From Equations of motion

The height of a jump is inversely proportional to the acceleration due to gravity. Let \( h_A \) be the height of the jump on planet A, and \( h_B \) be the height of the jump on planet B. Also, let \( g_A \) and \( g_B \) represent the acceleration due to gravity on planets A and B, respectively.

Given:
- \( g_A = 9g_B \)
- \( h_A = 2 \, \text{m} \)

The ratio of heights is:

\[
\frac{h_B}{h_A} = \frac{g_A}{g_B} = 9
\]

Thus, the height on planet B is:

\[
h_B = 9 \times h_A = 9 \times 2 = 18 \, \text{m}
\]

So, the height of the jump on planet B is \( 18 \, \text{m} \).

\[ g=\frac{GM}{R^{2}}= \frac{G\times \rho\frac{4}{3}\pi R^{3}}{R^{2}}=\frac{4G\rho\pi R}{3} \]

\[ \frac{g_{1}}{g_{2}}= \frac{R_{1}.\rho_{1}}{R_{2}.\rho_{2}}= \frac{4\times 1}{1\times 2}= \frac{2}{1} \]

The depth at which the acceleration due to gravity becomes \( \frac{1}{n} \) times the surface value is:

\[
d = \left( 1 - \frac{1}{n} \right) R
\]

where \( R \) is the radius of the Earth.

The weight of a body at height \( h \) is given by:

\[
W_h = W_0 \left( \frac{R}{R + h} \right)^2.
\]

To find \( h \) when \( W_h = \frac{1}{16} W_0 \):

\[
\frac{1}{16} = \left( \frac{R}{R + h} \right)^2.
\]

Taking the square root:

\[
\frac{1}{4} = \frac{R}{R + h}.
\]

Cross-multiplying gives:

\[
R + h = 4R ; h = 3R.
\]

Thus, the height is \( 3R \).

The acceleration due to gravity \( g' \) on the new planet can be calculated using the formula:

\[
g' = \frac{GM}{R^2}.
\]

For the new planet:

- Its radius \( R' = 3R \) (3 times the radius of Earth).
- Its mass \( M' \) is given by \( M' = \text{density} \times \text{volume} = \rho \times \frac{4}{3} \pi (R')^3 = \rho \times \frac{4}{3} \pi (3R)^3 = 27 \times \rho \times \frac{4}{3} \pi R^3 = 27M \) (mass is 27 times that of Earth).

Substituting into the formula:

\[
g' = \frac{G(27M)}{(3R)^2} = \frac{27GM}{9R^2} = 3g.
\]

Thus, \( g' = 3g \).

The height \( h \) at which gravity decreases by 36% (to 64% of its surface value) is found using:

\[
g_h = \frac{gR^2}{(R + h)^2}.
\]

Setting \( g_h = 0.64g \):

\[
0.64 = \frac{R^2}{(R + h)^2}.
\]

Taking the square root:

\[
0.8 = \frac{R}{R + h}.
\]

Cross-multiplying gives:

\[
0.8(R + h) = R \implies 0.8h = 0.2R \implies h = 0.25R=R/4
\]

Thus, the height is R/4.

To find the height \( h \) at which the weight of a body becomes \( \frac{1}{16} \) of its weight on the surface of the Earth, use the formula:

\[
W_h = W_0 \left( \frac{R}{R + h} \right)^2.
\]

Setting \( W_h = \frac{1}{16} W_0 \):

\[
\frac{1}{16} = \left( \frac{R}{R + h} \right)^2.
\]

Taking the square root:

\[
\frac{1}{4} = \frac{R}{R + h}.
\]

Cross-multiplying gives:

\[
R + h = 4R ; h = 4R - R = 3R.
\]

Thus, the height is \( 3R \).

The acceleration due to gravity \( g \) on the surface of a spherical planet can be calculated using the formula:

\[
g = \frac{GM_p}{R_p^2},
\]

where:
- \( G \) is the gravitational constant,
- \( M_p \) is the mass of the planet,
- \( R_p \) is the radius of the planet.

Given that the diameter \( D_p = 2R_p \), we can express the radius as \( R_p = \frac{D_p}{2} \).

Substituting this into the formula gives:

\[
g = \frac{GM_p}{\left(\frac{D_p}{2}\right)^2} = \frac{GM_p}{\frac{D_p^2}{4}} = \frac{4GM_p}{D_p^2}.
\]

Thus, the acceleration due to gravity experienced by the particle near the surface of the planet is:

\[
g = \frac{4GM_p}{D_p^2}.
\]

The change in the value of \( g \) at height \( h \) above the surface of the Earth and at depth \( d \) below the surface can be expressed using the following formulas:

1. **At height \( h \)**:
\[
g_h = g \left(1 - \frac{2h}{R}\right) \quad \text{(for small } h\text{)}
\]

The change in \( g \) is:
\[
\Delta g_h = g - g_h = g \frac{2h}{R} = \frac{2gh}{R}.
\]

2. **At depth \( d \)**:
\[
g_d = g \left(1 - \frac{d}{R}\right) \quad \text{(for small } d\text{)}
\]

The change in \( g \) is:
\[
\Delta g_d = g - g_d = g \frac{d}{R}.
\]

Setting the changes equal gives:

\[
\frac{2gh}{R} = \frac{g d}{R}.
\]

Cancelling \( g \) and \( R \) (assuming they are non-zero):

\[
2h = d.
\]

Thus, the relation between depth \( d \) and height \( h \) is:

\[
d = 2h.
\]

If the Earth's rotational motion were suddenly stopped, the value of \( g \) would effectively increase everywhere except at the poles due to the following reasons:

1. Centrifugal Force: The Earth's rotation creates a centrifugal force that slightly counteracts the force of gravity. At the equator, this effect is the greatest, reducing the apparent weight and thus the effective value of \( g \). When rotation stops, this centrifugal force disappears.

2. At the Poles: At the poles, there is no centrifugal force due to rotation since they are the axis of rotation. Therefore, the value of \( g \) would remain unchanged at the poles.

Consequently, \( g \) would increase at all places on Earth except at the poles, where it would stay the same.