Physics > Gravitation > 7.0 Kepler's law of planetary motion
Gravitation
1.0 Newton's law of gravitation
1.1 Characteristics of gravitational force
1.2 Universal gravitational constant
1.3 Principle of superposition of gravitation
1.4 Gravity
1.5 Acceleration due to gravity
1.6 Relation between $g$ and $G$
2.0 Variation of acceleration due to gravity
2.1 Variation of acceleration due to gravity $(g)$ due to shape of the earth
2.2 Variation of acceleration due to gravity $(g)$ due to altitude
2.3 Variation of acceleration due to gravity $(g)$ due to depth
2.4 Variation of acceleration due to gravity $(g)$ due to rotation of earth
3.0 Gravitational field
3.1 Gravitational field due to a point mass
3.2 Gravitational field due to a uniform solid sphere
3.3 Gravitational field due to a uniform spherical shell
3.4 Gravitatioal field due to a uniform circular ring at a point on its axis
4.0 Gravitational potential
4.1 Gravitational potential due to a point mass
4.2 Gravitational potential due to a uniform solid sphere
4.3 Gravitational potential due to a uniform thin spherical shell
4.4 Gravitational potential due to a uniform ring at a point on its centre
4.5 Relation between gravitational field and gravitational potential
5.0 Gravitational potential energy
5.1 Gravitational potential energy for a system of particles
5.2 Gravitational potential energy of a body on earth's surface
6.0 Satellites
6.1 Orbital speed of satellite
6.2 Time period of a satellite
6.3 Angular momentum of a satellite
6.4 Energy of a satellite
6.5 Types of satellite
6.6 Binding energy
6.7 Escape velocity
6.8 Weightlessness
7.0 Kepler's law of planetary motion
8.0 Problem solving technique
7.2 Kepler's second law
1.2 Universal gravitational constant
1.3 Principle of superposition of gravitation
1.4 Gravity
1.5 Acceleration due to gravity
1.6 Relation between $g$ and $G$
2.2 Variation of acceleration due to gravity $(g)$ due to altitude
2.3 Variation of acceleration due to gravity $(g)$ due to depth
2.4 Variation of acceleration due to gravity $(g)$ due to rotation of earth
3.2 Gravitational field due to a uniform solid sphere
3.3 Gravitational field due to a uniform spherical shell
3.4 Gravitatioal field due to a uniform circular ring at a point on its axis
4.2 Gravitational potential due to a uniform solid sphere
4.3 Gravitational potential due to a uniform thin spherical shell
4.4 Gravitational potential due to a uniform ring at a point on its centre
4.5 Relation between gravitational field and gravitational potential
5.2 Gravitational potential energy of a body on earth's surface
6.2 Time period of a satellite
6.3 Angular momentum of a satellite
6.4 Energy of a satellite
6.5 Types of satellite
6.6 Binding energy
6.7 Escape velocity
6.8 Weightlessness
This law is known as the law of areas.
The radius vector drawn from the sun to the planets sweeps out equal area in equal interval of time i.e. the areal velocity of the planet around the sun is constant.
Areal velocity is constant and it is given by, $$\frac{{d\overrightarrow A }}{{dt}} = {\text{constant (for a planet)}}$$
Consider a particle $P$ of mass $m$ moving with velocity $u$ as shown in the figure. In a small time interval $dt$, the vector from the sun $S$ to the planet $P$ turns through an angle $d\theta $.
The area swept out in this time interval is,
$dA =$ area of triangle as shown in figure
$$dA = \frac{1}{2} \times \left( {{\text{Base}}} \right) \times \left( {{\text{Height}}} \right)$$$$dA = \frac{1}{2} \times \left( {SP} \right) \times \left( {P'M} \right)$$$$dA = \frac{1}{2}\left( r \right)\left[ {v\sin \left( {\pi - \theta } \right)dt} \right]$$$$\frac{{dA}}{{dt}} = \frac{1}{2}rv\sin \theta $$ or $$\frac{{dA}}{{dt}} = \frac{{mvr\sin \theta }}{{2m}}$$
As we know,$$mvr\sin \theta = m\left| {\overrightarrow r \times \overrightarrow v } \right| = \overrightarrow L $$ So, $$\frac{{dA}}{{dt}} = \frac{L}{{2m}}$$ or $$\frac{{d\overrightarrow A }}{{dt}} = \frac{{\overrightarrow L }}{{2m}}$$
The above equation gives the relation between areal velocity $\left( {\frac{{d\overrightarrow A }}{{dt}}} \right)$ and angular momentum.
As we know,$$mvr\sin \theta = m\left| {\overrightarrow r \times \overrightarrow v } \right| = \overrightarrow L $$ So, $$\frac{{dA}}{{dt}} = \frac{L}{{2m}}$$ or $$\frac{{d\overrightarrow A }}{{dt}} = \frac{{\overrightarrow L }}{{2m}}$$
The above equation gives the relation between areal velocity $\left( {\frac{{d\overrightarrow A }}{{dt}}} \right)$ and angular momentum.
Kepler's second law follows from the law of conservation of angular momentum.
According to Kepler's second law, the speed of the planet is maximum when it is closest to the sun and is maximum when the planet is farthest from the sun.
As we know that the angular momentum is conserved about the sun. So, we can write,
$${L_P} = {L_A}$$$$m{v_P}\left[ {a\left( {1 - e} \right)} \right] = m{v_A}\left[ {a\left( {1 + e} \right)} \right]$$$$\frac{{{v_P}}}{{{v_A}}} = \frac{{1 + e}}{{1 - e}}$$ As $e<1$. So, $${v_P} > {v_A}$$
Therefore, velocity of planet is maximum at perhelion and minimum at the apihelion.
