Kepler's Second Law, also known as the **Law of Equal Areas**, states that a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time. This law implies that a planet moves faster when it is closer to the Sun and slower when it is farther away, resulting in an elliptical orbit.

### Connection to Angular Momentum Conservation:

1. **Angular Momentum Definition**:
Angular momentum (\( L \)) of a body moving around a point is given by:
\[
L = mvr
\]
where:
- \( m \) = mass of the body,
- \( v \) = tangential velocity,
- \( r \) = distance from the center of rotation (the Sun, in this case).

2. **Conservation of Angular Momentum**:
- In a system where no external torques act (like a planet orbiting the Sun), the angular momentum is conserved.
- This means that:
\[
L = mvr = \text{constant}
\]

3. **Implications for Planetary Motion**:
- As a planet moves in its elliptical orbit, its distance \( r \) from the Sun changes.
- To conserve angular momentum (\( L \)), if the planet is closer to the Sun (\( r \) decreases), its velocity \( v \) must increase. Conversely, when it is farther from the Sun (\( r \) increases), its velocity \( v \) must decrease.

4. **Area Swept Out**:
- The area swept out by the line segment joining the planet to the Sun in a time interval \( dt \) can be expressed in terms of angular momentum. The area (\( dA \)) swept out in time \( dt \) is:
\[
dA = \frac{1}{2} r^2 d\theta
\]
where \( d\theta \) is the angle subtended at the Sun during that time interval.

5. **Equal Areas in Equal Times**:
- Since angular momentum is conserved, the planet's motion adjusts in such a way that it sweeps out equal areas in equal times, which is the essence of Kepler's Second Law.

### Conclusion:
Kepler's Second Law reflects the conservation of angular momentum in planetary orbits. The relationship between distance from the Sun, velocity, and the areas swept out in equal time intervals illustrates that as a planet moves in its elliptical orbit, it adjusts its speed to conserve angular momentum, leading to the sweeping of equal areas in equal times.

The orbital period of a satellite depends only on the radius of its orbit and the mass of the Earth, not on the mass of the satellite itself. The period \( T \) is given by:

\[
T = 2\pi \sqrt{\frac{r^3}{GM}}
\]

Since the masses of the satellites do not appear in this formula, the periods of the two satellites will be the same if they are in the same orbit, regardless of their masses.

Therefore, the ratio of their periods of revolution is:

\[
\frac{T_1}{T_2} = 1
\]

So, the ratio is 1:1.

If the Earth's radius shrinks to half, but its mass remains the same, the orbital period of the satellite will not change.

The orbital period \( T \) of a satellite depends on the mass of the Earth \( M \) and the radius of the orbit \( r \), not the radius of the Earth itself. The formula for the period of a satellite in orbit is:

\[
T = 2\pi \sqrt{\frac{r^3}{GM}}
\]

Since the mass \( M \) of the Earth and the radius \( r \) of the satellite’s orbit (which is unaffected by the Earth shrinking) remain the same, the period \( T \) remains unchanged.

Thus, the new period of revolution will still be \( T \).

The energy required to move a satellite from one orbit to another can be found using the difference in total mechanical energy between the two orbits.

The total energy \( E \) of a satellite in orbit of radius \( r \) is:

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

For the initial orbit of radius \( r \), the energy is:

\[
E_1 = -\frac{GMm}{2r}
\]

For the final orbit of radius \( \frac{3r}{2} \), the energy is:

\[
E_2 = -\frac{GMm}{2 \times \frac{3r}{2}} = -\frac{GMm}{3r}
\]

The energy required to shift the satellite is the difference between the two energies:

\[
\Delta E = E_2 - E_1 = \left(-\frac{GMm}{3r}\right) - \left(-\frac{GMm}{2r}\right)
\]

\[
\Delta E = \frac{GMm}{2r} - \frac{GMm}{3r} = \frac{GMm}{6r}
\]

So, the energy required is:

\[
\Delta E = \frac{GMm}{6r}
\]

To find the number of years Mars takes to make one revolution around the Sun, we can use Kepler's Third Law, which states:

\[
T^2 \propto r^3
\]

where \( T \) is the orbital period and \( r \) is the average distance from the Sun.

Given:
- Let the average distance of Earth from the Sun be \( r_E \).
- The average distance of Mars from the Sun is \( r_M = 1.5 r_E \).

Using Kepler's Third Law:
1. For Earth:
\[
T_E^2 \propto r_E^3
\]

2. For Mars:
\[
T_M^2 \propto r_M^3 ; T_M^2 \propto (1.5 r_E)^3 = 1.5^3 r_E^3
\]

3. We know \( T_E \) (the period of Earth) is approximately **1 year**:
\[
T_M^2 = 1.5^3 T_E^2
\]
\[
T_M^2 = 1.5^3 \times 1^2 = 3.375
\]

4. Therefore,
\[
T_M = \sqrt{3.375} \approx 1.84 \, \text{years}
\]

Conclusion:
Mars takes approximately 1.84 years to make one revolution around the Sun.

The orbital velocity \( v \) at a height \( h \) above the Earth's surface is given by:

\[
v = \sqrt{\frac{GM}{R + h}}
\]

For the satellite orbiting just above the Earth's surface (i.e., \( h = 0 \)), the velocity is:

\[
v_0 = \sqrt{\frac{GM}{R}}
\]

For a satellite at an altitude of \( h = \frac{R}{2} \), the velocity \( v_h \) becomes:

\[
v_h = \sqrt{\frac{GM}{R + \frac{R}{2}}} = \sqrt{\frac{GM}{\frac{3R}{2}}} = \sqrt{\frac{2}{3}} v_0
\]

Thus, the orbital velocity at an altitude of \( \frac{R}{2} \) is:

\[
v_h = \sqrt{\frac{2}{3}} v_0
\]

The orbital speed \( v \) of a satellite is given by:

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

where \( r \) is the radius of the orbit.

Let the speeds of satellites A and B be \( v_A \) and \( v_B \), and their orbital radii be \( 4R \) and \( R \), respectively. Using the relation:

\[
v_A = \sqrt{\frac{GM}{4R}}, \quad v_B = \sqrt{\frac{GM}{R}}
\]

Given \( v_A = 3v \), we can write:

\[
3v = \sqrt{\frac{GM}{4R}}
\]

Now, the speed of satellite B is:

\[
v_B = \sqrt{\frac{GM}{R}} = 2 \times \sqrt{\frac{GM}{4R}} = 2 \times 3v = 6v
\]

So, the speed of satellite B is \( 6v \).

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 point where the gravitational force on the projectile is zero occurs when the gravitational forces from both planets are equal.

\[
\frac{G \cdot 4M \cdot m}{r^2} = \frac{G \cdot 9M \cdot m}{(6R - r)^2}
\]

Simplifying,

\[
\frac{4}{r^2} = \frac{9}{(6R - r)^2}
\]

Taking the square root:

\[
\frac{2}{r} = \frac{3}{6R - r}
\]

Cross-multiplying:

\[
2(6R - r) = 3r
\]

Solving:

\[
12R - 2r = 3r
\]

\[
5r = 12R
\]

\[
r = \frac{12R}{5}
\]

So, the distance from the lighter planet is \( \frac{12R}{5} \).

Gravitational potential \( V \) at the midpoint is the sum of the potentials due to both masses:

\[
V = -\frac{G \cdot 10^2}{0.5} - \frac{G \cdot 10^3}{0.5}
\]

Simplifying:

\[
V = -2G(10^2 + 10^3) = -2G \cdot 1100
\]

So, the gravitational potential at the midpoint is:

\[
V = -2200G
\]