On the Moon, the acceleration due to gravity is weaker (about \( \frac{1}{6} \) of that on Earth), affecting how a spring balance measures weight.

A spring balance relies on the gravitational force to measure weight based on the stretch of the spring, so it would give a lower reading on the Moon.

In contrast, a beam balance compares masses based on equilibrium, unaffected by gravity. Thus, it can be used without any change to measure weight accurately on the Moon.

To find the depth \( d \) at which the effective acceleration due to gravity is \( \frac{g}{4} \), we use the formula for gravity at depth:

\[
g_d = g \left(1 - \frac{d}{R}\right).
\]

Setting \( g_d = \frac{g}{4} \):

\[
\frac{g}{4} = g \left(1 - \frac{d}{R}\right).
\]

Dividing both sides by \( g \):

\[
\frac{1}{4} = 1 - \frac{d}{R}.
\]

Rearranging gives:

\[
\frac{d}{R} = 1 - \frac{1}{4} = \frac{3}{4}.
\]

Thus:

\[
d = \frac{3R}{4}.
\]

So, the depth at which the effective value of acceleration due to gravity is \( \frac{g}{4} \) is \( \frac{3R}{4} \).

\[ g= \frac{GMr}{R^{3}}  \] for internal point

\[ g= \frac{GM}{r^{2}}  \] for external point

Total gravitational potential at a point \( r = \frac{a}{2} \):

1. Potential due to the mass at the center:
The gravitational potential at a distance \( r \) from a point mass \( M \) is given by:
\[
V_{\text{center}} = -\frac{GM}{r}
\]
At \( r = \frac{a}{2} \), this becomes:
\[
V_{\text{center}} = -\frac{GM}{\frac{a}{2}} = -\frac{2GM}{a}
\]

2. **Potential due to the spherical shell**:
Inside a spherical shell, the gravitational potential is constant and equal to the potential at the surface, which is:
\[
V_{\text{shell}} = -\frac{GM}{a}
\]

### Total potential at \( r = \frac{a}{2} \):

The total gravitational potential is the sum of the potentials due to the mass at the center and the shell:
\[
V_{\text{total}} = V_{\text{center}} + V_{\text{shell}} = -\frac{2GM}{a} - \frac{GM}{a} = -\frac{3GM}{a}
\]

So, the gravitational potential at a distance \( \frac{a}{2} \) from the center is:
\[
V = -\frac{3GM}{a}
\]

Escape velocity is given by the formula \( v_e = \sqrt{2gR} \). The ratio of escape velocity of earth to moon is \( \frac{v_{earth}}{v_{moon}} = \sqrt{\frac{g_{earth}}{g_{moon}} \times \frac{R_{earth}}{R_{moon}}} = \sqrt{6 \times 10} = \sqrt{60} \approx 7.75 \approx 8 \).

Each particle has mass \(m = M/2\). The force is \(F = \frac{G m^2}{R^2} = \frac{GM^2}{4R^2}\). Acceleration of each is \(a = \frac{F}{m} = \frac{GM}{2R^2}\). Relative acceleration is \(a_{\text{rel}} = 2a = \frac{GM}{R^2}\).

Since the field is uniform, potential varies linearly with height. Potential difference at \(5.0\text{ m}\) is \(\Delta V = 4 \times \frac{5.0}{10} = 2\text{ J/kg}\. Work done is \(W = m \Delta V = 2.0 \times 2 = 4.0\text{ J}\).

The potential at the center of a shell of mass \(M\) and radius \(R\) is \(-\frac{GM}{R}\). By superposition, the total potential at the common center is \(V = -\frac{GM}{R} - \frac{Gm}{r} = -G\left[\frac{M}{R} + \frac{m}{r}\right]\).

Using energy conservation in the center-of-mass frame: \(\frac{1}{2} \mu v_{\text{rel}}^2 = \frac{GMm}{d}\), where \(\mu = \frac{Mm}{M+m}\) is the reduced mass. Substituting \(\mu\) yields \(v_{\text{rel}} = \sqrt{\frac{2G(M+m)}{d}}\).

The increase in potential energy is \(\Delta U = \frac{mgh}{1 + h/R}\). Since \(h = nR\), we get \(\Delta U = \frac{mg(nR)}{1 + n} = \left(\frac{n}{n+1}\right) mgR\).