$(a)$ Decreases, $(b)$ Decreases, $(c)$ Mass of the body, $(d)$ More

Acceleration due to gravity at depth $h$ is given by the relation:

$g _{h}=\left(1-\frac{2 h}{R_{ c }}\right) g$

Where,

$R_{e}=$ Radius of the Earth $g =$ Acceleration due to gravity

on the surface of the Earth

It is clear from the given relation that acceleration due to gravity decreases with an increase in height.

Acceleration due to gravity at depth $d$ is given by the relation:

$g _{d}=\left(1-\frac{d}{R_{e}}\right) g$

It is clear from the given relation that acceleration due to gravity decreases with an increase in depth.

Acceleration due to gravity of body of mass $m$ is given by the relation:

$g =\frac{ G M}{R^{2}}$

Where,

$G =$ Universal gravitational constant

$M=$ Mass of the Earth

$R=$ Radius of the Earth

Hence, it can be inferred that acceleration due to gravity is independent of the mass of the body.

Gravitational potential energy of two points $r_{2}$ and $r_{1}$ distance away from the centre of the Earth is respectively given by:

$V\left(r_{1}\right)=-\frac{ G m M}{r_{1}}$

$V\left(r_{2}\right)=-\frac{G m M}{r_{2}}$

Difference in potential energy, $V=V\left(r_{2}\right)-V\left(r_{1}\right)=-\operatorname{GmM}\left(\frac{1}{r_{2}}-\frac{1}{r_{1}}\right)$

Hence, this formula is more accurate than the formula $m g\left(r_{2}-r_{1}\right)$