**Question:**

Choose the correct alternative:

(a) Acceleration due to gravity increases/decreases with increasing altitude.

(b) Acceleration due to gravity increases/decreases with increasing depth. (assume the earth to be a sphere of uniform density).

(c) Acceleration due to gravity is independent of mass of the earth/mass of the body.

(d) The formula $-G M m\left(1 / r_{2}-1 / r_{1}\right)$ is more/less accurate than the formula $m g\left(r_{2}-r_{1}\right)$ for the difference of potential energy between two points $r_{2}$ and $r_{1}$ distance away from the centre of the earth.

**Solution:**

(a) Decreases

(b) Decreases

(c) Mass of the body

(d) More

Explanation:

(a) Acceleration due to gravity at depth *h* is given by the relation:

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

Where,

= 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.

(b) Acceleration due to gravity at depth *d* is given by the relation:

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

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

(c) Acceleration due to gravity of body of mass *m* is given by the relation:

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

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.

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

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

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

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

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