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Why is the total energy of an electron in an atom negative?

We know that $$E = -\frac{e^2}{8\pi\epsilon_0a_0}.$$ What does the negative sign in the above equation means?

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    $\begingroup$ You can see parallel to a planet moving around the sun. If the energy of the planet is negative, it's moving along the ellipse, if it's positive or zero, it's hyperbolic or parabolic motion. It's easy with planets, because $E = T + V$ and here the potential energy $V$ is stronger than kinetic energy $T$. Remember $V \sim -r^{-1}$. If the body has positive energy, it means it has energy to beat the potential of the sun and fly away. In atoms we don't assume parabolic, hyperbolic or elliptic motions, but the energy idea is the same. $\endgroup$
    – Jimmy Found
    Commented Apr 16, 2017 at 8:42

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It simply indicates that, to take the electron from its "orbit" around the nucleus and move it to infinity where the potential energy between the electron and the nucleus is $0$ requires us to supply at least $+E$. If we supply more than $+E$ the electron will have some kinetic energy left when it reaches infinity.

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We cannot measure the absolute value of energy of a Bohr orbit as mentioned by you.We can measure only the relative energy with respect to a fixed reference.In this case we fix our reference to the electron at an infinite distance (by infinite distance I do not mean that the electron is some light years away, it means that the electron is just far away from the influence of the nucleus of the atom) where the energy is assumed to be zero since there is no interaction. Now with respect to this energy level the electrons in the orbits are bound so they have lower energy which should be less than zero (i.e. negative energy as you said)

So the negative sign before the energy has a physical significance which tells us that the electron is bound to nucleus unlike the electron which is free having zero interaction energy.

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