^{2}\right)r\\&=2r\Omega _{0}\left(\Omega _{0}-{\frac {d}{dR}}{\theta }{\bigg |}_{R_{0}}\right)\\\end{aligned}}\,\!}

where ${\displaystyle \Omega _{0}=A-B}$, where A and B are the Oort constants. We may then write the condition for stability as:

${\displaystyle {{\frac {d}{dR}}{F_{R}}r\geq 4r(A-B)}\,\!}$

where ${\displaystyle r}$ is the distance across the orbitting body. Futhermore, the acceleration from self-gravity on the cluster is:

${\displaystyle g={F \over m_{*}}={GM \over r^{2}}={G{\frac {4}{3}}\pi \rho _{cc}r^{3} \over r^{2}}\,\!}$

Thus, the condition for stability is:

${\displaystyle {\rho _{cc}\geq {3 \over \pi G}A(A-B)}\,\!}$

If a cluster were hypothetically in the solar neighborhood, this would come out to be:

${\displaystyle \rho _{cc}\geq {3\cdot 235\cdot 15\cdot 25 \over 10^{6}}=8.4\cdot 10^{-2}{M_{\odot } \over pc^{3}}\,\!}$

In fact, as has been stated many times, the density in the solar neighborhood is ${\displaystyle 0.1{M_{\odot } \over pc^{3}}}$, so such objects in our neighborhood would be tidally stable.

For another case, suppose we are in an inertial frame (we are the galaxy). In this case, we have no Coriolis term:

{\displaystyle {\begin{aligned}{F_{R} \over m}&=-{\theta _{c}^{2} \over R}\\{\frac {d}{dR}}{F_{R}}&={\theta _{c}^{2} \over R^{2}}-2{\theta _{c} \over R}{\frac {d}{dR}}{\theta _{c}}\\\end{aligned}}\,\!}

where ${\displaystyle \theta _{c}}$ is the circular angular velocity of our galaxy around the center of mass. Using ${\displaystyle g={\frac {4}{3}}\pi \rho _{r}r^{2}{G \over r^{2}}}$, we have the condition for stability is:

${\displaystyle {\frac {d}{dR}}{F_{R}}r<{GM \over r^{2}}\,\!}$
${\displaystyle {\left({\theta _{c}^{2} \over R^{2}}-2{\theta _{c} \over R}{\frac {d}{dR}}{\theta _{c}}\right)r<{\frac {4}{3}}\pi G\rho _{cc}}\,\!}$

If we assume we have a flat rotation curve, then ${\displaystyle {\frac {d}{dR}}{\theta _{c}}=0}$, so we have

${\displaystyle {\rho _{cc}={3 \over 4\pi G}{\theta _{c}^{2} \over R^{2}}={3 \over 4\pi G}\Omega ^{2}}\,\!}$

For a dwarf galaxy at 50 kpc, this comes to:

${\displaystyle \rho _{gal}={3\cdot 235\cdot 220^{2} \over 4\pi (50\cdot 10^{3})^{2}}=1.1\cdot 10^{-3}{M_{\odot } \over pc^{3}}\,\!}$

We can estimate the density of a dwarf galaxy: it has ${\displaystyle 10^{6}}$ stars and has a diameter of about ${\displaystyle 1kpc}$, so we have ${\displaystyle \rho _{cc}\sim 10^{-3}{M_{\odot } \over pc^{3}}}$, so this galaxy would be on the border of stable. Anything closer (like the Sagittarius Dwarf) will get totally ruined. By the way, the “235” in the above equations is ${\displaystyle G}$ in units of km, s, and pc’s.