Solution:

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:
\[
A_1 : A_2 : A_3 = 2: 3 : 6
\]

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Solution:

Explanation:

1. **Kepler's Second Law**:
- Kepler's Second Law states that a line segment joining a planet to the Sun sweeps out equal areas during equal intervals of time. This means that the planet moves faster when it is closer to the Sun and slower when it is farther away.

2. **Orbital Mechanics**:
- When the planet is at perigee, it is closer to the Sun, resulting in a stronger gravitational pull, which increases its orbital speed.
- Conversely, when the planet is at apogee, it is farther from the Sun, leading to a weaker gravitational pull and a decrease in its speed.

Summary:
- Maximum Speed: At perigee (closest point).
- Minimum Speed: At apogee (farthest point).

If you have any specific context or example related to this concept, please let me know!

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Solution:

To find the time taken by a planet at a distance of \( 540 \times 10^{10} \) m to revolve around the Sun, we can use Kepler's Third Law:

\[
\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}
\]

 Given:
- Earth:
- \( T_1 = 1 \) year
- \( r_1 = 15 \times 10^{10} \) m
- Planet:
- \( r_2 = 540 \times 10^{10} \) m

 Calculation:
1. Set up the ratio:
\[
\frac{1^2}{T_2^2} = \left( \frac{15}{540} \right)^3
\]
\[
\frac{15}{540} = \frac{1}{36} \quad \Rightarrow \quad \left( \frac{1}{36} \right)^3 = \frac{1}{46656}
\]

2. Solve for \( T_2^2 \):
\[
T_2^2 = 46656 \quad \Rightarrow \quad T_2 = \sqrt{46656} = 216 \text{ years}
\]

The time taken by the planet to revolve around the Sun once is approximately 216 years.

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Solution:

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.

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Solution:

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.

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Solution:

To find the time period of a satellite at a distance \( 4R \) from the center of the Earth, we can use Kepler's third law of planetary motion, which states:

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

1. Given:
- Time period at Earth's surface (\( T_0 \)): \( 1.4 \) hours (or \( T_0 = 1.4 \times 3600 \) seconds).
- Radius of the Earth: \( R \).

2. At Distance \( 4R \):
- The new radius \( r = 4R \).

3. Using the relationship:
\[
\frac{T^2}{T_0^2} = \frac{r^3}{R^3}
\]

Substituting \( r = 4R \):
\[
\frac{T^2}{T_0^2} = \frac{(4R)^3}{R^3} = \frac{64R^3}{R^3} = 64
\]

4. Solving for \( T \):
\[
T^2 = 64 T_0^2 ; T = 8 T_0
\]

5. Substituting \( T_0 \):
\[
T = 8\sqrt{2}, \text{hours} \]

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Solution:

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:
\[
K = \frac{4 \pi^2}{GM}
\]

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Solution:

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.

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