To find the relation between \( G \) and \( K \), we can use both Kepler’s third law and Newton’s law of gravitation.

1. **Gravitational Force**:
From Newton's law of gravitation, the gravitational force between the Sun and a planet is:
\[
F = \frac{GMm}{r^2}
\]

2. **Centripetal Force**:
For a planet moving in a circular orbit, the gravitational force provides the necessary centripetal force. The centripetal force for a planet with mass \( m \) and orbital speed \( v \) is:
\[
F = \frac{mv^2}{r}
\]

Equating the two expressions for force:
\[
\frac{GMm}{r^2} = \frac{mv^2}{r}
\]
Simplifying, we get:
\[
v^2 = \frac{GM}{r}
\]

3. **Orbital Period**:
The orbital speed \( v \) is related to the period \( T \) by:
\[
v = \frac{2 \pi r}{T}
\]
Substituting into \( v^2 = \frac{GM}{r} \), we get:
\[
\left( \frac{2 \pi r}{T} \right)^2 = \frac{GM}{r}
\]
Simplifying:
\[
\frac{4 \pi^2 r^2}{T^2} = \frac{GM}{r}
\]
\[
T^2 = \frac{4 \pi^2 r^3}{GM}
\]

4. **Kepler’s Third Law**:
From Kepler's third law, we know:
\[
T^2 = Kr^3
\]

Comparing both expressions for \( T^2 \):
\[
K = \frac{4 \pi^2}{GM}
\]

### Conclusion:
The relation between \( G \) and \( K \) is:
\[
K = \frac{4 \pi^2}{GM}
\]

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

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

where \( R \) is the distance from the center of the Earth to the satellite and \( g \) is the acceleration due to gravity at the Earth's surface. Using the relation \( g = \frac{GM}{R_e^2} \), where \( R_e \) is the Earth's radius, we can rewrite the formula as:

\[
v = \sqrt{g \cdot \frac{R_e^2}{R}}
\]

Here, \( R = R_e + h \), where \( h = 0.25 \times 10^6 \) m is the height above the surface of the Earth.

Now, substituting values:

\[
R = 6.38 \times 10^6 + 0.25 \times 10^6 = 6.63 \times 10^6 \, \text{m}
\]

\[
v = \sqrt{9.8 \times \frac{(6.38 \times 10^6)^2}{6.63 \times 10^6}}
\]

Solving:

\[
v \approx 7.76 \, \text{km/s}
\]

So, the orbital speed of the satellite is approximately \( 7.76 \, \text{km/s} \).

Yes, the acceleration of the satellite \( S \) is always directed towards the center of the Earth. This is because the gravitational force, which provides the acceleration, always points towards the Earth's center, regardless of the satellite's position in its elliptical orbit. This centripetal acceleration is responsible for keeping the satellite in orbit.

According to **Kepler's second law** (law of areas), a planet sweeps out equal areas in equal times. This means the area swept out is proportional to the time taken.

In the problem, the area \( SCD \) is given to be twice the area \( SAB \). Therefore, the time taken to sweep these areas will also follow the same ratio.

Thus, the relation between the times is:

\[
t_1 = 2 t_2
\]

So, the time taken to move from \( C \) to \( D \) is twice the time taken to move from \( A \) to \( B \).

For a planet in an elliptical orbit:

- Total energy (E) is the sum of kinetic energy (T) and potential energy (U).
- In a bound orbit like an ellipse, E is always negative. This indicates that the planet is gravitationally bound to the star and cannot escape.
- Kinetic energy (T) is always positive.
- Potential energy (U) is negative due to the attractive gravitational force, and its magnitude is greater than T.
- Angular momentum (L) is constant for elliptical orbits.

Thus, E is always negative for elliptical orbits.

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 escape velocity \( V_e \) is the speed needed to escape Earth's gravitational pull. The energy conservation principle applies, where the total mechanical energy at the surface and at height \( h = 3R \) (where \( R \) is the Earth's radius) should be equal.

The total energy at the surface:
\[
E_1 = \frac{1}{2} m V_e^2 - \frac{G M m}{R}
\]

At height \( 3R \), the total energy is:
\[
E_2 = \frac{1}{2} m v^2 - \frac{G M m}{4R}
\]

Since \( E_1 = E_2 \), we equate the two:
\[
\frac{1}{2} m V_e^2 - \frac{G M m}{R} = \frac{1}{2} m v^2 - \frac{G M m}{4R}
\]

We know that the escape velocity is given by:
\[
V_e^2 = \frac{2 G M}{R}
\]

Substitute this into the equation:
\[
\frac{1}{2} m \frac{2 G M}{R} - \frac{G M m}{R} = \frac{1}{2} m v^2 - \frac{G M m}{4R}
\]

Simplifying:
\[
\frac{G M m}{R} - \frac{G M m}{R} = \frac{1}{2} m v^2 - \frac{G M m}{4R}
\]

This simplifies to:
\[
0 = \frac{1}{2} m v^2 - \frac{G M m}{4R}
\]

Solving for \( v^2 \):
\[
\frac{1}{2} m v^2 = \frac{G M m}{4R}
\]

\[
v^2 = \frac{2 G M}{4R} = \frac{G M}{2R}
\]

Thus, the velocity at height \( 3R \) is:
\[
v = \sqrt{\frac{G M}{2R}} = \frac{V_e}{\sqrt{2}}= 7.92 km/sec
\]

So, the velocity at height \( 3R \) is \( \frac{V_e}{\sqrt{2}} \).= 7.92 km/sec

To find the net gravitational force on the mass \( m \) at the centroid, we follow these steps:

1. Forces due to extra masses \( 2m \):
- Each of the two lower vertices has an extra mass \( 2m \).
- The gravitational force due to one \( 2m \) mass at a distance \( \frac{a}{\sqrt{3}} \) from the centroid is:
\[
F = \frac{G (2m) m}{\left( \frac{a}{\sqrt{3}} \right)^2} = \frac{6 G m^2}{a^2}
\]

2. Net force between the two extra masses:
- The angle between the two forces is \( 120^\circ \).
- The resultant force is given by the vector addition formula:
\[
F_{\text{resultant}} = \sqrt{F^2 + F^2 + 2 F F \cos 120^\circ}
\]
Since \( \cos 120^\circ = -\frac{1}{2} \):
\[
F_{\text{resultant}} = \sqrt{F^2 + F^2 - F^2} = F = \frac{6 G m^2}{a^2}
\]

3. Conclusion:
The net gravitational force on the mass \( m \) at the centroid is:
\[
F_{\text{resultant}} = \frac{6 G m^2}{a^2}
\]
This force is directed vertically downward.

When both \( r_1 \) and \( r_2 \) are greater than \( R \) (i.e., both are outside the sphere), the gravitational force at a distance \( r \) from the center of a uniform sphere is given by:

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

So, the forces \( F_1 \) and \( F_2 \) at distances \( r_1 \) and \( r_2 \) from the center are:

\[
F_1 = \frac{G M}{r_1^2}
\]
\[
F_2 = \frac{G M}{r_2^2}
\]

Now, taking the ratio \( \frac{F_1}{F_2} \):

\[
\frac{F_1}{F_2} = \frac{\frac{G M}{r_1^2}}{\frac{G M}{r_2^2}} = \frac{r_2^2}{r_1^2}
\]

Thus, the ratio of the gravitational forces is:

\[
\frac{F_1}{F_2} = \left( \frac{r_2}{r_1} \right)^2
\]