NEET Acceleration Due to Gravity and its variation MCQs & PYQs

From Newton's Third law of motion force applied by object A on B and Force applied by B on A will have equal magnitude.

The average density of the Earth (\( \rho \)) is directly proportional to the acceleration due to gravity (\( g \)) at the surface due to the relationship expressed by the formula:

\[
g = \frac{G M}{R^2},
\]

where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth,
- \( R \) is the radius of the Earth.

The mass \( M \) can be expressed in terms of density and volume:

\[
M = \rho V = \rho \left( \frac{4}{3} \pi R^3 \right).
\]

Substituting this into the equation for \( g \):

\[
g = \frac{G \left( \rho \cdot \frac{4}{3} \pi R^3 \right)}{R^2} = \frac{4\pi G \rho R}{3}.
\]

This shows that \( g \) is directly proportional to \( \rho \) (as \( R \) and \( G \) are constants), meaning that as the average density of the Earth increases, the value of \( g \) also increases.

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.

The time taken for an object to fall freely from a height \( h \) is given by the equation:

\[
h = \frac{1}{2} g t^2,
\]

where \( g \) is the acceleration due to gravity.

1. For Earth:
\[
h = \frac{1}{2} g_E t^2 ; t = \sqrt{\frac{2h}{g_E}}.
\]

2. For the Moon, where \( g_M \approx \frac{1}{6} g_E \):
\[
h = \frac{1}{2} g_M t_m^2 ; t_m = \sqrt{\frac{2h}{g_M}} = \sqrt{\frac{2h}{\frac{1}{6} g_E}} = \sqrt{12 \cdot \frac{2h}{g_E}}.
\]

This can be simplified using the time \( t \) from Earth:

\[
t_m = \sqrt{6} \cdot t = \sqrt{6} \cdot t.
\]

Thus, the time taken by the same body to reach the Moon's surface is approximately \( \sqrt{6} t \).

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

Acceleration due to gravity is \(g = \frac{GM}{R^2}\). Since mass \(M = \rho \times \frac{4}{3}\pi R^3\), we get \(g = \frac{G \left(\frac{4}{3}\pi R^3 \rho\right)}{R^2} = \frac{4}{3}\pi \rho GR\). Rearranging gives \(\rho = \frac{3g}{4\pi RG}\).

Acceleration due to gravity decreases with depth as \( g_d = g(1 - \frac{d}{R}) \) and with height as \( g_h = g(1 - \frac{2h}{R}) \). Since the formulas and rates of decrease are different, they decrease at different rates.

Acceleration due to gravity at the surface of a planet is given by \( g = \frac{4}{3} \pi G R \rho \). Therefore, \( \frac{g_1}{g_2} = \frac{R_1}{R_2} \times \frac{\rho_1}{\rho_2} = \frac{2}{3} \times \frac{3}{2} = 1 \).

Since acceleration due to gravity varies with height as \(g' = g \frac{R^2}{(R+h)^2}\), setting \(g' = g/9\) gives \(R+h = 3R\), which simplifies to \(h = 2R\).