Momentum of an ultrarelativistic electron I am aware of the relativistic equation:
$E^2 = (pc)^2 + (mc^2)^2$
And if we are dealing with a massless particle then $E = pc$
However I am doing some work in Astrophysics and have been told that the momentum of an ultra relativistic electron is $p =\frac E c$
I am confused as to why this is so seeing as though an electron does have a mass. 
 A: That is just an approximation. Of course electrons have mass, but for an ultrarelativistic electron you have that $p \gg m \implies \frac {m^2} {p^2} \approx 0$, so it is reasonable to make such an approximation. Explicitly you have:
$$E^2=p^2c^2+m^2c^4=p^2(c^2+ \frac {m^2}{p^2}) \approx p^2c^2 \implies p \approx \frac E c$$
A: As others have said, $p=\frac{E}{c}$ is an approximation in the ultra-relativistic case.
I will make this more explicit.
$p$ and $E$ are related to the spatial and temporal components of a 4-momentum vector $\tilde P$.
In terms of rapidities ($\theta$, where $v=c\tanh\theta$ for a timelike 4-momentum),
we have $p=mc\sinh\theta$ and $E=mc^2\cosh\theta$.
 So, for all timelike 4-momenta,
$$\frac{p}{E}=\frac{1}{c}\tanh\theta=\frac{1}{c}\left(\frac{v}{c}\right),$$
or equivalently,
$$p=\frac{E}{c}\tanh\theta=\frac{E}{c}\left(\frac{v}{c}\right),$$
As $v\rightarrow c$ but never reaching $c$ (that is, as $\theta\rightarrow\infty$) [while keeping $m$ fixed],
$$p \rightarrow \frac{E}{c}.$$ 
A: If you stipulate that the electron is ultrarelativistic then you are stipulating that its kinetic energy is far greater than its invariant energy $m_ec^2$.
Write the equation for the energy of the electron and assume $pc\gg m_ec^2$ so that we can approximate the radical as so
$$E = pc\,\sqrt{1 + \left(\frac{(m_ec^2)}{(pc)}\right)^2}\approx pc\left(1 + \frac{1}{2}\left(\frac{m_ec^2}{pc}\right)^2\right) = pc\left(1 + \frac{1}{2}\left(\frac{c}{\gamma v_e}\right)^2\right)$$
Now recall that $\gamma v_e$ becomes arbitrarily large as $v_e$ approaches $c$ and so, in the ultrarelativistic limit
$$\lim_{v_e \rightarrow c} E = pc$$
For example, if $v_e = 0.99999\,c$, then $\gamma = \frac{1}{\sqrt{1 - (0.99999)^2}} \approx 223.6$ and then
$$E = pc\,(1.00001000015)$$
and so this ultrarelativistic electron's momentum is well approximated as $p = \frac{E}{c}$ 
A: If a particle is ultrarelativistic it means that its momentum is way bigger than its mass. So in the equation
$$E^2=(pc)^2+(mc^2)^2$$
The second term is negligible and so you can approximate E=pc. The electron mass is very small (~0.5 MeV/c^2) so the ultrarelativistic regime is reached pretty soon. To convince yourself you can compute the energy of an electron with p=10 GeV considering or not its mass. You will see that including the mass in the calculation makes a barely visible difference at those energies.
