Introduction
Gravitation is one of the most fundamental forces in nature, responsible for the attraction between masses. It governs the motion of planets, stars, galaxies, and even light. In the context of JEE Advanced Physics, understanding gravitation is crucial for solving a variety of problems related to celestial mechanics, orbital dynamics, and gravitational fields.
Newton's Law of Gravitation
Newton's law of gravitation states that every point mass attracts every other point mass in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
Mathematical Formulation
The force of gravitation $F$ between two masses $m_1$ and $m_2$ separated by a distance $r$ is given by:
$$ F = G \frac{m_1 m_2}{r^2} $$
where $G$ is the gravitational constant, $G \approx 6.674 \times 10^{-11} , \text{N m}^2 \text{kg}^{-2}$.
NoteGravitational force is always attractive and acts along the line joining the two masses.
Vector Form
In vector form, the gravitational force $\vec{F}$ on mass $m_1$ due to $m_2$ is:
$$ \vec{F} = -G \frac{m_1 m_2}{r^2} \hat{r} $$
where $\hat{r}$ is the unit vector from $m_1$ to $m_2$.
Gravitational Field
The gravitational field $\vec{g}$ at a point in space is defined as the gravitational force experienced by a unit mass placed at that point.
Gravitational Field due to a Point Mass
For a point mass $M$, the gravitational field at a distance $r$ from the mass is:
$$ \vec{g} = -G \frac{M}{r^2} \hat{r} $$
ExampleConsider a mass of 5 kg placed 2 meters away from a 10 kg mass. The gravitational field at the location of the 5 kg mass due to the 10 kg mass is: $$ \vec{g} = -G \frac{10}{2^2} \hat{r} = -6.674 \times 10^{-11} \frac{10}{4} \hat{r} = -1.6685 \times 10^{-10} \hat{r} , \text{N/kg} $$
Gravitational Potential Energy
Gravitational potential energy $U$ is the energy associated with the position of a mass in a gravitational field.
Potential Energy of a Two-Mass System
For two masses $m_1$ and $m_2$ separated by a distance $r$, the gravitational potential energy is:
$$ U = -G \frac{m_1 m_2}{r} $$
NoteGravitational potential energy is negative because the force is attractive.
Gravitational Potential
The gravitational potential $V$ at a point in space is the gravitational potential energy per unit mass at that point.
$$ V = -G \frac{M}{r} $$
Kepler's Laws of Planetary Motion
Kepler's laws describe the motion of planets around the sun. They are empirical laws derived from observational data.
First Law: Law of Orbits
Every planet moves in an elliptical orbit with the sun at one of the two foci.
Second Law: Law of Areas
A line segment joining a planet and the sun sweeps out equal areas during equal intervals of time.
Third Law: Law of Periods
The square of the orbital period $T$ of a planet is directly proportional to the cube of the semi-major axis $a$ of its orbit.
$$ T^2 \propto a^3 $$
ExampleFor Earth, the semi-major axis $a = 1 , \text{AU}$ and the orbital period $T = 1 , \text{year}$. For Mars, $a = 1.524 , \text{AU}$. Using Kepler's third law: $$ T^2 \propto a^3 \Rightarrow T^2 = (1.524)^3 \Rightarrow T \approx 1.88 , \text{years} $$
Escape Velocity
Escape velocity is the minimum velocity required for an object to escape the gravitational influence of a celestial body without further propulsion.
Derivation
The escape velocity $v_e$ from a planet of mass $M$ and radius $R$ is derived from the conservation of energy principle:
$$ \frac{1}{2} m v_e^2 = G \frac{m M}{R} $$
Solving for $v_e$:
$$ v_e = \sqrt{\frac{2GM}{R}} $$
ExampleFor Earth, $M = 5.972 \times 10^{24} , \text{kg}$ and $R = 6.371 \times 10^6 , \text{m}$. The escape velocity is: $$ v_e = \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 5.972 \times 10^{24}}{6.371 \times 10^6}} \approx 11.2 , \text{km/s} $$
TipEscape velocity is independent of the mass of the object trying to escape.
Satellites and Orbital Motion
Satellites revolve around planets in specific orbits. The dynamics of these orbits can be understood through the principles of gravitation.
Orbital Velocity
The orbital velocity $v_o$ of a satellite in a circular orbit of radius $r$ around a planet of mass $M$ is:
$$ v_o = \sqrt{\frac{GM}{r}} $$
Geostationary Orbits
A geostationary orbit is one where a satellite appears stationary relative to a point on the equator of the planet. The orbital period is 24 hours.
Energy in Orbits
The total energy $E$ of a satellite in a circular orbit is the sum of its kinetic energy $K$ and potential energy $U$:
$$ K = \frac{1}{2} m v_o^2 = \frac{1}{2} G \frac{m M}{r} $$
$$ U = -G \frac{m M}{r} $$
$$ E = K + U = -\frac{1}{2} G \frac{m M}{r} $$
Common MistakeConfusing gravitational potential energy with gravitational potential. Remember, potential energy is for a system of two masses, while potential is per unit mass.
Summary
Gravitation is a fundamental force that influences the motion of celestial bodies and governs the dynamics of orbits and satellites. Understanding the principles of gravitational force, field, potential, and energy is crucial for solving problems in JEE Advanced Physics.
TipPractice problems involving different scenarios like varying distances, masses, and specific orbital conditions to strengthen your understanding of gravitation.
NoteAlways keep track of units and constants like $G$ to avoid calculation errors.