Introduction

From early observations, humans recognized that objects tend to fall towards the Earth. This tendency led to initial ideas about gravity.

Historically, Galileo Galilei was the first to scientifically determine that all objects, regardless of their mass, experience a constant acceleration towards the Earth (due to gravity, often denoted as $g$). He performed experiments, such as those with inclined planes, to verify this.

Simultaneously, observations of celestial bodies like stars and planets fascinated astronomers. Early models, like Ptolemy's geocentric model (Earth at the center, celestial bodies orbiting in circles or circles within circles), attempted to explain planetary motions but were complex.

Aryabhatta (5th century AD) proposed a heliocentric model (Sun at the center), which was later detailed by Nicolaus Copernicus (1473-1543). Copernicus's model suggested planets orbit the Sun in circles.

Tycho Brahe (1546-1601) meticulously recorded extensive naked-eye observations of planetary positions over many years.

His assistant, Johannes Kepler (1571-1640), analyzed this data and formulated three elegant laws describing planetary motion, moving beyond the idea of purely circular orbits. These laws were crucial in guiding Newton's later work on gravitation.

Kepler’s Laws

Kepler's three laws describe the motion of planets around the Sun:

Law Of Orbits

All planets move in elliptical orbits with the Sun situated at one of the foci of the ellipse.

An ellipse is a closed curve defined by two focal points ($F_1, F_2$). For any point P on the ellipse, the sum of the distances from the two foci is constant ($PF_1 + PF_2 = \text{constant}$). A circle is a special case of an ellipse where the two foci coincide at the center.

The point in the orbit closest to the Sun is called the perihelion, and the farthest point is the aphelion. The semi-major axis is half the distance between the perihelion and aphelion.

Law Of Areas

The line that joins any planet to the Sun sweeps equal areas in equal intervals of time.

This means a planet moves faster when it is nearer to the Sun and slower when it is farther away.

This law is a consequence of the conservation of angular momentum for a body moving under a central force (a force directed along the line joining the two bodies). Gravitation is a central force.

The area swept out by the position vector $\mathbf{r}$ in time $\Delta t$ is $\Delta A = \frac{1}{2} |\mathbf{r} \times \mathbf{v} \Delta t|$. So, the rate of sweeping area is $\frac{\Delta A}{\Delta t} = \frac{1}{2} |\mathbf{r} \times \mathbf{v}| = \frac{1}{2m} |\mathbf{r} \times m\mathbf{v}| = \frac{|\mathbf{L}|}{2m}$, where $\mathbf{L}$ is the angular momentum. For a central force, $\mathbf{L}$ is conserved ($\frac{d\mathbf{L}}{dt} = \mathbf{\tau} = \mathbf{r} \times \mathbf{F} = \mathbf{0}$ since $\mathbf{r}$ and $\mathbf{F}$ are parallel). Therefore, $\frac{\Delta A}{\Delta t}$ is constant.

Law Of Periods

The square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of its elliptical orbit.

where $T$ is the orbital period and $a$ is the semi-major axis of the ellipse. For circular orbits, $a$ is simply the radius $R$, so $T^2 \propto R^3$.

This law implies that planets farther from the Sun take longer to complete an orbit, and the relationship is specific.

Data for planets in our solar system confirm this law:

The nearly constant value of $T^2/a^3$ for all planets orbiting the Sun supports Kepler's third law.

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At the perihelion P and aphelion A, the velocity vector of the planet is perpendicular to the position vector from the Sun (assuming the Sun is at a focus S). This is evident from the symmetry of the ellipse at these extreme points. Also, the gravitational force is always directed towards the Sun (a central force), which means the torque about the Sun is zero. Therefore, the angular momentum ($\mathbf{L} = \mathbf{r} \times m\mathbf{v}$) of the planet about the Sun is conserved throughout its orbit.

At perihelion P, the angular momentum magnitude is $L_P = |\mathbf{r}_P \times m\mathbf{v}_P| = r_P (mv_P) \sin 90^\circ = m r_P v_P$.

At aphelion A, the angular momentum magnitude is $L_A = |\mathbf{r}_A \times m\mathbf{v}_A| = r_A (mv_A) \sin 90^\circ = m r_A v_A$.

By conservation of angular momentum, $L_P = L_A$:

$$ m r_P v_P = m r_A v_A $$ $$ r_P v_P = r_A v_A $$

Thus, the product of distance from the Sun and speed is constant at perihelion and aphelion. Since $r_A > r_P$ (aphelion is farther), the speed at aphelion must be less than the speed at perihelion ($v_A < v_P$).

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Regarding the time taken to traverse BAC and CPB:

According to Kepler's Second Law (Law of Areas), the line joining the planet to the Sun sweeps out equal areas in equal intervals of time. The area of the sector SBAC is the area swept by the radius vector from S as the planet moves from B to C, passing through A. The area of the sector SCPB is the area swept as the planet moves from C to B, passing through P.

By inspection of the diagram, the sector SBAC appears to have a significantly larger area than the sector SCPB. Since equal areas are swept in equal times, and Area(SBAC) $>$ Area(SCPB), the time taken to traverse BAC will be longer than the time taken to traverse CPB.

Universal Law Of Gravitation

Inspired by observations of falling objects on Earth and Kepler's laws of planetary motion, Isaac Newton proposed a single, universal law to explain both terrestrial gravity and the motion of celestial bodies.

Newton's key insight was to compare the acceleration of the Moon in its orbit around the Earth with the acceleration of an object falling to the Earth's surface (acceleration due to gravity $g$).

• The Moon is constantly accelerating towards the Earth (centripetal acceleration) to maintain its orbit. Its acceleration $a_m$ is given by $a_m = v_m^2 / R_m$, where $v_m$ is the Moon's orbital speed and $R_m$ is the distance to the Earth's center. $v_m = 2\pi R_m / T_m$, where $T_m$ is the Moon's orbital period ($\approx 27.3$ days). Calculating $a_m$ gives a value much smaller than $g$ at the Earth's surface ($g \approx 9.8$ m/s$^2$).
• Newton reasoned that the force causing the Moon's acceleration must be the same force causing objects to fall on Earth, but that this force must decrease with distance from the Earth.

He hypothesized that the gravitational force decreases as the inverse square of the distance. Comparing the Moon's acceleration $a_m$ at distance $R_m$ with $g$ at Earth's surface ($R_E$) leads to $a_m / g \approx (R_E / R_m)^2$. Using known values for $R_E$ and $R_m$ ($\approx 60 R_E$), this ratio is approximately $(1/60)^2 = 1/3600$. The calculated $a_m$ is indeed roughly $g/3600$, supporting the inverse square relationship.

Newton's Universal Law of Gravitation:

Every body in the universe attracts every other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

For two point masses $m_1$ and $m_2$ separated by a distance $r$, the magnitude of the gravitational force $\mathbf{F}$ is:

where $G$ is the universal gravitational constant. The force is always attractive, acting along the line joining the two point masses.

In vector form, the force $\mathbf{F}_{21}$ exerted on mass $m_1$ by mass $m_2$ is $\mathbf{F}_{21} = -G \frac{m_1 m_2}{r^2} \hat{\mathbf{r}}_{21}$, where $\hat{\mathbf{r}}_{21}$ is the unit vector from $m_2$ to $m_1$. Equivalently, $\mathbf{F}_{12} = -G \frac{m_1 m_2}{r^2} \hat{\mathbf{r}}_{12}$, where $\hat{\mathbf{r}}_{12}$ is the unit vector from $m_1$ to $m_2$. By Newton's Third Law, $\mathbf{F}_{12} = -\mathbf{F}_{21}$.

For a collection of point masses, the net gravitational force on any one mass is the vector sum of the forces exerted by all other masses individually (Principle of Superposition).

For extended objects with finite size, the force is calculated by integrating the forces between infinitesimal mass elements. For spherically symmetric bodies (like uniform spheres or spherical shells):

• A spherically symmetric body attracts a point mass outside it as if all its mass were concentrated at its center.
• A uniform spherical shell exerts zero net gravitational force on a point mass located anywhere *inside* the shell.

Example 8.2. 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. 8.5)

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The Gravitational Constant

The value of the universal gravitational constant $G$ was first measured experimentally by English scientist Henry Cavendish in 1798.

Cavendish's experiment used a sensitive torsion balance. The apparatus consists of a light rod with two small lead spheres at its ends, suspended by a fine wire. Two large lead spheres are brought near the small ones. The gravitational attraction between the large and small spheres creates a torque that twists the suspension wire.

The gravitational force between a large sphere of mass $M$ and a small sphere of mass $m$ separated by a distance $d$ is $F = G \frac{Mm}{d^2}$.

This force creates a torque $\tau_g = F \times L$ (where $L$ is the length of the rod). This torque is balanced by the restoring torque of the twisted wire, which is proportional to the angle of twist $\theta$, $\tau_{restoring} = \kappa \theta$, where $\kappa$ is the torsion constant of the wire.

At equilibrium, $\tau_g = \tau_{restoring}$.

By measuring $M, m, d, L, \theta$ and independently determining $\kappa$, Cavendish could calculate the value of $G$.

The currently accepted value of the universal gravitational constant is:

This value is extremely small, indicating that gravitational force is very weak compared to other fundamental forces like electromagnetic force, unless the masses involved are very large (like planets or stars).

Acceleration Due To Gravity Of The Earth

The acceleration experienced by an object due to the Earth's gravitational pull is called the acceleration due to gravity, denoted by $g$.

Assuming the Earth is a uniform solid sphere of mass $M_E$ and radius $R_E$, a point mass $m$ outside the Earth is attracted by the Earth as if the entire mass $M_E$ is concentrated at its center.

For a point mass $m$ on the surface of the Earth, the distance from the center is $R_E$. The gravitational force on $m$ is $F = G \frac{M_E m}{R_E^2}$.

According to Newton's Second Law, this force causes an acceleration $g = F/m$.

This formula relates the acceleration due to gravity $g$ at the Earth's surface to the Earth's mass $M_E$ and radius $R_E$.

Since $g$ and $R_E$ can be measured directly, the measurement of $G$ (by experiments like Cavendish's) allows us to calculate the mass of the Earth ($M_E = \frac{g R_E^2}{G}$). This is why Cavendish is sometimes credited with "weighing the Earth".

The acceleration due to gravity $g$ varies slightly with location on Earth due to factors like altitude, depth, Earth's rotation, and non-uniform mass distribution, but the value $\approx 9.8$ m/s$^2$ is commonly used.

Acceleration Due To Gravity Below And Above The Surface Of Earth

The acceleration due to gravity is not constant throughout space; it varies with distance from the Earth's center.

Acceleration Above The Surface

For a point mass $m$ at a height $h$ above the Earth's surface, its distance from the Earth's center is $r = R_E + h$. Assuming the Earth is a uniform sphere, the acceleration due to gravity $g(h)$ at this height is:

Since $g = \frac{G M_E}{R_E^2}$ is the acceleration at the surface ($h=0$), we can write:

For heights $h$ much smaller than the Earth's radius ($h \ll R_E$), we can use the binomial approximation $(1+x)^n \approx 1+nx$ for $|x| \ll 1$. Here $x=h/R_E$ and $n=-2$.

This shows that the acceleration due to gravity decreases linearly with increasing height for small heights above the surface.

Acceleration Below The Surface

For a point mass $m$ at a depth $d$ below the Earth's surface, its distance from the Earth's center is $r = R_E - d$.

Assuming the Earth is a uniform solid sphere of density $\rho$. When the mass $m$ is at a distance $r$ from the center (inside the Earth), only the mass contained within the sphere of radius $r$ contributes to the gravitational force on $m$. The outer shell (from radius $r$ to $R_E$) exerts zero net force on $m$ (as per the property of a spherical shell). The mass within the sphere of radius $r$ is $M_r = \rho \times \frac{4}{3}\pi r^3$. The total mass of Earth is $M_E = \rho \times \frac{4}{3}\pi R_E^3$. So $M_r = M_E \left(\frac{r}{R_E}\right)^3$.

The gravitational force on $m$ at distance $r$ is $F(r) = G \frac{M_r m}{r^2} = G \frac{M_E (r/R_E)^3 m}{r^2} = G \frac{M_E m}{R_E^3} r$.

The acceleration due to gravity $g(r)$ at distance $r$ from the center (or depth $d = R_E - r$) is:

At the surface, $r = R_E$, $g(R_E) = \frac{G M_E}{R_E^2} = g$. So, $g(r) = g \frac{r}{R_E}$.

Substituting $r = R_E - d$:

This shows that the acceleration due to gravity also decreases linearly with increasing depth below the surface.

At the center of the Earth ($d = R_E$, $r = 0$), $g(d=R_E) = g(r=0) = 0$.

Thus, the acceleration due to gravity is maximum at the surface and decreases as one moves up or down from the surface, reaching zero at the Earth's center.

Gravitational Potential Energy

The gravitational force is a conservative force, which means we can define a potential energy function associated with it.

For a particle of mass $m$ near the Earth's surface, where the gravitational force is approximately constant ($mg$) and vertically downwards, the work done in lifting the particle from height $h_1$ to $h_2$ is $W_{12} = mg(h_2 - h_1)$. We can define the gravitational potential energy $V(h) = mgh + C$, where $C$ is an arbitrary constant (the zero point). The work done is the change in potential energy: $W_{12} = V(h_2) - V(h_1)$.

For distances far from the Earth's surface where the gravitational force varies as $1/r^2$, the work done in moving a particle of mass $m$ from a distance $r_1$ to $r_2$ from the Earth's center ($r_2 > r_1$) is $W_{12} = \int_{r_1}^{r_2} \mathbf{F} \cdot d\mathbf{r}$. The force is $\mathbf{F} = -G \frac{M_E m}{r^2} \hat{\mathbf{r}}$ (directed inwards, $\hat{\mathbf{r}}$ is unit vector outwards). If lifting outwards, $d\mathbf{r} = dr \hat{\mathbf{r}}$.

We can associate a gravitational potential energy $V(r)$ at a distance $r$ from the Earth's center such that $W_{12} = V(r_1) - V(r_2)$. Comparing this with the result above, we can define:

It is conventional to choose the zero of potential energy at infinity, i.e., $V(\infty) = 0$. This implies $C=0$.

This formula is valid for $r \ge R_E$. The gravitational potential energy of a particle is negative because it is in a bound system (attracted to the Earth). To move it to infinity (where $V=0$) requires positive work.

For a system of multiple particles, the total gravitational potential energy is the sum of the potential energies for every pair of particles, calculated using the two-body formula: $V_{total} = \sum_{i

Gravitational potential is defined as the gravitational potential energy per unit mass. For a body of mass $M$, the gravitational potential at a distance $r$ is $\phi(r) = V(r)/m = -GM/r$ (with $\phi(\infty)=0$).

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Escape Speed

The escape speed ($v_e$) is the minimum initial speed required for an object to completely escape the gravitational influence of a celestial body, meaning it can reach an infinite distance away with zero kinetic energy remaining.

The principle of conservation of mechanical energy can be used to determine the escape speed. Consider an object of mass $m$ projected from the surface of a celestial body (like Earth) of mass $M$ and radius $R$.

Initial state (at the surface, distance $r_i = R$ from center, speed $v_i = v_e$):

• Initial kinetic energy $K_i = \frac{1}{2}mv_e^2$.
• Initial potential energy $V_i = -\frac{GMm}{R}$ (taking $V=0$ at infinity).
• Initial total mechanical energy $E_i = K_i + V_i = \frac{1}{2}mv_e^2 - \frac{GMm}{R}$.

Final state (at infinite distance, $r_f \to \infty$, with minimum speed to escape, so final speed $v_f = 0$):

• Final kinetic energy $K_f = \frac{1}{2}m(0)^2 = 0$.
• Final potential energy $V_f = -\frac{GMm}{\infty} = 0$.
• Final total mechanical energy $E_f = K_f + V_f = 0 + 0 = 0$.

By conservation of mechanical energy, $E_i = E_f$ (assuming only gravity does work):

Assuming $m \ne 0$, we can solve for $v_e$:

This is the escape speed from the surface of a celestial body of mass $M$ and radius $R$.

For Earth, using $M=M_E$ and $R=R_E$. We also know $g = GM_E/R_E^2$, so $GM_E = gR_E^2$.

Using $g \approx 9.8$ m/s$^2$ and $R_E \approx 6.37 \times 10^6$ m, the escape speed from Earth's surface is approximately $11.2$ km/s.

The escape speed depends on the mass and radius of the celestial body and the location from where it is projected (via $R$), but it does not depend on the mass of the object being projected or the direction of projection (as long as it's outwards).

The escape speed from the Moon is much lower ($\approx 2.3$ km/s) due to its smaller mass and radius, which explains why the Moon has no significant atmosphere (gas molecules exceed escape speed easily).

Example 8.4. 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. 8.10. 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.

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Earth Satellites

Earth satellites are objects orbiting the Earth under the influence of Earth's gravity. This motion is analogous to planets orbiting the Sun, and Kepler's laws apply (with Earth at one focus).

Consider a satellite of mass $m$ in a circular orbit at a distance $r = R_E + h$ from the Earth's center, where $R_E$ is Earth's radius and $h$ is the height above the surface. The speed of the satellite is $v$.

The necessary centripetal force for the circular orbit is $F_c = \frac{mv^2}{r} = \frac{mv^2}{R_E + h}$, directed towards the Earth's center.

This centripetal force is provided by the gravitational force between the Earth (mass $M_E$) and the satellite: $F_g = G \frac{M_E m}{r^2} = G \frac{M_E m}{(R_E + h)^2}$.

Equating these forces:

Assuming $m \ne 0$ and $R_E+h \ne 0$, we can solve for the orbital speed $v$:

The orbital speed decreases as the orbital radius (or height) increases.

The time period ($T$) of the satellite's circular orbit is the time taken to complete one revolution. The distance covered in one revolution is the circumference $2\pi r = 2\pi(R_E+h)$.

Squaring both sides: $T^2 = \frac{4\pi^2}{G M_E} (R_E + h)^3$. This is in the form $T^2 = k (R_E+h)^3$, where $k = \frac{4\pi^2}{G M_E}$ is a constant for all satellites orbiting the Earth. This confirms Kepler's third law for circular orbits around Earth.

For satellites very close to the Earth's surface ($h \ll R_E$), the orbital speed $v \approx \sqrt{\frac{G M_E}{R_E}} = \sqrt{gR_E}$ (since $g = GM_E/R_E^2$). Using $g \approx 9.8$ m/s$^2$ and $R_E \approx 6.4 \times 10^6$ m, $v \approx \sqrt{9.8 \times 6.4 \times 10^6} \approx \sqrt{62.72 \times 10^6} \approx 7.9$ km/s.

The time period for a low Earth orbit satellite is approximately $T \approx 2\pi \sqrt{R_E/g} \approx 2\pi \sqrt{(6.4 \times 10^6)/9.8} \approx 2\pi \sqrt{0.653 \times 10^6} \approx 2\pi \times 808 \approx 5078$ seconds, or about 84.6 minutes.

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Energy Of An Orbiting Satellite

An orbiting satellite possesses both kinetic energy due to its motion and gravitational potential energy due to its position in Earth's gravitational field.

Consider a satellite of mass $m$ in a circular orbit of radius $r = R_E + h$ around Earth (mass $M_E$). Its speed is $v = \sqrt{\frac{G M_E}{r}}$.

• Kinetic Energy (K): $K = \frac{1}{2} m v^2 = \frac{1}{2} m \left(\frac{G M_E}{r}\right) = \frac{G M_E m}{2r}$. Kinetic energy is always positive.
• Potential Energy (V): Taking the potential energy at infinity as zero, the potential energy at a distance $r$ is $V = -\frac{G M_E m}{r}$. Potential energy is negative for bound orbits.

Total Mechanical Energy (E) is the sum of kinetic and potential energy:

where $r = R_E + h$ for a circular orbit.

The total mechanical energy of a satellite in a circular orbit is negative. Its magnitude is half the magnitude of the potential energy and equal to the kinetic energy ($|E| = K = |V|/2$).

For elliptical orbits, the total energy is also negative and constant, given by $E = -\frac{G M_E m}{2a}$, where $a$ is the semi-major axis. Objects with zero or positive total energy are unbound and will escape Earth's gravitational influence (see Escape Speed section).

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Geostationary And Polar Satellites

Satellites can be placed in various types of orbits around the Earth, depending on their purpose.

Geostationary Satellites

• These satellites orbit the Earth in a circular orbit located in the equatorial plane.
• Their orbital period ($T$) is exactly equal to the Earth's rotational period, which is approximately 24 hours.
• Because they orbit in the same direction and with the same period as the Earth rotates, they appear to remain fixed over a specific point on the Earth's equator when viewed from the ground.
• The required orbital radius for a 24-hour period is found using $T^2 = \frac{4\pi^2}{G M_E} r^3$, where $r = R_E + h$. For $T=24$ hours, $r$ is approximately $42,200$ km from the Earth's center, which corresponds to a height of about $35,800$ km above the Earth's surface.
• Geostationary satellites are essential for communication and broadcast (TV, radio signals) as they can receive signals from one ground station and retransmit them to a wide area on Earth without needing complex tracking antennas.

Polar Satellites

• These satellites orbit the Earth in a north-south direction, passing over the Earth's poles.
• They are typically placed in low altitude orbits, around 500 to 800 km above the Earth's surface.
• Their orbital period is much shorter, typically around 100 minutes.
• As the Earth rotates beneath the polar satellite, the satellite's orbit allows it to pass over different regions with each pass. Over a day, the entire Earth can be viewed strip by strip.
• Polar satellites are primarily used for applications requiring global coverage and high resolution, such as remote sensing, meteorological imaging, and environmental monitoring.

Weightlessness

Weight is the force of gravity acting on an object. We typically perceive our weight when a support surface (like the ground or a weighing scale) exerts an equal and opposite force on us to prevent us from falling.

The phenomenon of weightlessness occurs when there is no support force counteracting gravity, or more precisely, when the object and its reference frame are in a state of free fall.

• If you stand on a spring balance, it measures the normal force it exerts on you, which is usually equal to your weight when at rest or moving at constant velocity.
• If the weighing scale is in a lift that is accelerating downwards, the reading on the scale is less than your weight (apparent weight $W_{apparent} = m(g-a)$).
• If the lift cable breaks and it is in free fall (accelerating downwards with acceleration $a=g$), the scale would exert zero normal force ($W_{apparent} = m(g-g) = 0$). The reading on the scale would be zero. You would feel weightless.

In an orbiting satellite, both the satellite and everything inside it (astronauts, objects) are continuously accelerating towards the Earth's center with the acceleration due to gravity ($g$ at that altitude) required for the orbit. They are effectively in a state of continuous free fall around the Earth.

Because they are in free fall, there is no support force counteracting gravity within the satellite's frame of reference (unless deliberately provided). Objects inside the satellite float freely.

Therefore, astronauts in orbit experience a state of weightlessness. It is crucial to understand that this is not because gravity is absent or negligible at that altitude; gravity is precisely what keeps the satellite in orbit. Weightlessness is an apparent phenomenon resulting from being in free fall in an orbiting frame of reference.

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Question 8.10. In the following two exercises, choose the correct answer from among the given ones:

The gravitational intensity at the centre of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see Fig 8.12)

(i) a

(ii) b

(iii) c

(iv) 0

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