Kepler’s Laws of Planetary Motion
Kepler’s Laws describe the motion of planets around the Sun. These laws were discovered by the German astronomer Johannes Kepler in the early 17th century after analysing extremely accurate observational data collected by Tycho Brahe.
Although Kepler formulated these laws empirically, they were later explained theoretically using Newton’s law of gravitation and the laws of motion. These laws apply not only to planets but also to satellites, comets, asteroids, and artificial spacecraft.
Kepler’s First Law (Law of Orbits)
Definition: Every planet moves around the Sun in an elliptical orbit, with the Sun located at one of the two foci of the ellipse.
An ellipse is a closed curve in which the sum of distances from any point on the curve to two fixed points (called foci) remains constant.
Because the Sun lies at one focus rather than the center, the distance between the planet and the Sun changes continuously during revolution.
The closest point of the orbit to the Sun is called perihelion, while the farthest point is called aphelion.
Mathematical form of the orbit:
\[ r=\frac{a(1-e^2)}{1+e\cos\theta} \]
where
- \(a\) = semi-major axis of ellipse
- \(e\) = eccentricity
- \(r\) = distance of planet from Sun
- \(\theta\) = angular position
Example
Earth's orbital eccentricity is approximately 0.0167, which is very small. Therefore the Earth's orbit appears almost circular although it is technically an ellipse.
Kepler’s Second Law (Law of Areas)
Definition: The line joining the planet and the Sun sweeps out equal areas in equal intervals of time.
This means the orbital speed of a planet is not constant.
- Planet moves faster near the Sun (perihelion)
- Planet moves slower far from the Sun (aphelion)
The law can be explained using the principle of conservation of angular momentum.
Since gravitational force acts along the line joining the planet and Sun, the torque about the Sun is zero:
\[ \tau = \vec r \times \vec F = 0 \]
Hence angular momentum remains constant:
\[ L = m r^2 \frac{d\theta}{dt} \]
The area swept in time \(dt\) is
\[ dA = \frac{1}{2} r^2 d\theta \]
Therefore,
\[ \frac{dA}{dt}=\frac{1}{2}r^2\frac{d\theta}{dt}=\text{constant} \]
Example
When Earth is closest to the Sun in early January, its orbital speed is slightly higher than when it is farthest from the Sun in July.
Kepler’s Third Law (Law of Periods)
Definition: The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit.
\[ T^2 \propto a^3 \]
For planets revolving around the same central body:
\[ \frac{T_1^2}{T_2^2}=\frac{a_1^3}{a_2^3} \]
Using Newton’s law of gravitation we obtain
\[ T^2=\frac{4\pi^2}{GM}a^3 \]
where
- \(T\) = orbital period
- \(a\) = semi-major axis
- \(M\) = mass of central body
Example
Jupiter is about 5.2 times farther from the Sun than Earth. According to Kepler’s third law, its orbital period becomes about 11.86 years, which matches astronomical observations.
Key Insights and Applications
- Explains planetary motion in the Solar System.
- Used to determine masses of stars and planets.
- Essential for designing satellite orbits.
- Forms the foundation of celestial mechanics.
Kepler’s laws were a major breakthrough in astronomy because they replaced the earlier belief that planets move in perfect circles. Later, Newton’s universal law of gravitation provided the theoretical basis explaining why these laws work.
Universal Law of Gravitation
The Universal Law of Gravitation, proposed by Sir Isaac Newton in 1687, explains the attractive force acting between any two masses in the universe. This law shows that the same force responsible for an apple falling on Earth also governs the motion of planets, satellites, stars, and galaxies.
Definition of Universal Law of Gravitation
Every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.
This force always acts along the line joining the centres of the two interacting masses.
Mathematical representation:
\[ F \propto m_1 m_2 \] \[ F \propto \frac{1}{r^2} \]
Combining the two proportionalities:
\[ F = G \frac{m_1 m_2}{r^2} \]
where
- \(F\) = gravitational force between two bodies
- \(m_1 , m_2\) = masses of the two bodies
- \(r\) = distance between their centres
- \(G\) = universal gravitational constant
Illustration of Gravitational Attraction
Reasoning Behind the Law
Experimental observations show that the gravitational force depends on two main factors:
- Amount of matter present in each body (mass)
- Distance separating the bodies
When the mass of either object increases, the gravitational pull increases proportionally. However, when the separation between them increases, the force decreases rapidly according to the inverse square law.
\[ F \propto \frac{m_1 m_2}{r^2} \]
Introducing the constant \(G\) converts the proportionality into equality:
\[ F = G \frac{m_1 m_2}{r^2} \]
Universal Gravitational Constant
The constant \(G\) is called the universal gravitational constant. Its value is
\[ G = 6.67 \times 10^{-11}\; \text{N m}^2 \text{kg}^{-2} \]
This value was first measured experimentally by Henry Cavendish using a torsion balance.
Evidence of Universality
The same gravitational law explains both terrestrial and celestial motions.
For a body of mass \(m\) orbiting a planet of mass \(M\), the gravitational force provides the centripetal force required for circular motion.
\[ F = \frac{mv^2}{r} \]
According to Newton's law of gravitation:
\[ \frac{mv^2}{r} = G\frac{Mm}{r^2} \]
Simplifying:
\[ v^2 = \frac{GM}{r} \]
This equation correctly predicts the orbital speeds of satellites and planets, confirming the universality of Newton’s law.
Characteristics of Gravitational Force
- Always attractive in nature.
- Acts along the line joining the centres of two masses.
- It is a central force.
- Obeys Newton’s Third Law.
- It is a long-range force with infinite range.
- Independent of the medium between bodies.
Example
Consider two bodies of masses \(5\,kg\) and \(10\,kg\) separated by a distance of \(2\,m\).
\[ F = G\frac{m_1m_2}{r^2} \]
\[ F = (6.67\times10^{-11})\frac{5\times10}{2^2} \]
\[ F = 4.17 \times 10^{-10} \; N \]
The extremely small value shows why gravitational attraction between small everyday objects is usually negligible.
Significance in Physics
The universal law of gravitation explains a wide range of physical phenomena, including planetary motion, tides in oceans, motion of satellites, and the formation of galaxies. In the NCERT Class XI syllabus, it forms the basis for topics such as acceleration due to gravity, satellite motion, gravitational potential energy, and escape velocity.
The Gravitational Constant (G)
The gravitational constant \(G\) is a fundamental constant of nature that determines the strength of gravitational interaction between two masses. In the universal law of gravitation, it sets the numerical scale of the gravitational force acting anywhere in the universe.
Definition of the Gravitational Constant
The gravitational constant \(G\) is defined as the magnitude of the gravitational force between two point masses of 1 kg each separated by a distance of 1 metre in vacuum.
Mathematical definition
\[ G=\frac{Fr^2}{m_1m_2} \]
where
- \(F\) = gravitational force
- \(m_1, m_2\) = interacting masses
- \(r\) = distance between their centres
The value of \(G\) is universal and independent of the nature, shape, or state of the bodies.
Role of G in the Law of Gravitation
According to Newton’s universal law of gravitation:
\[ F = G \frac{m_1 m_2}{r^2} \]
Here, \(G\) acts as the proportionality constant that converts the proportional relationship between force, mass, and distance into a precise physical equation.
Without \(G\), the law would only describe how gravitational force varies, but not its actual value.
Measurement of G (Cavendish Experiment)
The value of \(G\) was first measured by Henry Cavendish in 1798 using a torsion balance apparatus. The experiment measured the extremely small gravitational attraction between two small lead spheres.
Dimensional Formula of G
Starting from the gravitational force equation:
\[ F = G \frac{m_1 m_2}{r^2} \]
Rearranging,
\[ G = \frac{Fr^2}{m_1m_2} \]
Substituting fundamental dimensions:
\[ [F] = MLT^{-2} \]
\[ [G] = \frac{[MLT^{-2}]L^2}{M^2} \]
\[ [G] = M^{-1}L^3T^{-2} \]
SI Unit and Numerical Value of G
From the dimensional formula, the SI unit of \(G\) is
\[ \text{N m}^2 \text{kg}^{-2} \]
The experimentally measured value is
\[ G = 6.67\times10^{-11}\; \text{N m}^2 \text{kg}^{-2} \]
The very small magnitude of \(G\) indicates that gravitational force is extremely weak compared to other fundamental forces.
Example Calculation
Two bodies of mass \(2\,kg\) each are separated by a distance of \(1\,m\). Find the gravitational force between them.
\[ F = G\frac{m_1m_2}{r^2} \]
\[ F = (6.67\times10^{-11})\frac{2\times2}{1^2} \]
\[ F = 2.67\times10^{-10} \, N \]
The extremely small force shows why gravitational attraction between everyday objects is usually negligible.
Difference Between G and g
| Quantity | Meaning | Nature |
|---|---|---|
| G | Universal gravitational constant | Same everywhere in the universe |
| g | Acceleration due to gravity | Varies with planet, altitude and latitude |
Importance in Physics
The gravitational constant plays a crucial role in determining planetary masses, orbital motion, escape velocity, satellite dynamics, and large-scale structure of the universe.
In the NCERT Class XI Gravitation chapter, \(G\) forms the basis for studying gravitational acceleration, satellite motion, and gravitational potential energy.
Cavendish’s Experiment
Cavendish’s experiment (1798) is a historic physics experiment that measured the extremely small gravitational attraction between laboratory-sized masses. It provided the first accurate value of the universal gravitational constant \(G\) and allowed scientists to estimate the mass and average density of the Earth.
Definition
Cavendish’s experiment is an experimental technique that determines the value of the universal gravitational constant \(G\) by measuring the tiny gravitational force between known masses using a torsion balance.
Principle of the Experiment
The experiment is based on the gravitational attraction between two small lead spheres and two large lead spheres.
This gravitational attraction produces a small torque on a suspended rod. The rod twists slightly, and the angle of twist is measured.
At equilibrium:
Gravitational torque = Restoring torsional torque
Construction of the Apparatus
- A light horizontal rod suspended by a thin torsion wire.
- Two small lead spheres attached to the ends of the rod.
- Two large lead spheres placed near the small spheres.
- A mirror fixed to the wire to measure tiny angular deflections.
- A light beam reflected from the mirror onto a scale for precise measurement.
Theory and Mathematical Derivation
Let
- \(m\) = mass of small sphere
- \(M\) = mass of large sphere
- \(r\) = distance between their centres
- \(l\) = distance from suspension axis
Gravitational force between one pair of spheres:
\[ F = G \frac{Mm}{r^2} \]
Total gravitational torque:
\[ \tau_g = 2Fl \]
Restoring torque from the torsion wire:
\[ \tau_r = C\theta \]
At equilibrium:
\[ C\theta = 2Fl \]
Substituting the value of \(F\):
\[ C\theta = 2 \left(G\frac{Mm}{r^2}\right) l \]
Rearranging:
\[ G = \frac{C\theta r^2}{2Mml} \]
Determination of the Torsion Constant
The torsional constant \(C\) is determined using the time period of oscillation of the suspended system.
\[ T = 2\pi \sqrt{\frac{I}{C}} \]
Therefore,
\[ C = \frac{4\pi^2 I}{T^2} \]
Substituting this value into the equation allows the experimental determination of \(G\).
Result of the Experiment
The experiment gave the value
\[ G = 6.67 \times 10^{-11}\; \text{N m}^2 \text{kg}^{-2} \]
This extremely small value explains why gravitational forces between ordinary objects are usually negligible.
Significance of the Experiment
- First accurate measurement of the gravitational constant.
- Confirmed that gravity acts between ordinary objects.
- Allowed calculation of Earth's mass and average density.
- Provided strong experimental confirmation of Newton’s law of gravitation.
Acceleration Due to Gravity of the Earth
One of the most important consequences of the universal law of gravitation is the concept of acceleration due to gravity. It explains why all freely falling objects near the Earth's surface accelerate toward the Earth at nearly the same rate, regardless of their mass, provided air resistance is negligible.
Definition of Acceleration Due to Gravity
The acceleration due to gravity, denoted by \(g\), is defined as the acceleration produced in a body when it falls freely under the gravitational attraction of the Earth alone.
Its average value near the Earth's surface is approximately
\[ g \approx 9.8 \, \text{m s}^{-2} \]
This acceleration is always directed towards the centre of the Earth.
Gravitational Attraction of the Earth
Derivation of Acceleration Due to Gravity
Consider a body of mass \(m\) placed on the surface of the Earth.
- Mass of Earth = \(M\)
- Radius of Earth = \(R\)
According to the universal law of gravitation, the gravitational force acting on the body is
\[ F = G\frac{Mm}{R^2} \]
According to Newton's second law of motion,
\[ F = mg \]
Equating the two expressions,
\[ mg = G\frac{Mm}{R^2} \]
Cancelling \(m\) from both sides,
\[ g = G\frac{M}{R^2} \]
Thus, acceleration due to gravity depends only on the mass and radius of the Earth.
Independence of g from Mass of Falling Body
From the expression
\[ g = G\frac{M}{R^2} \]
we observe that the mass of the falling object does not appear in the formula.
Therefore, all bodies fall with the same acceleration in a gravitational field when air resistance is ignored.
Direction and Nature of g
Acceleration due to gravity is a vector quantity.
- Its direction is always towards the centre of the Earth.
- It acts vertically downward at the Earth's surface.
- It represents the strength of Earth's gravitational field.
Variation of g on the Earth
The value of \(g\) is not exactly the same everywhere on Earth. It varies slightly due to several factors:
- Altitude above Earth's surface
- Depth below Earth's surface
- Earth's rotation
- Non-spherical shape of the Earth
According to NCERT,
- \(g\) is maximum at the poles
- \(g\) is minimum at the equator
Relation Between Weight and g
The weight of a body is the gravitational force acting on it due to the Earth.
\[ W = mg \]
where
- \(W\) = weight
- \(m\) = mass of body
- \(g\) = acceleration due to gravity
Since weight depends on \(g\), any change in \(g\) leads to a change in weight, although the mass of the body remains constant.
Example
Find the weight of a body of mass \(5\,kg\) on Earth.
\[ W = mg \]
\[ W = 5 \times 9.8 \]
\[ W = 49\,N \]
Thus the weight of the body is 49 Newton.
Acceleration Due to Gravity Above and Below the Earth's Surface
Basic Definition
The acceleration due to gravity \(g\) is the acceleration produced in a body due to Earth's gravitational attraction.
\[ g = G\frac{M}{R^2} \]
where
- \(M\) = mass of Earth
- \(R\) = radius of Earth
- \(G\) = gravitational constant
This value represents the acceleration due to gravity at the Earth's surface.
Acceleration Due to Gravity Above the Earth's Surface
Consider a body at height \(h\) above Earth's surface. Its distance from the Earth's centre becomes \(R + h\).
Gravitational force acting on a body of mass \(m\):
\[ F = G\frac{Mm}{(R+h)^2} \]
Therefore acceleration due to gravity at height \(h\) is
\[ g_h = G\frac{M}{(R+h)^2} \]
Dividing by the surface value \(g = GM/R^2\)
\[ \frac{g_h}{g} = \left(\frac{R}{R+h}\right)^2 \]
Approximation for small height \(h << R\)
\[ g_h \approx g\left(1 - \frac{2h}{R}\right) \]
Acceleration Due to Gravity Below the Earth's Surface
Consider a body at depth \(d\) below the Earth's surface. Its distance from the centre becomes \(R-d\).
Only the mass of Earth enclosed within radius \(R-d\) contributes to the gravitational attraction.
Assuming uniform density of Earth,
\[ M_d = M\left(\frac{R-d}{R}\right)^3 \]
Gravitational force:
\[ F = G\frac{M_d m}{(R-d)^2} \]
After simplification,
\[ g_d = g\left(1 - \frac{d}{R}\right) \]
Thus acceleration due to gravity decreases linearly with depth.
Comparison of g Above and Below Earth's Surface
| Case | Formula | Nature of Variation |
|---|---|---|
| Above surface | \(g_h = g(R/(R+h))^2\) | Inverse square decrease |
| Below surface | \(g_d = g(1 - d/R)\) | Linear decrease |
Physical Interpretation
- As altitude increases, the distance from Earth's centre increases, reducing gravitational pull.
- Inside Earth, the effective mass causing gravity decreases.
- At the Earth's centre, gravitational forces from all directions cancel.
Gravitational Potential Energy
Gravitational potential energy is the energy possessed by a body due to its position in a gravitational field. It is an important concept that allows problems involving gravitational motion to be solved using energy conservation instead of forces.
Definition of Gravitational Potential Energy
The gravitational potential energy of a body at a point in a gravitational field is defined as the work done against the gravitational force in bringing the body from infinity to that point.
By convention,
\[ U(\infty) = 0 \]
Therefore the potential energy at any finite distance from Earth becomes negative.
Gravitational Potential Well
Derivation of Gravitational Potential Energy
Consider a body of mass \(m\) located at a distance \(r\) from the centre of the Earth (mass \(M\)).
Gravitational force acting on the body:
\[ F = G\frac{Mm}{r^2} \]
To move the body slowly outward through a small distance \(dr\), work done against gravity is
\[ dW = F\,dr \]
\[ dW = G\frac{Mm}{r^2}dr \]
Total work done in bringing the body from infinity to distance \(r\):
\[ U(r) = -\int_{\infty}^{r} G\frac{Mm}{r^2}dr \]
Solving the integral,
\[ U(r) = -\frac{GMm}{r} \]
This gives the gravitational potential energy of a body at distance \(r\).
Why Gravitational Potential Energy is Negative
Gravitational force is attractive. Therefore, when a body moves toward Earth, gravity does positive work.
Since potential energy at infinity is taken as zero, the potential energy at any finite distance becomes negative.
Gravitational Potential Energy Near Earth's Surface
For a body at small height \(h\) above Earth's surface,
\[ r = R + h \]
Substituting into the general equation,
\[ U = -\frac{GMm}{R+h} \]
Using binomial approximation for \(h << R\),
\[ U \approx -\frac{GMm}{R} + \frac{GMmh}{R^2} \]
Since
\[ g = \frac{GM}{R^2} \]
we obtain
\[ U \approx -\frac{GMm}{R} + mgh \]
The first term is constant near Earth's surface, so the change in potential energy becomes
\[ \Delta U = mgh \]
Relation Between Gravitational Potential and Potential Energy
Gravitational potential \(V\) is defined as potential energy per unit mass.
\[ V = \frac{U}{m} \]
Substituting \(U = -GMm/r\),
\[ V = -\frac{GM}{r} \]
Example
Calculate the gravitational potential energy of a \(2\,kg\) body at height \(10\,m\) above Earth's surface.
\[ U = mgh \]
\[ U = 2 \times 9.8 \times 10 \]
\[ U = 196\,J \]
Therefore the gravitational potential energy is 196 Joules.
Important Physical Points
- Gravitational potential energy depends on position in the gravitational field.
- Only changes in potential energy have physical significance.
- The concept simplifies problems involving satellites and escape velocity.
- Potential energy becomes zero at infinite distance.
Escape Speed
Definition of Escape Speed
Escape speed is defined as the minimum speed that must be given to a body at the surface of the Earth so that it can escape completely from Earth's gravitational field and reach infinity with zero residual speed.
In other words, the body moves infinitely far away and never returns under the influence of Earth's gravity alone.
Physical Idea Behind Escape Speed
When a body is projected upward from the Earth, its kinetic energy decreases because work is done against gravitational attraction.
If the initial kinetic energy is large enough, the body can move infinitely far away from Earth where gravitational potential energy becomes zero.
Escape speed corresponds to the case where the body just reaches infinity with zero kinetic energy.
Escape Motion Illustration
Derivation of Escape Speed
Consider a body of mass \(m\) projected upward from Earth's surface with initial speed \(v_e\).
- Mass of Earth = \(M\)
- Radius of Earth = \(R\)
Total mechanical energy at Earth's surface:
\[ E_{surface}=\frac{1}{2}mv_e^2-\frac{GMm}{R} \]
At infinity,
\[ KE = 0, \quad PE = 0 \]
\[ E_{\infty}=0 \]
Using conservation of mechanical energy,
\[ \frac{1}{2}mv_e^2-\frac{GMm}{R}=0 \]
Rearranging,
\[ v_e=\sqrt{\frac{2GM}{R}} \]
Relation Between Escape Speed and Acceleration Due to Gravity
Using
\[ g=\frac{GM}{R^2} \]
we obtain
\[ v_e=\sqrt{2gR} \]
This formula is convenient for numerical problems involving Earth.
Escape Speed at Height \(h\)
If a body is launched from height \(h\) above Earth's surface, the distance from Earth's centre becomes \(R+h\).
The escape speed becomes
\[ v_e=\sqrt{\frac{2GM}{R+h}} \]
Thus escape speed decreases with altitude.
Independence of Escape Speed from Mass
From the expression
\[ v_e=\sqrt{\frac{2GM}{R}} \]
we see that the mass of the escaping body does not appear in the formula.
Therefore escape speed is the same for all objects regardless of their mass.
Escape Speed from Earth
Using
- \(g = 9.8 \, m/s^2\)
- \(R = 6.4 \times 10^6 \, m\)
\[ v_e=\sqrt{2 \times 9.8 \times 6.4\times10^6} \]
\[ v_e \approx 11.2\,km/s \]
This is the escape speed from Earth's surface.
Important Physical Points
- Escape speed is a critical speed, not a force.
- It is independent of the mass of the escaping body.
- It decreases with increasing altitude.
- Air resistance and Earth's rotation are neglected in theoretical calculation.
Earth Satellites
Earth satellites are one of the most important practical applications of the laws of gravitation. Their motion demonstrates how Earth's gravitational force provides the centripetal force necessary for an object to move in a circular orbit around the Earth.
Definition of an Earth Satellite
An Earth satellite is a body that revolves around the Earth under the influence of Earth's gravitational attraction.
- Natural satellites – e.g., the Moon
- Artificial satellites – man-made satellites used for communication, navigation, weather observation, and scientific research
Motion of a Satellite Around Earth
Principle of Satellite Motion
For a satellite moving in a circular orbit, the gravitational force between the Earth and the satellite provides the necessary centripetal force.
\[ F_{gravity} = F_{centripetal} \]
This balance allows the satellite to remain in a stable orbit.
Orbital Velocity of an Earth Satellite
Consider a satellite of mass \(m\) revolving around the Earth in a circular orbit of radius \(r\).
Gravitational force acting on the satellite:
\[ F = G\frac{Mm}{r^2} \]
Centripetal force required for circular motion:
\[ F = \frac{mv^2}{r} \]
Equating the two,
\[ G\frac{Mm}{r^2}=\frac{mv^2}{r} \]
Simplifying,
\[ v=\sqrt{\frac{GM}{r}} \]
Time Period of a Satellite
The time period \(T\) of a satellite is the time taken to complete one revolution around the Earth.
\[ T=\frac{2\pi r}{v} \]
Substituting \(v=\sqrt{GM/r}\),
\[ T=2\pi\sqrt{\frac{r^3}{GM}} \]
This result agrees with Kepler’s third law.
Energy of a Satellite
Kinetic energy of satellite:
\[ K=\frac{1}{2}mv^2=\frac{GMm}{2r} \]
Gravitational potential energy:
\[ U=-\frac{GMm}{r} \]
Total mechanical energy:
\[ E=K+U=-\frac{GMm}{2r} \]
The negative sign indicates the satellite is gravitationally bound to Earth.
Geostationary Satellites
A geostationary satellite is a satellite that appears stationary relative to the Earth's surface.
Conditions for geostationary orbit:
- Orbital period = 24 hours
- Orbit lies in Earth's equatorial plane
- Motion is in the same direction as Earth's rotation
Height of geostationary orbit ≈ 36,000 km.
Example
Find the orbital velocity of a satellite close to Earth's surface.
\[ v=\sqrt{gR} \]
\[ v=\sqrt{9.8 \times 6.4\times10^6} \]
\[ v\approx7.9\,km/s \]
Important Physical Points
- Orbital velocity depends only on Earth’s mass and orbital radius.
- Satellite motion is independent of satellite mass.
- Satellites closer to Earth move faster.
- Astronauts experience weightlessness due to continuous free fall.
Example–1
Let the speed of the planet at the perihelion \(P\) in Fig. 7.1(a) be \(v_P\) and the Sun–planet distance \(SP\) be \(r_P\). Relate \((r_P, v_P)\) to the corresponding quantities at the aphelion \((r_A, v_A)\). Will the planet take equal times to traverse arcs \(BAC\) and \(CPB\)?
Brief Theory
The gravitational force between the Sun and a planet always acts along the line joining them. Therefore the torque about the Sun is zero.
When torque is zero, the angular momentum of the planet remains conserved. This conservation leads directly to Kepler’s second law, which states that the line joining the planet and the Sun sweeps out equal areas in equal intervals of time.
At perihelion the planet is closest to the Sun, whereas at aphelion it is farthest from the Sun. The speed of the planet changes in such a way that angular momentum remains constant.
Solution Map
- Apply conservation of angular momentum of the planet.
- Write angular momentum expressions at perihelion and aphelion.
- Relate \(r_P v_P\) and \(r_A v_A\).
- Use Kepler’s second law to compare traversal times.
Solution
Let the speed of the planet at perihelion \(P\) be \(v_P\) and the Sun–planet distance be \(r_P\).
At aphelion \(A\), the corresponding speed and distance are \(v_A\) and \(r_A\).
The gravitational force between the Sun and the planet acts along the line joining them. Hence the torque about the Sun is zero.
Therefore the angular momentum of the planet about the Sun remains constant.
\[ L = mrv \]
Applying conservation of angular momentum at perihelion and aphelion,
\[ L_P = L_A \]
\[ mr_P v_P = mr_A v_A \]
Cancelling the common factor \(m\),
\[ r_P v_P = r_A v_A \]
Hence,
\[ \frac{v_P}{v_A} = \frac{r_A}{r_P} \]
Since \(r_P < r_A\), it follows that
\[ v_P > v_A \]
Thus the planet moves faster at perihelion than at aphelion.
Time Taken Along the Arcs
According to Kepler’s second law, the line joining the planet and the Sun sweeps out equal areas in equal intervals of time.
In the given figure, the regions \(BAC\) and \(CPB\) represent equal areas swept by the radius vector.
Therefore the planet takes equal times to traverse the arcs \(BAC\) and \(CPB\).
Example–2
Three equal masses of \(m\) kg each are fixed at the vertices of an equilateral triangle \(ABC\).
(a) What is the force acting on a mass \(2m\) placed at the centroid \(G\) of the triangle? (b) What is the force if the mass at vertex \(A\) is doubled?
Given: \(AG = BG = CG = 1\,\text{m}\).
Brief Theory
The gravitational force between two masses is given by
\[ F = \frac{G m_1 m_2}{r^2} \]
Gravitational force is a vector quantity. Therefore when multiple masses act on a body, the forces must be added using vector addition.
In symmetric situations like an equilateral triangle, equal forces may cancel each other due to symmetry.
Solution Map
- Calculate gravitational force on \(2m\) due to each vertex mass.
- Represent forces as vectors.
- Use symmetry to determine the resultant.
Geometry of the Problem
(a) Force on mass \(2m\) at centroid
Force on mass \(2m\) at \(G\) due to mass \(m\) at vertex \(A\):
\[ F = \frac{G(2m)(m)}{1^2} \]
\[ F = 2Gm^2 \]
The same magnitude of force acts due to masses at \(B\) and \(C\).
These three forces are separated by angles of \(120^\circ\). Because they are equal in magnitude and symmetrically arranged, their vector sum becomes zero.
Net force on \(2m\) at \(G\) = 0
(b) When mass at \(A\) is doubled
Mass at \(A\) becomes \(2m\).
Force on \(2m\) at \(G\) due to \(A\):
\[ F_{GA} = \frac{G(2m)(2m)}{1^2} \]
\[ F_{GA} = 4Gm^2 \]
Forces due to masses at \(B\) and \(C\) remain
\[ F_{GB} = F_{GC} = 2Gm^2 \]
The horizontal components of the forces from \(B\) and \(C\) cancel each other.
Their vertical components combine to give
\[ 2Gm^2 \]
This acts opposite to the force from \(A\).
Hence the resultant force becomes
\[ F_R = 2Gm^2 \]
directed towards vertex \(A\).
Example–3
Find the gravitational potential energy of a system of four particles placed at the vertices of a square of side \(l\). Also obtain the gravitational potential at the centre of the square.
Brief Theory
The gravitational potential energy between two masses \(m_1\) and \(m_2\) separated by distance \(r\) is
\[ U = -\frac{G m_1 m_2}{r} \]
For a system of many particles, the total potential energy is obtained by adding the energies of all distinct pairs of particles.
Solution Map
- Count all interacting particle pairs.
- Pairs along the sides of the square.
- Pairs along the diagonals of the square.
- Add all pair energies to obtain total potential energy.
- Calculate gravitational potential at the centre due to all four masses.
Geometry of the System
Total Potential Energy of the System
Four masses produce interactions in pairs.
Number of unique pairs:
\[ \frac{4 \times 3}{2} = 6 \]
These consist of:
- 4 side pairs (distance \(l\))
- 2 diagonal pairs (distance \(\sqrt{2}l\))
Potential energy of one side pair:
\[ U_{side} = -\frac{Gm^2}{l} \]
Total contribution from four sides:
\[ U_{sides} = -\frac{4Gm^2}{l} \]
Potential energy of one diagonal pair:
\[ U_{diag} = -\frac{Gm^2}{\sqrt{2}l} \]
Total contribution from two diagonals:
\[ U_{diagonals} = -\frac{2Gm^2}{\sqrt{2}l} \]
Simplifying,
\[ U_{diagonals} = -\frac{\sqrt{2}Gm^2}{l} \]
Therefore total potential energy:
\[ U = -\frac{Gm^2}{l}(4+\sqrt{2}) \]
Potential at the Centre of the Square
Distance from the centre of the square to each vertex:
\[ r = \frac{l}{\sqrt{2}} \]
Gravitational potential due to one mass:
\[ V_1 = -\frac{Gm}{r} \]
Substituting \(r = l/\sqrt{2}\),
\[ V_1 = -\frac{\sqrt{2}Gm}{l} \]
Total potential due to four masses:
\[ V_{centre} = 4V_1 \]
\[ V_{centre} = -\frac{4\sqrt{2}Gm}{l} \]
Example–4
Two uniform solid spheres of equal radii \(R\), but masses \(M\) and \(4M\), have a centre-to-centre separation \(6R\). A projectile of mass \(m\) is projected from the surface of the sphere of mass \(M\) towards the centre of the second sphere.
Obtain an expression for the minimum speed required so that the projectile just reaches the surface of the second sphere.
Brief Theory
When two gravitational fields act simultaneously, a point may exist where the gravitational forces due to the two masses cancel each other. This point is called the neutral point.
For minimum speed problems, the projectile must reach this point with zero velocity, because it represents the maximum potential energy barrier along the path.
The solution therefore involves:
- Finding the neutral point
- Applying conservation of mechanical energy
Solution Map
- Find the position of the neutral point using equality of forces.
- Compute total gravitational potential energy at the starting point.
- Compute potential energy at the neutral point.
- Apply conservation of mechanical energy.
Geometry of the System
Position of the Neutral Point
Let the neutral point \(N\) be at distance \(r\) from the centre of mass \(M\).
At this point gravitational forces balance:
\[ \frac{GMm}{r^2}=\frac{G(4M)m}{(6R-r)^2} \]
Simplifying,
\[ \frac{1}{r^2}=\frac{4}{(6R-r)^2} \]
\[ 6R-r=2r \]
\[ r=2R \]
Thus the neutral point lies \(2R\) from the centre of the sphere of mass \(M\).
Conservation of Mechanical Energy
Initial position: surface of sphere \(M\)
- Distance from \(M\) = \(R\)
- Distance from \(4M\) = \(5R\)
Initial energy:
\[ E_i=\frac12 mv^2-\frac{GMm}{R}-\frac{4GMm}{5R} \]
At neutral point \(N\):
- Distance from \(M\) = \(2R\)
- Distance from \(4M\) = \(4R\)
Since minimum speed condition implies velocity at \(N=0\),
\[ E_n=-\frac{GMm}{2R}-\frac{4GMm}{4R} \]
\[ E_n=-\frac{3GMm}{2R} \]
Final Result
Using conservation of energy \(E_i=E_n\):
\[ \frac12 mv^2-\frac{GMm}{R}-\frac{4GMm}{5R} =-\frac{3GMm}{2R} \]
Simplifying,
\[ v^2=\frac{3GM}{5R} \]
\[ \boxed{v=\sqrt{\frac{3GM}{5R}}} \]
Example–5
The planet Mars has two moons, Phobos and Deimos.
(i) Phobos has a period of 7 hours 39 minutes and an orbital radius
\(9.4 \times 10^{3}\) km. Calculate the mass of Mars.
(ii) Assume that Earth and Mars move in circular orbits around the Sun.
The Martian orbit is \(1.52\) times the orbital radius of Earth.
Find the length of the Martian year in days.
Brief Theory
For a satellite moving in a circular orbit around a planet, gravitational force provides the centripetal force.
\[ T^2 = \frac{4\pi^2}{GM}R^3 \]
This relation can be used to determine the mass of the planet.
Kepler’s third law also relates the orbital period and orbital radius of planets around the Sun:
\[ \frac{T_1^2}{T_2^2}=\frac{R_1^3}{R_2^3} \]
Solution Map
- Convert orbital period into seconds.
- Apply satellite motion formula to determine mass of Mars.
- Use Kepler’s third law to determine Martian year.
Orbital Motion of a Moon Around Mars
(i) Mass of Mars
Orbital period of Phobos:
\[ T = 7\,\text{hours} + 39\,\text{minutes} \]
\[ T = (7\times3600 + 39\times60) \]
\[ T = 2.754\times10^4\,s \]
Orbital radius:
\[ R = 9.4\times10^3\,km = 9.4\times10^6\,m \]
Using
\[ M=\frac{4\pi^2R^3}{GT^2} \]
Substituting values:
\[ M \approx 6.5\times10^{23}\,kg \]
(ii) Length of Martian Year
Given:
\[ R_M = 1.52\,R_E \]
Using Kepler’s third law:
\[ \frac{T_M^2}{T_E^2}=\frac{R_M^3}{R_E^3} \]
\[ T_M^2 = T_E^2(1.52)^3 \]
Taking \(T_E = 365\) days:
\[ T_M = 365(1.52)^{3/2} \]
\[ T_M \approx 684\,days \]
Final Results
- Mass of Mars \( \approx 6.5\times10^{23}\,kg \)
- Length of Martian year \( \approx 684\,days \)
Example–6
Weighing the Earth: You are given the following data:
\(g = 9.81\,\text{m s}^{-2}\), \(R_E = 6.37\times10^6\,\text{m}\), Earth–Moon distance \(R = 3.84\times10^8\,\text{m}\), Time period of Moon's revolution \(T = 27.3\) days.
Obtain the mass of the Earth \(M_E\) in two different ways.
Brief Theory
The mass of Earth can be determined using two independent physical ideas:
- Using the relation between surface gravity and Earth’s mass.
- Using the orbital motion of the Moon around the Earth.
Both methods arise from Newton’s law of gravitation.
Solution Map
- Method 1: Use \(g = GM_E/R_E^2\).
- Method 2: Use orbital motion of Moon \(T^2 = 4\pi^2R^3/(GM_E)\).
- Compare the values obtained.
Earth–Moon System
Method 1: Using Surface Gravity
The relation between surface gravity and Earth’s mass is
\[ g = \frac{GM_E}{R_E^2} \]
Rearranging,
\[ M_E = \frac{gR_E^2}{G} \]
Substituting values:
\[ M_E = \frac{9.81(6.37\times10^6)^2} {6.67\times10^{-11}} \]
\[ M_E \approx 5.97\times10^{24}\,\text{kg} \]
Method 2: Using Orbital Motion of the Moon
For a satellite orbiting Earth,
\[ T^2 = \frac{4\pi^2R^3}{GM_E} \]
Rearranging,
\[ M_E = \frac{4\pi^2R^3}{GT^2} \]
Convert period into seconds:
\[ T = 27.3\times24\times3600 \]
Substituting values:
\[ M_E = \frac{4\pi^2(3.84\times10^8)^3} {6.67\times10^{-11}(27.3\times24\times3600)^2} \]
\[ M_E \approx 6.02\times10^{24}\,\text{kg} \]
Final Result
Both methods give approximately the same value:
\(M_E \approx 6\times10^{24}\,\text{kg}\)
Example–7
Express the constant \(k\) of Eq. (7.38) in days and kilometres. Given \(k = 10^{-13}\,\text{s}^2\,\text{m}^{-3}\).
The Moon is at a distance \(3.84\times10^{5}\,\text{km}\) from the Earth. Obtain its time-period of revolution in days.
Brief Theory
For satellites orbiting the same central body, the orbital period and orbital radius are related by
\[ T^{2}=kR^{3} \]
where \(k\) is a constant depending only on the mass of the central body. To compute the period in specific units, the constant \(k\) must be expressed in compatible units.
Solution Map
- Convert \(k\) from \(s^{2}m^{-3}\) to \(day^{2}km^{-3}\).
- Substitute the Moon’s orbital radius.
- Compute the orbital period.
Earth–Moon Orbit
Converting the Constant \(k\)
Given:
\[ k = 10^{-13}\,\text{s}^2\,\text{m}^{-3} \]
Use conversion factors:
\(1\,\text{day}=86400\,\text{s}\) \(1\,\text{km}=10^{3}\,\text{m}\)
Therefore
\[ k = 10^{-13} \left(\frac{1\,\text{day}}{86400\,\text{s}}\right)^2 \left(\frac{10^3\,\text{m}}{1\,\text{km}}\right)^3 \]
\[ k \approx 1.33\times10^{-14}\, \text{day}^2\,\text{km}^{-3} \]
Time Period of the Moon
Distance of the Moon from Earth:
\[ R = 3.84\times10^{5}\,\text{km} \]
Using
\[ T^{2}=kR^{3} \]
Substitute values:
\[ T^{2}= 1.33\times10^{-14} (3.84\times10^{5})^{3} \]
Evaluating,
\[ T \approx 27.3\,\text{days} \]
Final Result
- \(k \approx 1.33\times10^{-14}\,day^{2}\,km^{-3}\)
- Time period of Moon \( \approx 27.3\,days \)
Example–8
A \(400\,\text{kg}\) satellite is in a circular orbit of radius \(2R_E\) about the Earth. How much energy is required to transfer it to a circular orbit of radius \(4R_E\)?
Also determine the changes in the kinetic energy and potential energy.
Brief Theory
For a satellite in a circular orbit of radius \(r\), the energies are
\[ E = -\frac{GM_E m}{2r} \]
\[ K = \frac{GM_E m}{2r} \]
\[ U = -\frac{GM_E m}{r} \]
Increasing orbital radius makes total energy **less negative**, meaning external energy must be supplied.
Solution Map
- Compute total energy in the initial orbit.
- Compute total energy in the final orbit.
- The difference gives the required energy.
- Use KE and PE formulas to determine their changes.
Orbital Transfer
Energy Required for Orbit Transfer
Initial orbit radius:
\(r_i = 2R_E\)
Initial total energy:
\[ E_i = -\frac{GM_E m}{4R_E} \]
Final orbit radius:
\(r_f = 4R_E\)
Final total energy:
\[ E_f = -\frac{GM_E m}{8R_E} \]
Energy required:
\[ \Delta E = E_f - E_i \]
\[ \Delta E = \frac{GM_E m}{8R_E} \]
Using \(GM_E = gR_E^2\),
\[ \Delta E = \frac{g m R_E}{8} \]
Substituting values:
\(g=9.81\,\text{m s}^{-2}\), \(m=400\,\text{kg}\), \(R_E=6.37\times10^6\,\text{m}\)
\[ \Delta E \approx 3.13\times10^{9}\,\text{J} \]
Changes in Kinetic and Potential Energy
Initial kinetic energy:
\[ K_i = \frac{GM_E m}{4R_E} \]
Final kinetic energy:
\[ K_f = \frac{GM_E m}{8R_E} \]
\[ \Delta K = -\frac{GM_E m}{8R_E} \]
Initial potential energy:
\[ U_i = -\frac{GM_E m}{2R_E} \]
Final potential energy:
\[ U_f = -\frac{GM_E m}{4R_E} \]
\[ \Delta U = \frac{GM_E m}{4R_E} \]
Final Results
- Energy supplied: \(3.13\times10^9\,\text{J}\)
- Increase in potential energy: \(+\frac{GM_E m}{4R_E}\)
- Decrease in kinetic energy: \(-\frac{GM_E m}{8R_E}\)
Gravitation – All Important Formulas in One Place
This quick revision sheet summarizes all the key formulas from the chapter Gravitation.
Universal Gravitation
\[F = G\frac{m_1 m_2}{r^2}\]
Force between two massesAcceleration Due to Gravity
\[g = \frac{GM}{R^2}\]
\(g \approx 9.8\,m\,s^{-2}\)Variation with Height
\[g_h = g\left(\frac{R}{R+h}\right)^2\]
g decreases with altitudeVariation with Depth
\[g_d = g\left(1-\frac{d}{R}\right)\]
g = 0 at Earth centreGravitational PE
\[U = -\frac{GMm}{r}\]
Near surface: \(U\approx mgh\)Escape Velocity
\[v_e = \sqrt{\frac{2GM}{R}}\]
Earth ≈ 11.2 km/sOrbital Velocity
\[v = \sqrt{\frac{GM}{r}}\]
Near Earth ≈ 7.9 km/sSatellite Time Period
\[T = 2\pi\sqrt{\frac{r^3}{GM}}\]
Circular orbitSatellite Total Energy
\[E = -\frac{GMm}{2r}\]
Bound orbit energyKinetic Energy
\[K = \frac{GMm}{2r}\]
Satellite kinetic energyPotential Energy
\[U = -\frac{GMm}{r}\]
Gravitational potential energyKepler’s Third Law
\[T^2 \propto r^3\]
Planetary motion law- \(F = Gm_1m_2/r^2\)
- \(g = GM/R^2\)
- \(v = \sqrt{GM/r}\)
- \(E = -GMm/(2r)\)
Top 10 Conceptual Mistakes Students Make in Gravitation
Many students lose marks in gravitation because of conceptual misunderstandings rather than difficult mathematics. Avoid the following common mistakes.
1. Confusing Mass and Weight
- Mass is constant everywhere.
- Weight depends on \(g\).
2. Assuming g is Constant Everywhere
The value of \(g\) changes with:
- Altitude
- Depth
- Latitude
3. Forgetting the Negative Sign of Potential Energy
Gravitational potential energy is always negative because gravity is attractive.
\[ U = -\frac{GMm}{r} \]4. Thinking Satellites Have No Gravity Acting
Satellites remain in orbit because gravity continuously provides centripetal force.
5. Confusing Orbital Velocity with Escape Velocity
- Orbital velocity ≈ 7.9 km/s
- Escape velocity ≈ 11.2 km/s
6. Ignoring Earth's Radius in Satellite Problems
Orbit radius: \[ r = R_E + h \] NOT simply \(h\).7. Assuming Energy of Satellite is Positive
Satellite energy is always negative for bound orbit. \[ E = -\frac{GMm}{2r} \]8. Misinterpreting Weightlessness
Weightlessness occurs because objects are in **free fall**, not because gravity is absent.9. Forgetting Pair Counting in Multi-Particle Potential Energy
Number of interacting pairs: \[ \frac{n(n-1)}{2} \]10. Forgetting That Escape Velocity is Independent of Mass
\[ v_e = \sqrt{\frac{2GM}{R}} \] Mass of projectile cancels.10 Most Important NCERT Gravitation Questions for Exams
The following questions represent the most frequently tested concepts from the NCERT Class XI chapter Gravitation. Students preparing for school exams, JEE, NEET, and other competitive exams should practice these questions carefully.
1. Derive the Expression for Escape Velocity
Starting from conservation of energy, derive the escape velocity of a body from the Earth’s surface.
\[v_e = \sqrt{\frac{2GM}{R}}\]2. Derive the Expression for Acceleration Due to Gravity
Using Newton’s law of gravitation, obtain the expression for acceleration due to gravity near the Earth’s surface.
\[g = \frac{GM}{R^2}\]3. Show that Orbital Velocity of a Satellite is Independent of Its Mass
Starting from centripetal force and gravitational force, derive the formula for orbital velocity of a satellite.
\[v = \sqrt{\frac{GM}{r}}\]4. Derive the Total Energy of a Satellite in Circular Orbit
Show that the total mechanical energy of a satellite in circular orbit is negative and depends only on orbital radius.
\[E = -\frac{GMm}{2r}\]5. Show That Acceleration Due to Gravity Decreases with Height
Derive the expression for \(g\) at a height \(h\) above the Earth’s surface.
\[ g_h = g\left(\frac{R}{R+h}\right)^2 \]6. Show That Acceleration Due to Gravity Decreases with Depth
Assuming Earth has uniform density, derive the expression for \(g\) at depth \(d\) below the Earth’s surface.
\[ g_d = g\left(1-\frac{d}{R}\right) \]7. Find the Gravitational Potential Energy of Two Masses
Derive the expression for gravitational potential energy of two particles separated by distance \(r\).
U = -\frac{GMm}{r}8. Prove Kepler’s Third Law Using Newton’s Law of Gravitation
Show that the square of the orbital period of a planet is proportional to the cube of the orbital radius.
\[ T^2 \propto r^3 \]9. Explain the Working Principle of Cavendish Experiment
Describe how Cavendish measured the gravitational constant \(G\) using a torsion balance.
10. Derive the Time Period of a Satellite
Starting from orbital velocity, derive the time period of a satellite moving in a circular orbit around the Earth.
\[ T = 2\pi\sqrt{\frac{r^3}{GM}} \]- Newton’s law of gravitation
- Conservation of energy
- Circular motion
- Kepler’s laws
Gravitation – Quick Concept Revision (2-Minute Exam Checklist)
Before an exam, students should quickly revise the following key ideas. If you understand every point in this checklist, you have mastered the core concepts of the chapter Gravitation.
1. Universal Law of Gravitation
Every mass attracts every other mass with a force proportional to their masses and inversely proportional to the square of the distance.
\[F = G\frac{m_1 m_2}{r^2}\]- Force always acts along the line joining the masses.
- Always attractive.
2. Acceleration Due to Gravity
\[g = \frac{GM}{R^2}\]- Independent of mass of falling body.
- Average value near Earth surface: \(9.8\,m\,s^{-2}\).
3. Variation of g
With height
\[ g_h = g\left(\frac{R}{R+h}\right)^2 \]With depth
\[ g_d = g\left(1-\frac{d}{R}\right) \]4. Gravitational Potential Energy
\[U = -\frac{GMm}{r}\]- Always negative because gravity is attractive.
- Near Earth surface: \(U \approx mgh\).
5. Escape Velocity
\[v_e = \sqrt{\frac{2GM}{R}}\]- Independent of mass of projectile.
- Escape velocity of Earth ≈ 11.2 km/s.
6. Orbital Velocity of Satellite
\[v = \sqrt{\frac{GM}{r}}\]- Near Earth surface ≈ 7.9 km/s.
- Decreases with increasing orbital radius.
7. Satellite Energy Relations
Total energy of satellite:
\[ E = -\frac{GMm}{2r} \]Kinetic and potential energies:
\[ K = \frac{GMm}{2r}, \quad U = -\frac{GMm}{r} \]- Total energy is always negative for bound orbit.
8. Kepler’s Third Law
\[ T^2 \propto r^3 \]- Planets farther from Sun move more slowly.
9. Weightlessness
- Occurs when objects are in free fall.
- Gravity is still acting.
10. Key Numerical Relations to Remember
- \(v_e = \sqrt{2gR}\)
- \(r = R_E + h\)
- \(T = 2\pi\sqrt{\frac{r^3}{GM}}\)
- \(E = -GMm/(2r)\)
- \(\color{blue}F = Gm_1m_2/r^2\)
- \(\color{blue}g = GM/R^2\)
- \(\color{blue}v = \sqrt{GM/r}\)
- \(\color{blue}v_e = \sqrt{2GM/R}\)
- \(\color{blue}E = -GMm/(2r)\)
Interactive Concept Map of Gravitation (Visual Learning Section)
This concept map shows how the main ideas of the Gravitation chapter are connected. Students can use this visual summary to revise the entire chapter quickly.
Core Formula Behind the Entire Chapter
Almost every topic in the Gravitation chapter originates from Newton’s law of gravitation.
F = G\frac{m_1 m_2}{r^2}From Gravitation to Orbital Motion
When gravitational force provides centripetal force, satellites and planets move in orbit.
v = \sqrt{\frac{GM}{r}}Energy in Gravitational Systems
U = -\frac{GMm}{r}Gravitational potential energy is negative because gravitational interaction is attractive.
- Start from Newton's Law of Gravitation.
- Understand how it explains planetary motion and satellites.
- Use energy concepts to derive escape velocity.
- Apply the same law to explain variation of \(g\).
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