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The gravitational potential energy at infinity os supposed to be zero. Since body always moves towards lower potential, the gravitational potential is taken as negative so that gravitational potential energy can decrease on moving towards earth.

On the other hand,for near earth surface, potential is taken to be zero at earth surface so that body at height $h$ with $PE= mgh$ decreases and becomes zero at earth surface.

Can we take potential energy at height $h$ to be $mgh$ instead of $mgh$ and potential at earth surface to be towards negative infinity for sake of uniformity?

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Your first mistake is in the first word "the". There is no absolute "gravitational potential energy". Every specific choice of "potential energy" is always relative to some point (in the state space). If you choose that to be the point-at-infinity, then your gravitational potential energy (relative to that point) is negative on earth and increases as you move away from the earth, reaching $0$ at infinity. If you choose instead that point to be on the surface of the earth, then your (relative) gravitational potential energy is zero at the surface (by definition).

Your second mistake is in not understanding how the gravitational potential energy based on the assumption of a point mass is derived in Newtonian mechanics. The force on one point mass is $\frac{G·m_1·m_2}{r^2}$ towards the other point mass, where $m_1,m_2$ are the masses and $r$ is the distance between their centres of mass. One antiderivative with respect to $r$ is $-\frac{G·m_1·m_2}{r}$, but that is clearly not the only antiderivative. You can validly choose to use any antiderivative to define the gravitational potential energy; nothing requires you to make it equal to $0$ when $r = ∞$. But, it is convenient to make it so just because it is more elegant.

In fact, once there are more than two masses it is not so simplistic anymore. For example, at any point P between stationary point masses X,Y with masses $m_1,m_2$, the total gravitational force on an object at P with mass $m_3$ is $\frac{G·m_1·m_3}{r^2}-\frac{G·m_2·m_3}{(d-r)^2}$ towards X, where $r$ is your distance from X and $d$ is the distance from X to Y. As before, you can choose any antiderivative with respect to $r$ to define the total gravitational potential energy of P, and it is convenient to use $-\frac{G·m_1·m_3}{r}-\frac{G·m_2·m_3}{d-r}$ (i.e. no extra constant symbol), but that has nothing to do with being zero at infinity or whatever, since it is negative for every $r$ in the interval $(0,d)$, and it is maximum at some $r∈(0,d)$ such that $\frac{G·m_1·m_3}{r^2} = \frac{G·m_2·m_3}{(d-r)^2}$.

Also note that gravitational potential energy as derived above is not the right quantity to consider if you are talking about objects moving along with the earth's rotation (i.e. rotating reference frame), because you also need to add in the apparent centrifugal force due to the rotation. Just for instance, in the rotating reference frame centred at the earth that makes the moon stationary (assuming a circular orbit), the earth's gravitational force on the moon exactly cancels out the centrifugal force. The total potential energy likewise has to include both the gravitational potential energy as well as the effective potential energy due to the centrifugal force. (See for example the lagrangian points.)

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