The mass/energy of an $H$-atom and the gravitational force between it and another particle of mass $m$ The gravitational force between an $H$-atom and another particle of mass $m$ will be given by Newton's law: 
$$F=\frac{GMm}{r^2}$$
The question is, what is $M$ here?  I thought the answer would be $$M=m_{proton} + m_{electron}.$$
But the answer given is $$M=m_{proton}+m_{electron}-\frac{B}{c^2}$$ where $B$ is the Binding Energy.
I am not able to understand why this extra term came into play. Is this in some way because of the separation between the proton and electron, that is, the $H$-atom not being a point-mass?
 A: Suppose you start with a proton and an electron separated by a large distance. The mass of this system is just $m_p + m_e$.
Now let the proton and electron fall towards each other under their mututal electrostatic attraction. As they fall they will speed up, so by the time the proton and electron are about one hydrogen atom radius apart they're moving with a high speed. Note that we haven't added or removed any energy, so the mass/energy of the system is still $m_p + m_e$.
The trouble is that this won't form a hydrogen atom because the proton and electron will just speed past each other and fly away again. To form a hydrogen atom we have to take the kinetic energy of the electron and proton out of the system so we can bring them to a stop. Let's call the kinetic energy $E_k$. This energy has a mass given by Einstein's famous equation $E = mc^2$, so the mass of our atom is the mass we started with less the energy we've taken out:
$$ m_H = m_p + m_e - \frac{E_k}{c^2} $$
And $E_k$ is just the binding energy. That's why we have to subtract off the binding energy.
Or you could look at the problem the other way round. Start with a hydrogen atom of mass $m_H$. To split the electron and proton apart we have to add energy. In fact the amount of energy we have to add is just the ionisation energy, $E_i$, and this adds a mass of $E_i/c^2$.
Having split the atom we now just have a separate proton and electron, with a combined mass of $m_p + m_e$, so we have:
$$ m_H + \frac{E_i}{c^2} = m_p + m_e $$
and a quick rearrangement once again gives us:
$$ m_H  = m_p + m_e - \frac{E_i}{c^2} $$
A: Because E = mc^2, B/c^2 = m. The binding energy ends up adding mass to the system of the atom.
