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 \( 18 \, \text{m} \).

The gravitational force between two masses \( m_1 \) and \( m_2 \) is:

\[
F = G \frac{m_1 \cdot m_2}{r^2}
\]

To maximize \( F \), we set \( m_1 = x \) and \( m_2 = 100 - x \), then maximize the product \( m_1 \cdot m_2 = x(100 - x) \).

This product is maximized when \( x = 50 \). So, the masses should be split equally.

The ratio of masses is:

\[
\frac{m_1}{m_2} = \frac{50}{50} = 1
\]

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

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

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 escape velocity \( V_e \) from the Earth's surface is given by:

\[
V_e = \sqrt{\frac{2GM}{R}}
\]

At a height \( h = R \) (i.e., the platform is at a distance \( 2R \) from the Earth's center), the escape velocity \( V \) becomes:

\[
V = \sqrt{\frac{2GM}{2R}} = \frac{V_e}{\sqrt{2}}
\]

So, the factor \( F \) is:

\[
F = \frac{V}{V_e} = \frac{1}{\sqrt{2}}
\]

According to **Kepler's Second Law** (the Law of Equal Areas), a line segment joining a planet and the Sun sweeps out equal areas in equal times. Therefore, the area swept out in a given time is proportional to the time interval.

### Given:
- Area \( A_1 \) is swept in 2 days.
- Area \( A_2 \) is swept in 3 days.
- Area \( A_3 \) is swept in 6 days.

### Area Swept Proportions:
Since the area swept is proportional to the time taken:
\[
\frac{A_1}{A_2} = \frac{2 \text{ days}}{3 \text{ days}} \quad \Rightarrow \quad A_1 = \frac{2}{3} A_2
\]

\[
\frac{A_2}{A_3} = \frac{3 \text{ days}}{6 \text{ days}} \quad \Rightarrow \quad A_2 = \frac{1}{2} A_3
\]

### Relating Areas:
Substituting the relationship of \( A_2 \) in \( A_1 \):
\[
A_1 = \frac{2}{3} \left(\frac{1}{2} A_3\right) = \frac{1}{3} A_3
\]

### Final Relationships:
Now, we can express the areas in terms of \( A_3 \):
- \( A_1 = \frac{1}{3} A_3 \)
- \( A_2 = \frac{1}{2} A_3 \)

### Conclusion:
The relationship between the areas \( A_1 \), \( A_2 \), and \( A_3 \) can be summarized as:
\[
A_1 : A_2 : A_3 = 1 : \frac{3}{2} : 3
\]
This can also be expressed as:
\[
A_1 : A_2 : A_3 = 2: 3 : 6
\]

The total mechanical energy \( E \) of a satellite in a circular orbit is the sum of its kinetic energy \( K \) and potential energy \( U \).

1. **Kinetic Energy (K):**

\[
K = \frac{1}{2} m v^2
\]

2. **Potential Energy (U):**

\[
U = -\frac{GMm}{r}
\]

For a satellite orbiting closely around the Earth, its velocity \( v \) is related to the Earth's gravitational constant by:

\[
v = \sqrt{\frac{GM}{r}}
\]

Substitute \( v^2 = \frac{GM}{r} \) into the kinetic energy expression:

\[
K = \frac{1}{2} m \left( \frac{GM}{r} \right) = \frac{GMm}{2r}
\]

Now, the **total energy** is:

\[
E = K + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}
\]

Thus, the total energy of the satellite is:

\[
E = -\frac{GMm}{2r}= -\frac{1}{2}mv^{2}
\]

According to Newton’s third law, for every action, there is an equal and opposite reaction.

The Earth exerts a gravitational force on the Moon, and the Moon exerts an equal and opposite gravitational force on the Earth. This reaction force is the gravitational force on Earth by the Moon.

The initial gravitational force between two bodies is:

\[
F = G \frac{m \cdot m}{r^2}
\]

After transferring 25% of mass from one body to the other, the new masses become \( 0.75m \) and \( 1.25m \), and the separation becomes \( \frac{r}{2} \). The new gravitational force is:

\[
F' = G \frac{(0.75m)(1.25m)}{\left(\frac{r}{2}\right)^2}
\]

Simplifying:

\[
F' = G \frac{0.9375 m^2}{\frac{r^2}{4}} = 4 \times 0.9375 \times \frac{G m^2}{r^2} = 15F/4
\]

So, the new gravitational force is:

\[
F' = 3.75F= 15F/4
\]

If the universal constant of gravitation \( G \) were decreasing uniformly with time, the gravitational force between the satellite and the Earth would weaken. However, the satellite's angular momentum would remain conserved because angular momentum depends on the mass, velocity, and radius of orbit, and not directly on \( G \).

According to the law of conservation of angular momentum:

\[
L = m v r
\]

where \( L \) is angular momentum, \( m \) is the satellite's mass, \( v \) is its velocity, and \( r \) is the orbital radius. Since no external torque acts on the system, angular momentum is conserved even if \( G \) changes.

However, the satellite’s orbital parameters like its velocity and orbital radius might change over time due to the weakening gravitational force, but the total angular momentum will remain constant.