I know that in relativistic condition the increase in kinetic energy of an object increases its relativistic mass as $$m=\frac{m_0}{(1-v^2/c^2)^{1/2}},$$ and mass is another form of energy.
So my question is if that object have same amount of potential energy instead of kinetic energy then can we say that its relativistic mass is increased?
The short-short answer: "Potential energy always belongs to system rather than to a single object, and the system's mass is increased when you add potential energy to the system but the component parts do not change their masses."
I know, we often say that if you raise a book off the lab bench the book gains potential energy (attributing the energy to the book). But that's a short hand, because the motion of the book alone is not sufficient for the energy to change: it must make that motion in the presence of the planet. If you take the book and the bench into deep space and perform the same action ("lift" the book 1 meter "above" the bench) nothing special happens.
As an aside: the term "relativistic mass" isn't necessary here as the invariant mass of the system is increased. You'll find that despite its popularity in pop-sci sources and secondary-school text-books most physicists in sub-fields that use relativity all the time organize the discipline with a different set of definitions that don't include "relativistic mass" and instead recognize only one mass (called the "invariant mass" when you need to be painfully clear). Personally, I feel very strongly that calling $\gamma m$ the "relativistic mass" only encourages sloppy thinking about relativity and would encourage you to find a source that doesn't use it.
In that context it is worth noting that most of the mass of ordinary matter is binding energy due to the strong force, so most of the "ordinary mass" around you is of exactly the kind you are asking about.
To add to Statics' answer. You can think of potential energy due to other than the gravitational interaction (e.g. electromagnetic) as the energy of the field itself. Not only this additional mass exists but you can also know where in the system it is located.
Edit: A word of caution. If you calculate the energy of the field of a point charge, you will get infinity. That's because EM fields can act on the charge they were produced by. It makes sense then: for example the electrostatic potential is $~ \frac{1}{r} = \frac{1}{0} = \infty$. This is not very relevant to reality, because to model electrons we need quantum mechanics, but you can get limited success with modelling them classically as balls of finite radius instead of point charges. I believe, the Feynman lectures have a discussion about that.
Disregarding your choice of word in your question i.e. "relativistic", I can go ahead and answer. Energy and mass are the same thing, proportioned by the constant speed of light. To visualize this you could rewrite the formulae $$E = mc^2$$ to
$$m = E / c^2 = (m_0c^2 + V) / c^2 = m_0 + V/c^2$$
Where V is the potential energy. Of course you can incorporate your "relativistic" mass as a function of speed ( as you have written in your question) and it will still be true. I just skipped it here to simplify.
The potential energy is not limited to type of potential energy... It could be gravitational energy, or a compressed spring. In all cases the reference of frame need to be considered and not violated.
In special relativity the concept of potential is not defined. In general relativity potential is related to gravitational fields. To describe a potential of a massive star one could use the Schwarzschild metric.
Assuming that the object in a potential field is not moving the total energy $E$, which should stay conserved, is its rest mass $m_0$ plus the potential plus the kinetic energy. By defining the relativistic mass as the total energy of an object at rest one could see the dependence on the potential.
Derivation
The Schwarzschild metric can be writtten as $$\rm{d}s^2=-(1-\frac{2M}{r})dt^2 + (1-\frac{2M}{r})^{-1}dr^2 - r^2(d\theta^2+\sin^2\theta d\phi^2)$$ with $M$ the mass of the star, $r$ the distance from the star and angles $\theta$ and $\phi$. The metric is independent of time, hence there is a Killing vector $K^\mu=(1,0,0,0)$ related to the momentum dual vector $p_\mu$ such that $K^\mu \cdot p_\mu=const=p_0 =p_t=E$ - the total conserved energy of the object as seen at infinity.
The momentum mass relation reads $p^\mu \cdot p_\mu=-m^2_0$. Or in components: $$p^\mu \cdot p_\mu = p_\nu \cdot p_\mu\, g^{\mu\nu}=p_t p_t g^{tt} + p_r p_r g^{rr} + p_\theta p_\theta g^{\theta\theta} +p_\phi p_\phi g^{\phi\phi}=-m^2_0.$$Let the object be at rest: $p_r=p_\theta=p_\phi=0$. Hence one obtains $$p_t p_t g^{tt}=-E^2(1-\frac{2M}{r})^{-1}=-m^2_0.$$
$E$ is the total energy of the object at rest in a grav potential and can aswell be defined as the relativistic mass which would be dependent on the distance from the star or on the potential $$m=m_0\sqrt{(1-\frac{2M}{r})}.$$
The relativistic mass of an object at rest at infinity would just be its rest mass.
I think that the apparent mass increase is a property of the frame of reference. If there are two inertial frames and their relative velocity is $v$ and the $\gamma$ parameter is $\frac{1}{\sqrt{1-(v/c)^2}}$ then the mass $m_0$ in any of the rest frame is perceived as $\gamma m_0$ in the in other frame.
It should be noted that this type of argument is very crude and not considered accurate in modern terminology but for the sake of argument it will work.
Rise/fall of potential energy is also relative to the reference level, Simply by changing the reference level the potential energy of an object can go up or down. Hence in my opinion the change in potential energy can not increase the so called "relativistic mass".
Also from another point of view since $v=0 ,\ \gamma=1$.
hope this will help
