GRAVITATION-Notes
Physics - Notes
Kepler’s Laws
Kepler’s laws are a set of three fundamental principles that describe the motion of planets around the Sun. These laws were formulated from systematic astronomical observations and later found to be a natural consequence of Newton’s law of gravitation. They provide a bridge between observational astronomy and classical mechanics and form an essential part of the NCERT Class XI chapter on Gravitation.
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.
This law emphasizes that planetary motion is not perfectly circular. The ellipse represents a closed curve in which the sum of distances from any point on the curve to the two fixed foci remains constant. The presence of the Sun at one focus explains the varying distance of a planet from the Sun during its revolution.
Mathematical Representation: The polar equation of an ellipse with the Sun at the focus is
\[\begin{aligned} \boxed{\;r = \frac{a(1 - e^2)}{1 + e\cos\theta}\;} \end{aligned}\]
where \(r\) is the distance of the planet from the Sun, \(a\) is the semi-major axis, \(e\) is the eccentricity of the orbit, and \(\theta\) is the angular position of the planet.
Physical Aspect: The value of eccentricity determines how stretched the orbit is. For most planets, \(e\) is small, which is why their orbits appear nearly circular, though they are strictly elliptical.
Kepler’s Second Law (Law of Areas)
Definition: The line joining a planet and the Sun sweeps out equal areas in equal intervals of time.
This law describes the changing speed of a planet as it moves along its orbit. A planet moves faster when it is closer to the Sun and slower when it is farther away.
Derivation using Angular Momentum:
The gravitational force between the Sun and a planet acts along the line joining them. Hence, the torque about the Sun is zero.
\[\begin{aligned} \boldsymbol{\tau} &= \mathbf{r} \times \mathbf{F} &= 0 \end{aligned}\]
Since torque is zero, angular momentum remains conserved.
\[\begin{aligned} \frac{d\mathbf{L}}{dt} &= 0 \\ \Rightarrow \quad \mathbf{L} &= m r^2 \frac{d\theta}{dt} = \text{constant} \end{aligned}\]
The area swept in a small time interval \(dt\) is
\[\begin{aligned} dA = \frac{1}{2} r^2 d\theta \end{aligned}\]
Dividing both sides by \(dt\),
\[\begin{aligned} \frac{dA}{dt} = \frac{1}{2} r^2 \frac{d\theta}{dt} = \text{constant} \end{aligned}\]
Thus, equal areas are swept in equal times, proving Kepler’s second law.
Important Aspect: This law is a direct consequence of conservation of angular momentum and holds true for any motion under a central force.
Kepler’s Third Law (Law of Periods)
Definition: The square of the time period of revolution of a planet is directly proportional to the cube of the semi-major axis of its orbit.
Mathematically,
\[\begin{aligned} T^2 \propto a^3 \end{aligned}\]
Derivation using Newton’s Law of Gravitation:
Consider a planet of mass \(m\) revolving around the Sun of mass \(M\) in a nearly circular orbit of radius \(r\). The gravitational force provides the necessary centripetal force.
\[\begin{aligned} \frac{G M m}{r^2} = \frac{m v^2}{r} \end{aligned}\]
Simplifying,
\[\begin{aligned} v^2 = \frac{G M}{r} \end{aligned}\]
The orbital speed is related to the time period \(T\) by
\[\begin{aligned} v = \frac{2\pi r}{T} \end{aligned}\]
Substituting and simplifying,
\[\begin{aligned} \frac{4\pi^2 r^2}{T^2} = \frac{G M}{r} \end{aligned}\]
Rearranging,
\[\begin{aligned} T^2 = \frac{4\pi^2}{G M} r^3 \end{aligned}\]
This shows that \(T^2\) is proportional to \(r^3\), confirming Kepler’s third law.
Physical Interpretation: All planets orbiting the same central body obey the same proportionality constant, which depends only on the mass of the central body.
Key Insights and Applications
Kepler’s laws explain the non-uniform motion of planets, the shape of planetary orbits, and the relation between orbital size and time period. They apply not only to planets but also to satellites, comets, and artificial spacecraft. These laws laid the foundation for Newton’s theory of gravitation and remain crucial for understanding celestial mechanics.
In the NCERT syllabus, Kepler’s laws help students visualize gravitational motion, connect force with orbital dynamics, and develop a deeper appreciation of how universal physical laws govern motion on a cosmic scale.
UNIVERSAL LAW OF GRAVITATION
The Universal Law of Gravitation is a fundamental law of nature that explains the attractive force acting between any two masses in the universe. It establishes that gravitation is not restricted to the Earth alone but is a universal interaction governing 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 two masses.
Mathematical statement:
\[ F \propto m_1 m_2 \] \[ F \propto \frac{1}{r^2} \]
Combining both proportionalities,
\[ F = G \frac{m_1 m_2}{r^2} \]
where \(m_1\) and \(m_2\) are the masses of the two bodies, \(r\) is the distance between their centres, and \(G\) is the universal gravitational constant.
Derivation of the Law
The gravitational force depends on two basic physical factors: the amount of matter present and the separation between the bodies. Experimental observations show that when the mass of either body is doubled, the gravitational pull also doubles, indicating direct proportionality with mass.
Similarly, increasing the distance between the bodies reduces the force rapidly, following an inverse square dependence.
Combining these results mathematically:
\[ F \propto \frac{m_1 m_2}{r^2} \]
Introducing a proportionality constant \(G\) to convert this relation into an equality,
\[ F = G \frac{m_1 m_2}{r^2} \]
This equation represents the universal law of gravitation in its complete mathematical form.
Proof of Universality
The same gravitational law successfully explains a wide range of natural phenomena. The falling of an apple toward the Earth and the motion of the Moon around the Earth are governed by the same interaction. The centripetal force required for the Moon’s circular motion is provided by Earth’s gravitational attraction.
For a body of mass \(m\) orbiting a mass \(M\):
\[ F = \frac{m v^2}{r} \]
According to the law of gravitation,
\[ \frac{m v^2}{r} = G \frac{M m}{r^2} \]
Simplifying,
\[ v^2 = \frac{G M}{r} \]
This result correctly predicts orbital speeds of planets and satellites, confirming that the same gravitational law applies universally, from terrestrial objects to celestial bodies.
Nature and Characteristics of Gravitational Force
Gravitational force is always attractive in nature and acts along the line joining the centres of two interacting masses. It is a central force and obeys Newton’s third law, meaning the forces on the two masses are equal in magnitude and opposite in direction.
The force is long-ranged and extends infinitely, though its magnitude decreases rapidly with distance. Unlike electromagnetic forces, gravitational force does not depend on the medium between the interacting bodies.
Important Physical Aspects
The universal gravitational constant \(G\) has the same value everywhere in the universe, highlighting the universal nature of the law. Its small numerical value explains why gravitational force is noticeable only when at least one of the interacting masses is very large, such as planets or stars.
The law forms the foundation for understanding weight, free fall, motion of planets, tides, satellite dynamics, and the large-scale structure of the universe.
Significance in NCERT Context
In the NCERT Class XI syllabus, the universal law of gravitation serves as the cornerstone for subsequent topics such as acceleration due to gravity, motion of satellites, Kepler’s laws, and gravitational potential energy. It unifies diverse physical phenomena under a single mathematical framework.
The law demonstrates how simple physical principles, when expressed mathematically, can explain motions ranging from falling objects on Earth to the revolution of planets around the Sun.
The gravitational constant
The gravitational constant is a fundamental constant of nature that quantifies the intrinsic strength of gravitational interaction between two masses. While the universal law of gravitation describes how gravitational force depends on mass and distance, the constant \(G\) sets the absolute scale of this force throughout the universe.
Definition of the Gravitational Constant
The gravitational constant \(G\) is defined as the magnitude of the gravitational force acting between two point masses of one kilogram each, placed one metre apart in vacuum.
Mathematical definition:
\[ G = \frac{F r^2}{m_1 m_2} \]
Here, \(F\) is the gravitational force between masses \(m_1\) and \(m_2\) separated by a distance \(r\). The value of \(G\) is universal and does not depend on the nature, shape, or state of the interacting bodies.
Role of G in the Universal Law of Gravitation
According to the universal law of gravitation, the force of attraction between two masses is given by
\[ F = G \frac{m_1 m_2}{r^2} \]
In this expression, \(G\) acts as the proportionality constant that converts the proportional relationship between force, masses, and distance into an exact physical law. Without \(G\), the law would describe only how force varies, not its actual magnitude.
Dimensional Formula of G
To obtain the dimensions of \(G\), we start from the gravitational force equation and express each quantity in fundamental units.
\[ F = G \frac{m_1 m_2}{r^2} \]
Rearranging,
\[ G = \frac{F r^2}{m_1 m_2} \]
Substituting dimensions,
\[ [G] = \frac{[M L T^{-2}] \cdot L^2}{M^2} \]
Simplifying,
\[ [G] = M^{-1} L^{3} T^{-2} \]
This dimensional formula is essential for verifying gravitational equations and understanding how gravitational effects scale with mass, distance, and time.
SI Unit and Numerical Value of G
From its dimensional formula, the SI unit of the gravitational constant is derived as
\[ \text{unit of } G = \text{N m}^2 \text{ kg}^{-2} \]
Experimentally, the value of the gravitational constant is found to be
\[ G = 6.67 \times 10^{-11}\ \text{N m}^2 \text{ kg}^{-2} \]
The extremely small value of \(G\) indicates that gravitational force is very weak compared to other fundamental forces, becoming noticeable only when at least one of the interacting masses is very large.
Proof of the Weak Nature of Gravitation
Consider two ordinary objects of small mass placed a short distance apart. Even though they attract each other gravitationally, the resulting force is too small to produce any observable motion. This weakness directly arises from the small numerical value of \(G\).
In contrast, when one of the masses is comparable to that of the Earth or a planet, the product \(G m_1 m_2\) becomes sufficiently large, leading to measurable gravitational effects such as weight, orbital motion, and tides.
Important Physical Aspects of G
The gravitational constant is truly universal. Its value remains the same everywhere in the universe, irrespective of location, medium, temperature, or time. This universality distinguishes it from quantities like acceleration due to gravity, which vary from place to place.
The constant \(G\) plays a crucial role in determining the mass of celestial bodies, predicting planetary orbits, calculating escape velocity, and understanding the large-scale structure of the universe.
Significance in the NCERT Framework
In NCERT Class XI Physics, the gravitational constant serves as a foundational concept linking terrestrial and celestial phenomena. It allows students to move from qualitative ideas of attraction to precise quantitative predictions.
By introducing \(G\), gravitation becomes a measurable and testable interaction, reinforcing the idea that the same physical laws govern motion on Earth as well as motion in the cosmos.
Cavendish’s Experiment
Cavendish’s experiment is a landmark experiment in classical physics that provided the first accurate measurement of the gravitational constant. More importantly, it demonstrated that gravitational attraction between ordinary laboratory-sized masses is real and measurable, thereby confirming the universal nature of Newton’s law of gravitation.
Definition
Cavendish’s experiment is an experimental method used to determine the value of the universal gravitational constant \(G\) by measuring the extremely weak gravitational force between known masses using a sensitive torsion balance.
The experiment indirectly allows the mass and mean density of the Earth to be calculated, which is why it is often described as the experiment that “weighed the Earth.”
Principle of the Experiment
The experiment is based on the gravitational attraction between two small masses and two large masses. This attraction produces a tiny torque in a suspended system. The angular twist produced is measured, and from this the gravitational force is calculated.
The restoring torque provided by the torsion wire balances the gravitational torque, leading to a stable angular deflection.
Construction and Arrangement
The apparatus consists of a light horizontal rod suspended at its center by a thin torsion wire. Two small lead spheres of equal mass are attached at the ends of the rod. Two much heavier lead spheres are positioned close to the small spheres, producing a gravitational pull.
A small mirror fixed to the suspension wire reflects a beam of light onto a distant scale, allowing minute angular deflections of the rod to be measured with high precision.
Theory and Derivation
Let each small sphere have mass \(m\) and each large sphere have mass \(M\). Let the distance between the centers of the interacting spheres be \(r\).
The gravitational force acting between one pair of masses is
\[ F = G \frac{M m}{r^2} \]
This force produces a torque about the suspension axis. If the perpendicular distance from the axis to the line of action of the force is \(l\), then the total gravitational torque due to both sides is
\[ \tau_g = 2 F l \]
The torsion wire provides a restoring torque proportional to the angular twist \(\theta\),
\[ \tau_r = C \theta \]
where \(C\) is the torsional constant of the wire.
At equilibrium, the gravitational torque balances the restoring torque,
\[ C \theta = 2 F l \]
Substituting the value of \(F\),
\[ C \theta = 2 \left( G \frac{M m}{r^2} \right) l \]
Rearranging to obtain \(G\),
\[ G = \frac{C \theta r^2}{2 M m l} \]
Since all quantities on the right-hand side can be measured experimentally, the value of the gravitational constant can be determined.
Determination of the Torsional Constant
The torsional constant \(C\) of the wire is found by measuring the time period \(T\) of small oscillations of the suspended system. If \(I\) is the moment of inertia of the rod and attached masses, then
\[ T = 2\pi \sqrt{\frac{I}{C}} \]
Solving for \(C\),
\[ C = \frac{4\pi^2 I}{T^2} \]
This experimentally obtained value of \(C\) is used in the main equation to calculate \(G\).
Proof of the Universal Nature of Gravitation
Cavendish’s experiment confirmed that gravitational attraction exists even between small masses placed a few centimetres apart. This proved that gravity is not limited to astronomical bodies but is a universal interaction acting between all forms of matter.
The experimentally obtained value of \(G\) was found to be constant, independent of the size, shape, or material of the masses used.
Important Aspects and Significance
The experiment provided the first accurate value of the gravitational constant, enabling the calculation of Earth’s mass and average density. It established gravitation as a measurable laboratory force despite its extreme weakness.
ACCELERATION DUE TO GRAVITY OF THE EARTH
Acceleration due to gravity is one of the most important consequences of the universal law of gravitation discussed in the NCERT textbook. It explains why all freely falling bodies near the Earth’s surface accelerate downward at nearly the same rate, irrespective of their masses, 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.
It is always directed towards the centre of the Earth and its average value near the Earth’s surface is approximately \(9.8 \, \text{m s}^{-2}\).
Derivation of Acceleration Due to Gravity
Consider a body of mass \(m\) placed on the surface of the Earth. Let the mass of the Earth be \(M\) and its radius be \(R\).
According to the universal law of gravitation, the gravitational force acting on the body is
\[ F = G \frac{M m}{R^2} \]
This gravitational force is responsible for producing acceleration in the body. According to Newton’s second law of motion,
\[ F = m g \]
Equating the two expressions for force,
\[ m g = G \frac{M m}{R^2} \]
Cancelling the mass \(m\) from both sides,
\[ g = G \frac{M}{R^2} \]
This expression gives the acceleration due to gravity at the Earth’s surface in terms of the Earth’s mass and radius.
Proof that g is Independent of the Mass of the Body
From the derived expression
\[ g = G \frac{M}{R^2} \]
it is clear that the acceleration due to gravity depends only on the mass and radius of the Earth. The mass of the falling body does not appear in this expression.
Hence, all bodies, irrespective of their masses or composition, fall with the same acceleration near the Earth’s surface when air resistance is neglected. This conclusion is consistent with experimental observations and forms a key concept emphasized in the NCERT textbook.
Direction and Nature of g
The acceleration due to gravity is a vector quantity. Its direction at any point on the Earth’s surface is vertically downward, pointing towards the Earth’s centre.
Since gravity is an attractive force, the acceleration produced by it always acts inward, pulling objects toward the Earth.
Variation of g on the Earth
Although \(g\) is often taken as constant, its value varies slightly from place to place due to factors such as the Earth’s rotation, its non-spherical shape, and changes in altitude and depth.
According to NCERT, \(g\) is maximum at the poles and minimum at the equator, mainly due to the Earth’s rotation and equatorial bulge.
Relation Between Weight and g
The weight of a body is the gravitational force acting on it due to the Earth. It is given by
\[ W = m g \]
Since weight depends on \(g\), any variation in the value of \(g\) leads to a corresponding change in the weight of the body, even though its mass remains constant.
ACCELERATION DUE TO GRAVITY BELOW AND ABOVE THE SURFACE OF EARTH
Basic Definition
Acceleration due to gravity, denoted by \(g\), is the acceleration produced in a body solely due to the Earth’s gravitational attraction. At the Earth’s surface, it is given by
\[ g = G\frac{M}{R^2} \]
where \(M\) is the mass of the Earth and \(R\) is its radius. This expression serves as the reference value for studying variations of \(g\) above and below the surface.
Acceleration Due to Gravity Above the Earth’s Surface
Consider a body at a height \(h\) above the Earth’s surface. The distance of the body from the Earth’s centre becomes \(R+h\). According to the universal law of gravitation, the gravitational force acting on a body of mass \(m\) at this height is
\[ F = G\frac{M m}{(R+h)^2} \]
This force produces an acceleration \(g_h\) in the body. Using Newton’s second law,
\[ m g_h = G\frac{M m}{(R+h)^2} \]
Cancelling \(m\),
\[ g_h = G\frac{M}{(R+h)^2} \]
Dividing this expression by the surface value \(g = G M / R^2\),
\[ \frac{g_h}{g} = \left(\frac{R}{R+h}\right)^2 \]
This result shows that acceleration due to gravity decreases with increase in height. For heights much smaller than the Earth’s radius, this decrease is gradual, which explains why \(g\) is nearly constant close to the surface.
Important aspect: At very large distances from the Earth, such as in space, the value of \(g\) becomes very small, leading to conditions commonly described as weightlessness.
Acceleration Due to Gravity Below the Earth’s Surface
Now consider a body at a depth \(d\) below the Earth’s surface. Its distance from the Earth’s centre is \(R-d\).
According to NCERT, only the mass of the Earth enclosed within this radius contributes to the gravitational attraction. Assuming the Earth to be a uniform sphere, the enclosed mass \(M_d\) is proportional to the volume of the sphere of radius \(R-d\).
\[ M_d = M \left(\frac{R-d}{R}\right)^3 \]
The gravitational force on a body of mass \(m\) at depth \(d\) is
\[ F = G\frac{M_d m}{(R-d)^2} \]
Substituting the value of \(M_d\),
\[ F = G\frac{M m}{R^3}(R-d) \]
The corresponding acceleration due to gravity \(g_d\) is
\[ g_d = G\frac{M}{R^3}(R-d) \]
Using \(g = G M / R^2\), we obtain
\[ g_d = g\left(1 - \frac{d}{R}\right) \]
This expression shows that the acceleration due to gravity decreases linearly with depth and becomes zero at the centre of the Earth.
Physical interpretation: At the Earth’s centre, gravitational pulls from all directions balance each other, resulting in zero net acceleration.
Comparison of g Above and Below the Surface
Above the Earth’s surface, the value of \(g\) decreases according to an inverse square law with distance from the centre. Below the surface, it decreases linearly with depth under the assumption of uniform density.
These two different variations highlight the role of Earth’s mass distribution and distance in determining gravitational effects.
GRAVITATIONAL POTENTIAL ENERGY
Gravitational potential energy is a measure of the energy possessed by a body due to its position in a gravitational field. In the NCERT treatment of gravitation, this concept naturally follows the universal law of gravitation and provides a powerful way to analyze motion under gravity using energy methods 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, assuming the body is brought slowly without any change in kinetic energy.
By convention, the gravitational potential energy of a body at infinity is taken as zero.
Derivation of Gravitational Potential Energy
Consider a body of mass \(m\) placed at a distance \(r\) from the centre of the Earth, whose mass is \(M\). The gravitational force acting on the body is
\[ F = G\frac{M m}{r^2} \]
This force is directed radially inward, toward the Earth’s centre. To move the body slowly outward through a small distance \(dr\), an equal and opposite force must be applied.
The work done \(dW\) against gravity in moving the body through distance \(dr\) is
\[ dW = F \, dr = G\frac{M m}{r^2} \, dr \]
The total work done in bringing the body from infinity to distance \(r\) is obtained by integrating,
\[ U(r) = - \int\limits_{\infty}^{r} G\frac{M m}{r^2} \, dr \]
Evaluating the integral,
\[ U(r) = - G M m \int\limits_{\infty}^{r} \frac{dr}{r^2} \]
\[ U(r) = - G M m \left[ -\frac{1}{r} \right]_{\infty}^{r} \]
\[ U(r) = -\frac{G M m}{r} \]
This expression gives the gravitational potential energy of a body at a distance \(r\) from the Earth’s centre.
Proof of the Negative Sign of Gravitational Potential Energy
The gravitational force is attractive in nature. As a result, work is done by gravity when a body moves toward the Earth, reducing the energy that must be supplied externally.
Since the reference level of zero potential energy is chosen at infinity, the potential energy at any finite distance from the Earth must be negative. This negative sign indicates that energy must be supplied to remove the body completely from the Earth’s gravitational influence.
Gravitational Potential Energy Near the Earth’s Surface
For a body at a small height \(h\) above the Earth’s surface, where \(h \ll R\), the gravitational potential energy expression can be simplified using approximation.
The distance from the Earth’s centre is \(r = R + h\). Substituting into the general expression,
\[ U = -\frac{G M m}{R + h} \]
Using binomial approximation for small \(h/R\),
\[ U \approx -\frac{G M m}{R} + \frac{G M m h}{R^2} \]
Since \(g = G M / R^2\),
\[ U \approx -\frac{G M m}{R} + m g h \]
The term \(-GMm/R\) is constant for all points near the surface. Therefore, the change in gravitational potential energy near the Earth’s surface is
\[ \Delta U = m g h \]
This familiar expression is widely used in solving problems involving motion near the Earth.
Important Physical Aspects
Gravitational potential energy depends on the relative position of interacting masses and the choice of reference level. Only changes in potential energy have physical significance.
The concept provides a convenient energy-based approach to studying free fall, motion of satellites, and escape velocity, reducing complex force-based problems to simpler energy calculations.
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 the Earth’s gravitational field and reach infinity with zero residual speed.
In this context, “escape” means that the body never returns to the Earth under the influence of gravity alone.
Physical Idea Behind Escape Speed
When a body is projected upward from the Earth, its kinetic energy decreases as it moves away because work is done against gravitational attraction. If the initial kinetic energy is sufficiently large, the body can reach an infinite distance where the gravitational pull of the Earth becomes negligible.
Escape speed corresponds to the special case where the body just manages to reach infinity with zero kinetic energy.
Derivation of Escape Speed
Consider a body of mass \(m\) projected vertically upward from the Earth’s surface with speed \(v_e\). Let the mass of the Earth be \(M\) and its radius be \(R\).
The total mechanical energy of the body at the Earth’s surface is the sum of its kinetic energy and gravitational potential energy.
\[ E_{\text{surface}} = \frac{1}{2} m v_e^2 - \frac{G M m}{R} \]
When the body reaches infinity, both the gravitational potential energy and the kinetic energy become zero.
\[ E_{\infty} = 0 \]
According to the law of conservation of mechanical energy,
\[ E_{\text{surface}} = E_{\infty} \]
Substituting the expressions,
\[ \frac{1}{2} m v_e^2 - \frac{G M m}{R} = 0 \]
Rearranging,
\[ \frac{1}{2} m v_e^2 = \frac{G M m}{R} \]
Cancelling the mass \(m\),
\[ v_e = \sqrt{\frac{2 G M}{R}} \]
This expression gives the escape speed from the surface of the Earth.
Relation Between Escape Speed and Acceleration Due to Gravity
Using the relation \(g = G M / R^2\), the expression for escape speed can be rewritten as
\[ v_e = \sqrt{2 g R} \]
This form is especially useful in numerical problems where the value of \(g\) is known.
Proof that Escape Speed is Independent of Mass
From the derived expression,
\[ v_e = \sqrt{\frac{2 G M}{R}} \]
it is clear that the escape speed depends only on the mass and radius of the Earth. The mass of the escaping body does not appear in the formula.
Therefore, all bodies, irrespective of their masses, require the same escape speed from the Earth’s surface when air resistance is neglected.
Numerical Value for the Earth
Substituting the values of \(g = 9.8 \, \text{m s}^{-2}\) and \(R = 6.4 \times 10^6 \, \text{m}\),
\[ v_e = \sqrt{2 \times 9.8 \times 6.4 \times 10^6} \]
\[ v_e \approx 11.2 \, \text{km s}^{-1} \]
This is the escape speed from the surface of the Earth.
Important Physical Aspects
Escape speed is not a force but a critical speed. Once a body is given this speed, no additional energy is required for it to escape the Earth’s gravitational field.
It assumes the absence of air resistance and that the Earth does not rotate. In practical situations, rockets must exceed this speed to account for energy losses.
EARTH SATELLITES
Earth satellites are practical applications of the laws of gravitation and circular motion discussed in the NCERT textbook. Their motion clearly demonstrates how gravitational force can act as the necessary centripetal force to keep an object moving continuously around the Earth without falling onto it.
Definition of an Earth Satellite
An Earth satellite is a body that revolves around the Earth under the influence of the Earth’s gravitational attraction. Satellites may be natural, such as the Moon, or artificial, such as communication and weather satellites launched by humans.
In NCERT Class XI, the study of Earth satellites focuses mainly on artificial satellites moving in circular or nearly circular orbits around the Earth.
Basic Principle of Satellite Motion
A satellite remains in orbit because the gravitational force of the Earth provides the exact centripetal force required for circular motion. The satellite is continuously falling toward the Earth, but due to its tangential velocity, it keeps missing the Earth.
This delicate balance between gravitational attraction and forward motion results in a stable orbit.
Derivation of Orbital Velocity of an Earth Satellite
Consider a satellite of mass \(m\) revolving in a circular orbit of radius \(r\) around the Earth. Let the mass of the Earth be \(M\).
The gravitational force acting on the satellite is
\[ F = G\frac{M m}{r^2} \]
This force provides the required centripetal force for circular motion,
\[ F = \frac{m v^2}{r} \]
Equating the two expressions,
\[ G\frac{M m}{r^2} = \frac{m v^2}{r} \]
Cancelling \(m\) and simplifying,
\[ v = \sqrt{\frac{G M}{r}} \]
This expression gives the orbital speed of a satellite at a distance \(r\) from the Earth’s centre.
Time Period of an Earth Satellite
The time period \(T\) of a satellite is the time taken to complete one full revolution around the Earth.
The circumference of the orbit is \(2\pi r\). Therefore,
\[ T = \frac{2\pi r}{v} \]
Substituting the expression for orbital velocity,
\[ T = 2\pi \sqrt{\frac{r^3}{G M}} \]
This relation shows that the square of the time period of a satellite is proportional to the cube of the radius of its orbit, in agreement with Kepler’s third law.
Energy of an Earth Satellite
The total mechanical energy of a satellite in a circular orbit is the sum of its kinetic energy and gravitational potential energy.
\[ K = \frac{1}{2} m v^2 = \frac{G M m}{2r} \]
\[ U = -\frac{G M m}{r} \]
Hence, the total energy is
\[ E = K + U = -\frac{G M m}{2r} \]
The negative sign indicates that the satellite is bound to the Earth by gravity.
Proof that Satellite Motion is Independent of Mass
From the expressions for orbital velocity and time period,
\[ v = \sqrt{\frac{G M}{r}}, \qquad T = 2\pi \sqrt{\frac{r^3}{G M}} \]
it is evident that neither quantity depends on the mass of the satellite. Therefore, satellites of different masses placed in the same orbit will have identical speeds and time periods.
Important Physical Aspects
The height of a satellite above the Earth determines its orbital speed and time period. Satellites closer to the Earth move faster and complete revolutions in shorter times compared to those at greater heights.
The concept of Earth satellites explains weightlessness experienced by astronauts, as both the satellite and the astronaut are in continuous free fall under gravity.
Significance in the NCERT Context
Earth satellites form a crucial link between gravitation and real-world applications such as communication, navigation, weather forecasting, and scientific observation.
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 BAC and CPB ?
Solution
Let the speed of the planet at perihelion \(P\) be denoted by \(v_{P}\) and the Sun–planet distance at perihelion by \(r_{P}\). At aphelion, let the corresponding speed and Sun–planet distance be \(v_{A}\) and \(r_{A}\), respectively.
The gravitational force on the planet is always directed towards the Sun, so the torque about the Sun is zero. Hence the angular momentum of the planet about the Sun remains conserved throughout its orbital motion.
Using conservation of angular momentum at perihelion and aphelion, \[ \begin{aligned} L_{P} &= m r_{P} v_{P},\\ L_{A} &= m r_{A} v_{A},\\ L_{P} &= L_{A} \\ \Rightarrow m r_{P} v_{P} &= m r_{A} v_{A}. \end{aligned} \] Cancelling the common factor \(m\), we obtain \[ \begin{aligned} r_{P} v_{P} &= r_{A} v_{A},\\[4pt] \frac{v_{P}}{v_{A}} &= \frac{r_{A}}{r_{P}}. \end{aligned} \] Thus, the speed of the planet is greater at perihelion than at aphelion because \(r_{P} < r_{A}\).
According to Kepler’s second law, the line joining the planet and the Sun sweeps out equal areas in equal intervals of time. The regions \(BAC\) and \(CPB\) in the elliptical orbit have equal areas; 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 the vertex A is doubled ?
Take AG = BG = CG = 1 m (see Fig. 7.5)
Solution
Three equal masses of \(m\) kg each are placed at the vertices \(A,B,C\) of an equilateral triangle. A mass \(2m\) is placed at the centroid \(G\). The distance of each vertex from the centroid is given as \(AG = BG = CG = 1\,\text{m}\).
The gravitational force between two point masses \(m_1\) and \(m_2\) separated by distance \(r\) is \[ F = \dfrac{G m_1 m_2}{r^2}. \] For the force on the mass \(2m\) at \(G\) due to a mass \(m\) at any vertex (say \(A\)), \[ \begin{aligned} F_{GA} &= \dfrac{G (2m)(m)}{1^2} \\ &= 2Gm^2. \end{aligned} \] Thus the magnitude of each of the three forces on the mass \(2m\) at \(G\) is \(2Gm^2\).
Resolving these three forces along the \(x\)- and \(y\)-axes, let the force from \(A\) be along the positive \(y\)-axis, while the forces from \(B\) and \(C\) make angles of \(30^\circ\) with the negative and positive \(x\)-axes respectively. Then \[ \begin{aligned} \vec F_{GA} &= 2Gm^2\,\hat{\jmath},\\ \vec F_{GB} &= 2Gm^2\left(-\cos 30^\circ\,\hat{\imath} - \sin 30^\circ\,\hat{\jmath}\right),\\ \vec F_{GC} &= 2Gm^2\left(\phantom{-}\cos 30^\circ\,\hat{\imath} - \sin 30^\circ\,\hat{\jmath}\right). \end{aligned} \] The resultant force on \(2m\) at \(G\) is \[ \begin{aligned} \vec F_R &= \vec F_{GA} + \vec F_{GB} + \vec F_{GC} \\ &= 2Gm^2\left[\hat{\jmath} - \cos 30^\circ\,\hat{\imath} - \sin 30^\circ\,\hat{\jmath} + \cos 30^\circ\,\hat{\imath} - \sin 30^\circ\,\hat{\jmath}\right] \\ &= 2Gm^2\left[\hat{\jmath} - \dfrac{1}{2}\hat{\jmath} - \dfrac{1}{2}\hat{\jmath}\right] \\ &= 2Gm^2 \cdot 0 \\ &= 0. \end{aligned} \] Hence, for three equal masses at the vertices, the net gravitational force on the mass \(2m\) at the centroid is zero.
Now, if the mass at vertex \(A\) is doubled from \(m\) to \(2m\), the mass at \(G\) is still \(2m\). The gravitational force on \(2m\) at \(G\) due to the new mass \(2m\) at \(A\) is \[ \begin{aligned} F_{GA} &= \dfrac{G (2m)(2m)}{1^2} \\ &= 4Gm^2. \end{aligned} \] The forces due to \(B\) and \(C\) remain the same as before, each of magnitude \(2Gm^2\).
Writing all forces in component form again, we have \[ \begin{aligned} \vec F_{GA} &= 4Gm^2\,\hat{\jmath},\\ \vec F_{GB} &= 2Gm^2\left(-\cos 30^\circ\,\hat{\imath} - \sin 30^\circ\,\hat{\jmath}\right),\\ \vec F_{GC} &= 2Gm^2\left(\phantom{-}\cos 30^\circ\,\hat{\imath} - \sin 30^\circ\,\hat{\jmath}\right). \end{aligned} \] Adding these three vectors, \[ \begin{aligned} \vec F_R &= \vec F_{GA} + \vec F_{GB} + \vec F_{GC} \\ &= 4Gm^2\,\hat{\jmath} + 2Gm^2\left(-\cos 30^\circ\,\hat{\imath} - \sin 30^\circ\,\hat{\jmath}\right) + 2Gm^2\left(\cos 30^\circ\,\hat{\imath} - \sin 30^\circ\,\hat{\jmath}\right) \\ &= 4Gm^2\,\hat{\jmath} + 2Gm^2\left[-\sin 30^\circ - \sin 30^\circ\right]\hat{\jmath} \\ &= 4Gm^2\,\hat{\jmath} - 4Gm^2\left(\dfrac{1}{2}\right)\hat{\jmath} \\ &= 2Gm^2\,\hat{\jmath}. \end{aligned} \] Therefore, when the mass at vertex \(A\) is doubled, the net gravitational force on the mass \(2m\) at \(G\) is \[ \vec F_R = 2Gm^2\,\hat{\jmath}, \] directed along the positive \(y\)-axis (towards \(A\)).
Example-3
Find the potential energy of a system of four particles placed at the vertices of a square of side l. Also obtain the potential at the centre of the square.
Solution
Consider four equal masses \(m\) placed at the vertices of a square of side \(l\). The potential energy of the system is the sum of the gravitational potential energies of all distinct pairs of masses. For any pair of point masses \(m_1\) and \(m_2\) separated by distance \(r\), the gravitational potential energy is \[ U = -\dfrac{G m_1 m_2}{r}. \]
Along each side of the square the separation between adjacent particles is \(l\). There are four such sides, so the contribution from all side pairs is \[ \begin{aligned} U_{\text{sides}} &= 4\left(-\dfrac{G m \cdot m}{l}\right) \\ &= -\dfrac{4G m^{2}}{l} \end{aligned} \] Across each diagonal, the separation between opposite vertices is \(\sqrt{2}\,l\). There are two diagonals, hence \[ \begin{aligned} U_{\text{diagonals}} &= 2\left(-\dfrac{G m \cdot m}{\sqrt{2}\,l}\right) \\ &= -\dfrac{2G m^{2}}{\sqrt{2}\,l} \\ &= -\dfrac{\sqrt{2}G m^{2}}{l} \end{aligned} \]
Therefore, the total potential energy of the four–particle system is \[ \begin{aligned} U_{\text{total}} &= U_{\text{sides}} + U_{\text{diagonals}} \\ &= -\dfrac{4G m^{2}}{l} - \dfrac{\sqrt{2}G m^{2}}{l} \\ &= -\dfrac{G m^{2}}{l}\left(4 + \sqrt{2}\right) \end{aligned} \] (If a numerical value is desired, \(\sqrt{2} \approx 1.414\), so \(4 + \sqrt{2} \approx 5.414\).)
Next, the potential at the centre of the square is found by summing the potentials due to each mass. The distance from the centre to any vertex is half of the diagonal. The diagonal of a square of side \(r\) is \[ d = \sqrt{2}\,r, \] so the distance from the centre to a corner is \[ \begin{aligned} r &= \dfrac{d}{2} = \dfrac{\sqrt{2}}{2}r = \dfrac{l}{\sqrt{2}} \end{aligned} \] The potential at the centre due to one mass is \[ \begin{aligned} V_1 &= -\dfrac{G m}{r} = -\dfrac{G m}{r/\sqrt{2}} \\ &= -\dfrac{\sqrt{2}G m}{r} \end{aligned} \] Since there are four identical masses, the total potential at the centre is \[ \begin{aligned} V_{\text{centre}} &= 4V_1 \\ &= 4\left(-\dfrac{\sqrt{2}G m}{r}\right) \\ &= -\dfrac{4\sqrt{2}G m}{r} \end{aligned} \]
Example-4
Two uniform solid spheres of equal radii R, but mass M and 4 M have a centre to centre separation 6 R, as shown in Fig. 7.10. The two spheres are held fixed. A projectile of mass m is projected from the surface of the sphere of mass M directly towards the centre of the second sphere. Obtain an expression for the minimum speed v of the projectile so that it reaches the surface of the second sphere.
Solution
Two uniform solid spheres of masses \(M\) and \(4M\) have equal radii \(R\) and their centres are separated by a distance \(6R\). A projectile of mass \(m\) is projected from the surface of the sphere of mass \(M\) directly towards the centre of the sphere of mass \(4M\). The aim is to find the minimum initial speed \(v\) such that the projectile just reaches the surface of the second sphere.
Let the line joining the centres be the \(x\)-axis, and let the centre of the sphere of mass \(M\) be at \(O\) and that of mass \(4M\) at \(O'\). The separation \(OO'\) is \(6R\). Let \(N\) be the point on the line \(OO'\) where the net gravitational force on the projectile is zero, called the neutral point. Let the distance of \(N\) from \(O\) be \(r\).
At the neutral point \(N\), the magnitudes of the gravitational attractions due to the two spheres on the projectile must be equal: \[ \begin{aligned} \dfrac{GMm}{r^{2}} &= \dfrac{G(4M)m}{(6R - r)^{2}} \end{aligned} \] Cancelling the common factors and rearranging, \[ \begin{aligned} \dfrac{1}{r^{2}} &= \dfrac{4}{(6R - r)^{2}},\\ (6R - r)^{2} &= 4r^{2},\\ 6R - r &= \pm 2r \end{aligned} \] Taking the physically meaningful solution, \[ \begin{aligned} 6R - r &= 2r \Rightarrow 6R = 3r \Rightarrow r = 2R \end{aligned} \] The other solution \(6R - r = -2r \Rightarrow r = -6R\) is not acceptable. Thus the neutral point \(N\) is at a distance \(ON = 2R\) from the centre of the sphere of mass \(M\).
Next, the mechanical energy of the projectile is considered. At the surface of the sphere of mass \(M\), the projectile is at a distance \(R\) from the centre \(O\), and at a distance \(5R\) from the centre \(O'\) (since \(OO' = 6R\)). Its total mechanical energy at this point is \[ \begin{aligned} E_{i} &= \dfrac{1}{2}mv^{2} - \dfrac{GMm}{R} - \dfrac{G(4M)m}{5R} \end{aligned} \] At the neutral point \(N\), the projectile has minimum speed at the threshold case. For the smallest initial speed that still allows the projectile to move past \(N\) and reach the second sphere, the speed at \(N\) can be taken as zero. Hence the total mechanical energy at \(N\) is purely potential. The distances from the centres at \(N\) are \(ON = 2R\) and \(O'N = 4R\). Therefore, \[ \begin{aligned} E_{n} &= -\dfrac{GMm}{2R} - \dfrac{G(4M)m}{4R} \\ &= -\dfrac{GMm}{2R} - \dfrac{GMm}{R} \\ &= -\dfrac{3}{2}\dfrac{GMm}{R} \end{aligned} \]
By conservation of mechanical energy, the total energy at the surface of the first sphere must equal the total energy at the neutral point: \[ \begin{aligned} E_{i} &= E_{n},\\ \dfrac{1}{2}mv^{2} - \dfrac{GMm}{R} - \dfrac{4GMm}{5R} &= -\dfrac{3GMm}{2R} \end{aligned} \] Dividing by \(m\) and simplifying, \[ \begin{aligned} \dfrac{1}{2}v^{2} &= -\dfrac{3GM}{2R} + \dfrac{GM}{R} + \dfrac{4GM}{5R} \end{aligned} \] Combining the terms in the bracket, \[ \begin{aligned} \dfrac{1}{2}v^{2} &= \dfrac{GM}{R}\left(1 + \dfrac{4}{5} - \dfrac{3}{2}\right) \\ &= \dfrac{GM}{R}\left(\dfrac{10 + 8 - 15}{10}\right) \\ &= \dfrac{GM}{R}\left(\dfrac{3}{10}\right) \end{aligned} \] Thus, \[ \begin{aligned} v^{2} &= \dfrac{2GM}{R}\left(\dfrac{3}{10}\right) = \dfrac{3GM}{5R},\\ v &= \sqrt{\dfrac{3GM}{5R}} \end{aligned} \] This is the minimum speed required at the surface of the sphere of mass \(M\) for the projectile to just reach the surface of the sphere of mass \(4M\).
Example-5
The planet Mars has two moons, phobos and delmos.
(i) phobos has a period 7 hours, 39 minutes and an orbital radius of 9.4 ×103 km. Calculate the mass of
mars.
(ii) Assume that earth and mars move in circular orbits around the sun, with the martian orbit being
1.52
times the orbital radius of the earth. What is the length of the martian year in days ?
Solution
The planet Mars has two moons, Phobos and Deimos. For Phobos, the orbital period is \(7\) hours \(39\) minutes and the orbital radius is \(9.4 \times 10^{3}\,\text{km}\). First, the time period is converted entirely into seconds. \[ \begin{aligned} T &= 7 \times 60 + 39 \ \text{minutes} \\ &= 420 + 39 \ \text{minutes} \\ &= 459 \ \text{minutes} \\ &= 459 \times 60 \ \text{s} \\ &= 2.754 \times 10^{4} \ \text{s} \end{aligned} \] The orbital radius in SI units is \[ R = 9.4 \times 10^{3} \ \text{km} = 9.4 \times 10^{6} \ \text{m}. \]
For a moon of negligible mass orbiting a planet of mass \(M_{M}\) in a circular orbit of radius \(R\), the relation between the period and the radius is \[ \begin{aligned} T^{2} &= \dfrac{4\pi^{2}}{G M_{M}} R^{3},\\ M_{M} &= \dfrac{4\pi^{2} R^{3}}{G T^{2}} \end{aligned} \] Substituting the values, \[ \begin{aligned} M_{M} &= \dfrac{4\pi^{2} \left(9.4 \times 10^{6}\right)^{3}}{6.67 \times 10^{-11} \left(459 \times 60\right)^{2}} \\ &\approx 6.5 \times 10^{23} \ \text{kg} \end{aligned} \] Thus, the mass of Mars is approximately \(6.5 \times 10^{23}\,\text{kg}\).
Next, assume that both Earth and Mars move in circular orbits around the Sun. Let \(R_{ES}\) and \(R_{MS}\) be the orbital radii of Earth and Mars respectively, and let \(T_{E}\) and \(T_{M}\) be their orbital periods. The martian orbit is given to be \(1.52\) times the orbital radius of the Earth: \[ R_{MS} = 1.52\,R_{ES} \] By Kepler’s third law, \[ \begin{aligned} \dfrac{T_{M}^{2}}{T_{E}^{2}} &= \dfrac{R_{MS}^{3}}{R_{ES}^{3}},\\ T_{M}^{2} &= T_{E}^{2} \left(1.52\right)^{3}. \end{aligned} \] Taking \(T_{E} = 365\) days, \[ \begin{aligned} T_{M} &= T_{E} \left(1.52\right)^{3/2} \\ &= 365 \left(1.52\right)^{3/2} \ \text{days} \\ &\approx 365 \times 1.87 \ \text{days} \\ &\approx 6.8 \times 10^{2} \ \text{days}. \end{aligned} \] Therefore, the martian year is about \(684\) days long.
Example-6
Weighing the Earth : You are given the following data: g = 9.81 ms–2, RE = 6.37×106 m, the distance to the moon R = 3.84×108 m and the time period of the moon’s revolution is 27.3 days. Obtain the mass of the Earth ME in two different ways.
Solution
Given data are: \(g = 9.81 \,\text{m s}^{-2}\), \(R_{E} = 6.37 \times 10^{6} \,\text{m}\), the Earth–Moon distance \(R = 3.84 \times 10^{8} \,\text{m}\), and the time period of the Moon’s revolution \(T_{m} = 27.3\) days. The mass of the Earth \(M_{E}\) is obtained in two independent ways.
First, using the relation between the acceleration due to gravity at the Earth’s surface and the mass of the Earth, \[ \begin{aligned} g &= \dfrac{G M_{E}}{R_{E}^{2}},\\[4pt] M_{E} &= \dfrac{g R_{E}^{2}}{G} \end{aligned} \] Substituting numerical values, \[ \begin{aligned} M_{E} &= \dfrac{9.81 \times \left(6.37 \times 10^{6}\right)^{2}}{6.67 \times 10^{-11}} \\ &\approx 5.97 \times 10^{24} \,\text{kg} \end{aligned} \]
Second, the mass of the Earth is obtained from the orbital motion of the Moon using Kepler’s third law. For a satellite of negligible mass in circular orbit of radius \(R\) around the Earth, \[ \begin{aligned} T^{2} &= \dfrac{4\pi^{2} R^{3}}{G M_{E}},\\[4pt] M_{E} &= \dfrac{4\pi^{2} R^{3}}{G T^{2}} \end{aligned} \] The orbital radius is \(R = 3.84 \times 10^{8} \,\text{m}\) and the time period in seconds is \[ \begin{aligned} T &= 27.3 \times 24 \times 60 \times 60 \ \text{s} \end{aligned} \] Therefore, \[ \begin{aligned} M_{E} &= \dfrac{4\pi^{2} \left(3.84 \times 10^{8}\right)^{3}}{6.67 \times 10^{-11} \left(27.3 \times 24 \times 60 \times 60\right)^{2}} \\ &\approx 6.02 \times 10^{24} \,\text{kg} \end{aligned} \] Both methods give nearly the same value, confirming that the mass of the Earth is about \(6 \times 10^{24} \,\text{kg}\).
Example-7
Express the constant k of Eq. (7.38) in days and kilometres. Given k = 10–13 s2 m–3. The moon is at a distance of 3.84×105 km from the earth. Obtain its time-period of revolution in days.
Solution
The constant \(k\) in Eq. (7.38) is given as \(k = 10^{-13} \,\text{s}^{2}\,\text{m}^{-3}\). It is to be expressed in units of days and kilometres so that \(T\) is in days and the distance is in kilometres.
The relation is \[ T^{2} = k \left(R_{E} + h\right)^{3}. \] To convert \(k\) from \(\text{s}^{2}\,\text{m}^{-3}\) to \(\text{day}^{2}\,\text{km}^{-3}\), use \(1\,\text{day} = 24 \times 60 \times 60\,\text{s}\) and \(1\,\text{km} = 10^{3}\,\text{m}\). Thus \[ \begin{aligned} k &= 10^{-13} \,\text{s}^{2}\,\text{m}^{-3} \\ &= 10^{-13} \left(\dfrac{1\,\text{day}}{24 \times 60 \times 60\,\text{s}}\right)^{2} \left(\dfrac{10^{3}\,\text{m}}{1\,\text{km}}\right)^{3} \\ &= 10^{-13} \dfrac{10^{9}}{\left(24 \times 60 \times 60\right)^{2}} \,\text{day}^{2}\,\text{km}^{-3} \end{aligned} \] Evaluating the numerical factor, \[ \begin{aligned} k &\approx 1.33 \times 10^{-4} \dfrac{1}{\left(24 \times 60 \times 60\right)^{2}} \,\text{day}^{2}\,\text{km}^{-3} \\ &\approx 1.33 \times 10^{-14} \,\text{day}^{2}\,\text{km}^{-3} \end{aligned} \] So, in days and kilometres, \(k \approx 1.33 \times 10^{-14} \,\text{day}^{2}\,\text{km}^{-3}\).
The distance of the moon from the Earth is \(R = 3.84 \times 10^{5}\,\text{km}\). Using the relation \(T^{2} = kR^{3}\) with \(T\) in days and \(R\) in kilometres, \[ \begin{aligned} T^{2} &= 1.33 \times 10^{-14} \left(3.84 \times 10^{5}\right)^{3}, \\ T &= \sqrt{1.33 \times 10^{-14} \left(3.84 \times 10^{5}\right)^{3}} \end{aligned} \] Writing it more explicitly, \[ \begin{aligned} T &= \sqrt{1.33 \times 10^{-14} \times 3.84^{3} \times 10^{15}} \\ &\approx 27.3 \ \text{days} \end{aligned} \] Hence, the time period of revolution of the moon is \(27.3\) days.
Example-8
A 400 kg satellite is in a circular orbit of radius 2RE about the Earth. How much energy is required to transfer it to a circular orbit of radius 4RE? What are the changes in the kinetic and potential energies ?
Solution
The mass of the satellite is \(m = 400 \,\text{kg}\). It is initially in a circular orbit of radius \(2R_{E}\) around the Earth and is transferred to a circular orbit of radius \(4R_{E}\). The total mechanical energy of a satellite of mass \(m\) in a circular orbit of radius \(r\) is \[ E = -\dfrac{G M_{E} m}{2r}. \]
For the initial orbit of radius \(r_{i} = 2R_{E}\), \[ \begin{aligned} E_{i} &= -\dfrac{G M_{E} m}{2(2R_{E})} \\ &= -\dfrac{G M_{E} m}{4R_{E}} \end{aligned} \] For the final orbit of radius \(r_{f} = 4R_{E}\), \[ \begin{aligned} E_{f} &= -\dfrac{G M_{E} m}{2(4R_{E})} \\ &= -\dfrac{G M_{E} m}{8R_{E}} \end{aligned} \]
The energy required to transfer the satellite from the first orbit to the second is the increase in total mechanical energy: \[ \begin{aligned} \Delta E &= E_{f} - E_{i} \\ &= -\dfrac{G M_{E} m}{8R_{E}} + \dfrac{G M_{E} m}{4R_{E}} \\ &= \dfrac{G M_{E} m}{8R_{E}} \end{aligned} \] Using \(G M_{E} = g R_{E}^{2}\) \[ \begin{aligned} \Delta E &= \dfrac{g R_{E}^{2} m}{8R_{E}} \\ &= \dfrac{g m R_{E}}{8} \end{aligned} \] Substituting \(g = 9.81 \,\text{m s}^{-2}\), \(m = 400 \,\text{kg}\) and \(R_{E} = 6.37 \times 10^{6} \,\text{m}\), \[ \begin{aligned} \Delta E &= \dfrac{9.81 \times 400 \times 6.37 \times 10^{6}}{8} \\ &\approx 3.13 \times 10^{9} \\text{J} \end{aligned} \] This is the energy that must be supplied to the satellite.
The gravitational potential energy of the satellite in a circular orbit of radius \(r\) is \[ U = -\dfrac{G M_{E} m}{r} \] and its kinetic energy is \[ K = \dfrac{G M_{E} m}{2r} \] Initially, at \(r_{i} = 2R_{E}\) \[ \begin{aligned} U_{i} &= -\dfrac{G M_{E} m}{2R_{E}}\\ K_{i} &= \dfrac{G M_{E} m}{4R_{E}} \end{aligned} \] Finally, at \(r_{f} = 4R_{E}\) \[ \begin{aligned} U_{f} &= -\dfrac{G M_{E} m}{4R_{E}}\\ K_{f} &= \dfrac{G M_{E} m}{8R_{E}} \end{aligned} \] Hence the changes in potential and kinetic energies are \[ \begin{aligned} \Delta U &= U_{f} - U_{i} \\\\&= -\dfrac{G M_{E} m}{4R_{E}} + \dfrac{G M_{E} m}{2R_{E}} \\\\&= \dfrac{G M_{E} m}{4R_{E}},\\[4pt] \Delta K &= K_{f} - K_{i} \\\\&= \dfrac{G M_{E} m}{8R_{E}} - \dfrac{G M_{E} m}{4R_{E}} \\\\&= -\dfrac{G M_{E} m}{8R_{E}} \end{aligned} \] Clearly, \[ \Delta U = 2\Delta E \quad \text{and} \quad \Delta K = -\Delta E, \] that is, the potential energy increases by twice the supplied energy while the kinetic energy decreases by an equal amount to keep the total energy change equal to \(\Delta E = 3.13 \times 10^{9} \,\text{J}\).
