Gravitation

🧠 Concepts Map of the deep dives

This chapter breaks into the concepts below. Each has its own tab with interactive 3D, a full derivation, and a JEE worked example. The Contents tab has the complete prose.

📖 Contents Full chapter — read straight through

The complete chapter in reading order. Each section is narratable; the concept tabs add interactive 3D and worked examples on top of this text.

Kepler's Three Laws of Planetary Motion 🔉✕⇢deep concept

Long before Newton, a nobleman called Tycho Brahe spent his entire lifetime recording observations of the planets with the naked eye. His compiled data were analysed later by his assistant Johannes Kepler, who could extract from the data three elegant laws that now go by the name of Kepler's laws. These laws were known to Newton and enabled him to make a great scientific leap in proposing his universal law of gravitation. The historical order matters for a JEE student: Kepler's laws are not consequences derived from a force law that was already in hand. They are a compact summary of what the planets are actually observed to do. Newton's achievement was to run the logic backwards, reading the force law out of the three laws, and this chapter treats the laws in exactly that spirit. 🔉✕⇢

The first is the law of orbits: all planets move in elliptical orbits with the Sun situated at one of the foci of the ellipse. This law was a deviation from the earlier Copernican model which allowed only circular orbits. To picture the ellipse precisely, recall how it is drawn. Select two points $F_1$ and $F_2$; take a length of a string and fix its ends at $F_1$ and $F_2$ by pins; with the tip of a pencil stretch the string taut and then draw a curve by moving the pencil keeping the string taut throughout. The closed curve you get is called an ellipse. Clearly, for any point $T$ on the ellipse, the sum of the distances from $F_1$ and $F_2$ is a constant, exactly equal to the fixed length of the string. The two points $F_1$ and $F_2$ are called the focii. 🔉✕⇢

Join the points $F_1$ and $F_2$ and extend the line to intersect the ellipse at points $P$ and $A$. The midpoint of the line $PA$ is the centre of the ellipse $O$, and the length $PO = AO$ is called the semi-major axis of the ellipse, written $a$. The closest point $P$ is called the perihelion and the farthest point $A$ the aphelion, and the semi-major axis is half the distance $AP$. For a circle, the two focii merge into one and the semi-major axis becomes the radius of the circle. The circle is therefore a special case of the ellipse, obtained in the limit where the two foci coincide at the centre. This is why a circular orbit is not a contradiction of the first law; it is the degenerate member of the same family of closed curves. 🔉✕⇢

For quantitative work it helps to name the two extreme distances measured from the Sun, which sits at one focus $S$. At perihelion the planet is nearest the Sun, at a distance $r_P$; at aphelion it is farthest, at a distance $r_A$. Because the Sun is at a focus and the centre $O$ is at the midpoint of $PA$, these two distances straddle the semi-major axis symmetrically, so that $r_P + r_A = 2a$. The semi-major axis is thus the arithmetic mean of the perihelion and aphelion distances, $a = (r_P + r_A)/2$. For our Earth the orbit is so close to circular that the ratio of the semi-minor to semi-major axis is $b/a = 0.99986$; the orbits of all planets except Mercury and Mars are very close to being circular, which is exactly why the earlier circular models were serviceable for so long. 🔉✕⇢

The second is the law of areas: the line that joins any planet to the Sun sweeps equal areas in equal intervals of time. This law comes from the observations that planets appear to move slower when they are farther from the Sun than when they are nearer. In one small interval of time $\Delta t$ the planet moves through a small arc and the line joining it to the Sun sweeps out a small shaded area $\Delta A$. The content of the law is that this swept area, divided by the time taken, is the same everywhere on the orbit — near perihelion the planet covers a long arc in the interval, near aphelion only a short arc, but the two thin triangular slivers of area are equal. 🔉✕⇢

To see the law of areas as a theorem rather than a mere observation, consider the small area swept in time $\Delta t$. If the planet is at position $\mathbf{r}$ measured from the Sun and moves by $\Delta \mathbf{r}$ in that interval, the swept area is the area of the thin triangle with those two sides, $\Delta A = \tfrac{1}{2}\,|\mathbf{r} \times \Delta \mathbf{r}|$. Dividing by $\Delta t$ and passing to the limit gives the areal velocity, the rate at which area is swept: $\dfrac{dA}{dt} = \dfrac{1}{2}\,|\mathbf{r} \times \mathbf{v}|$, where $\mathbf{v}$ is the velocity of the planet. This is the quantity the second law asserts to be constant in time. 🔉✕⇢

Now introduce the angular momentum. For a planet of mass $m$ the linear momentum is $\mathbf{p} = m\mathbf{v}$, and the angular momentum about the Sun is $\mathbf{L} = \mathbf{r} \times \mathbf{p} = m\,(\mathbf{r} \times \mathbf{v})$. Comparing with the areal velocity above gives the clean relation $\dfrac{dA}{dt} = \dfrac{|\mathbf{L}|}{2m}$. The rate of sweeping area is the magnitude of the angular momentum divided by twice the mass. So the law of areas is completely equivalent to the statement that the angular momentum $\mathbf{L}$ of the planet stays constant in both magnitude and direction as it moves around the orbit. 🔉✕⇢

Why should the angular momentum be conserved? The torque on the planet about the Sun is $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$, and by the rotational form of Newton's second law $\boldsymbol{\tau} = d\mathbf{L}/dt$. The gravitational force on the planet points straight along the line joining it to the Sun — it is a central force. For a central force $\mathbf{F}$ is parallel (or antiparallel) to $\mathbf{r}$, so the cross product $\mathbf{r} \times \mathbf{F}$ vanishes: the torque about the Sun is zero. With zero torque, $d\mathbf{L}/dt = 0$ and the angular momentum is conserved. Hence the areal velocity is constant and equal areas are swept in equal times. 🔉✕⇢

The JEE-relevant subtlety, stressed in the points to ponder, is this: angular momentum conservation leads to Kepler's second law, but it is not special to the inverse square law of gravitation. It holds for any central force whatsoever. Whether the attraction fell off as $1/r^2$ or as some other function of $r$, so long as the force stays directed along the line to the centre the torque is zero and the law of areas survives. The second law therefore tests only the centrality of the force, not its detailed strength. It is the first and third laws that carry the specific fingerprint of the inverse square dependence. 🔉✕⇢

The constancy of $\mathbf{L}$ has an immediate consequence for the speed. Because $\mathbf{L} = m\,(\mathbf{r}\times\mathbf{v})$ has fixed magnitude, and at both perihelion and aphelion the velocity is perpendicular to the radius (the planet is momentarily neither approaching nor receding), the magnitude there is simply $L = m\, r_P v_P = m\, r_A v_A$. Therefore $r_P v_P = r_A v_A$, so the speed $v_P$ at perihelion and the speed $v_A$ at aphelion obey $\dfrac{v_P}{v_A} = \dfrac{r_A}{r_P}$. Since $r_P \lt r_A$, we get $v_P \gt v_A$: the planet moves fastest at perihelion, its closest approach, and slowest at aphelion, exactly matching the observation that planets appear to move slower when farther from the Sun. 🔉✕⇢

The third is the law of periods: the square of the time period $T$ of revolution of a planet is proportional to the cube of the semi-major axis of its orbit, $T^2 \propto a^3$. The heart of this law, and the reason it is so powerful, is that the constant of proportionality is the same for all the planets going around the same Sun. It does not depend on the mass of the planet, nor on how elliptical its particular orbit happens to be — only on the semi-major axis. Since the orbits of most planets are very close to circular, we can derive the law cleanly for the case of a circular orbit and then quote the fact that it holds for the ellipse with $a$ in place of the radius. 🔉✕⇢

Take a planet of mass $m$ in a circular orbit of radius $r$ about the Sun of mass $M$. The centripetal force required to hold it on the circle at speed $v$ is directed towards the centre and has magnitude $\dfrac{mv^2}{r}$. This centripetal force is provided entirely by the gravitational force, which for two point masses (or spheres whose masses act as if concentrated at their centres) is $\dfrac{GMm}{r^2}$. Equating the force that is needed to the force that is available, and cancelling the mass $m$ of the planet, gives $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$, so that $v^2 = \dfrac{GM}{r}$. The orbital speed is fixed purely by the mass of the central body and the radius; the planet's own mass has dropped out. 🔉✕⇢

In one complete orbit the planet traverses the circumference $2\pi r$ at speed $v$, so its time period is $T = \dfrac{2\pi r}{v}$. Substituting $v = \sqrt{GM/r}$ gives $T = \dfrac{2\pi r}{\sqrt{GM/r}} = 2\pi\,\dfrac{r^{3/2}}{\sqrt{GM}}$. Squaring both sides yields $T^2 = \dfrac{4\pi^2}{GM}\,r^3$. This is precisely Kepler's law of periods: $T^2$ is proportional to $r^3$, with the constant of proportionality $\dfrac{4\pi^2}{GM}$. Because this constant contains only $G$ and the mass $M$ of the Sun, it is the same for every planet in orbit about that Sun, exactly as the third law demands. Writing it as $T^2 = K_S\, r^3$, the constant $K_S = 4\pi^2/GM$ is the same for all planets in circular orbits. 🔉✕⇢

The same derivation applies word for word to earth satellites, which are objects that revolve around the earth; their motion is very similar to the motion of planets around the Sun, and hence Kepler's laws of planetary motion are equally applicable to them. For a satellite of mass $m$ in a circular orbit at distance $(R_E + h)$ from the centre of the earth of mass $M_E$, the centripetal force $\dfrac{mV^2}{(R_E+h)}$ is provided by the gravitational force $\dfrac{GM_E m}{(R_E+h)^2}$, giving $V^2 = \dfrac{GM_E}{(R_E+h)}$. The time period then works out to $T = \dfrac{2\pi(R_E+h)^{3/2}}{\sqrt{GM_E}}$, and squaring gives $T^2 = k\,(R_E+h)^3$ with $k = \dfrac{4\pi^2}{GM_E}$ — Kepler's law of periods as applied to the motion of satellites around the earth. 🔉✕⇢

For an elliptical orbit the same equation holds provided we replace the circular radius by the semi-major axis of the ellipse, with the earth (or the Sun) at one of the foci. This is the rigorous general statement: $T^2 = \dfrac{4\pi^2}{GM}\,a^3$. The Moon is the only natural satellite of the earth, with a near circular orbit and a time period of approximately $27.3$ days, which is also roughly equal to the rotational period of the Moon about its own axis. That the same constant $4\pi^2/GM_E$ governs both an artificial satellite skimming the surface and the distant Moon is a striking vindication of the universality built into the third law. 🔉✕⇢

The universality of the constant makes the third law a measuring instrument. Because $T^2 = \dfrac{4\pi^2}{GM}\,a^3$, once the period $T$ and semi-major axis $a$ of any one satellite are measured, the mass of the central body follows immediately as $M = \dfrac{4\pi^2 a^3}{G T^2}$. This is how the mass of a planet like Mars is found from the period and orbital radius of a moon such as Phobos, and how the mass of the Sun is inferred from the Earth's own orbit. The planet whose mass is being weighed never has to be visited; its gravitational grip on a satellite betrays it. 🔉✕⇢

The third law also lets us compare two bodies without ever invoking $G$ or the central mass explicitly. For two planets orbiting the same Sun, taking the ratio kills the common constant: $\dfrac{T_1^2}{T_2^2} = \dfrac{a_1^3}{a_2^3}$. This is the form used to find the length of the martian year from the ratio of the Mars-Sun distance to the Earth-Sun distance. With the martian orbital radius being $1.52$ times that of the earth, $T_M = (1.52)^{3/2}\times 365 \approx 684$ days. A single dimensionless ratio of distances fixes the ratio of periods, a testament to how much geometry alone dictates in this problem. 🔉✕⇢

It is worth being explicit about what cancels and what survives in these derivations, because JEE questions probe exactly this. In $v^2 = GM/r$ the orbiting body's mass $m$ cancelled; that is why all satellites at a given radius have the same speed and period regardless of their own mass, and why an astronaut and the space station orbit together. In $T^2 = 4\pi^2 r^3/GM$ the surviving mass is that of the central body $M$, not the orbiting one. So the third-law constant is a property of the attractor — the Sun for planets, the earth for its satellites — and this is the deep reason it is common to all bodies orbiting that one attractor. 🔉✕⇢

A structural point often examined: the second law fixes the shape of motion along a given orbit, the first law fixes the shape of the orbit itself, and the third law fixes how orbits of different sizes are timed relative to one another. The second law alone, following from central-force angular-momentum conservation, cannot distinguish an inverse-square attraction from any other central pull. But demanding in addition that the closed orbits be ellipses with the Sun at a focus, and that the periods scale as $T^2\propto a^3$, singles out the inverse-square force uniquely. The three laws together are far more restrictive than any one of them. 🔉✕⇢

This is the road Newton travelled in reverse. Suppose, as the third law asserts for the readily analysed circular case, that $T^2 = C\,r^3$ for some constant $C$ common to all planets. The speed on a circle is $v = 2\pi r/T$, so the centripetal acceleration is $a_c = \dfrac{v^2}{r} = \dfrac{4\pi^2 r}{T^2}$. Substituting $T^2 = C\,r^3$ gives $a_c = \dfrac{4\pi^2 r}{C\,r^3} = \dfrac{4\pi^2}{C}\,\dfrac{1}{r^2}$. The acceleration of a planet, and hence the force per unit mass driving it, falls off as $1/r^2$. Kepler's third law, read as an equation about acceleration, secretes the inverse-square law inside it. 🔉✕⇢

Multiplying that acceleration by the planet's mass $m$ gives the force $F = m\,a_c = \dfrac{4\pi^2 m}{C}\,\dfrac{1}{r^2}$, so the force on the planet is proportional to its own mass and to $1/r^2$. Newton's third law then demands an equal and opposite reaction on the Sun, which by the same reasoning must be proportional to the mass of the Sun. For the force to be symmetric between the two bodies it must therefore be proportional to the product of the two masses, giving $F = \dfrac{G M m}{r^2}$ with a single universal constant $G$. Thus the law of areas gave Newton the direction of the force (central), and the law of periods gave him its magnitude (inverse square, product of masses) — the complete universal law of gravitation, read out of Kepler's data. 🔉✕⇢

The universal constant $G$ that appears here is not something the astronomy can supply, since the planetary ratios always cancel it. Its value entering the universal law of gravitation must be determined experimentally, and this was first done by the English scientist Henry Cavendish in 1798. Two small lead spheres on a bar are suspended by a fine wire; two large lead spheres brought close by attract the small ones with equal and opposite force, producing not a net force but only a torque on the bar. The suspended wire twists until the restoring torque of the wire equals the gravitational torque, and observing the angle of twist enables one to calculate $G$. The currently accepted value is $G = 6.67\times10^{-11}\ \mathrm{N\,m^2/kg^2}$. 🔉✕⇢

It is precisely because Cavendish measured $G$ that Kepler's third law could be turned into a weighing scale for the earth. The acceleration due to gravity at the surface is $g = \dfrac{GM_E}{R_E^2}$, so with $g$ readily measurable, $R_E$ a known quantity, and $G$ from Cavendish's experiment, one estimates the mass of the earth as $M_E = \dfrac{g R_E^2}{G} \approx 5.97\times10^{24}$ kg. The very same mass comes out of applying the third law to the Moon as a satellite of the earth, $M_E = \dfrac{4\pi^2 R^3}{G T^2}$, and the two methods agree to within one per cent. This is the concrete meaning of the popular statement that Cavendish weighed the earth. 🔉✕⇢

A final caution for problem solving. In these derivations the Sun, the earth and the planets have been treated as point masses, or equivalently as spheres whose entire mass may be taken as concentrated at the centre. That step is legitimate only because the force of attraction between a spherically symmetric body of uniform density and a point mass situated outside it is just as if the entire mass were concentrated at the centre — a result proved separately in this chapter. For a spherically symmetric body the force on an external particle is therefore central, and Kepler's whole framework applies. The three laws, the areal velocity $dA/dt = L/2m$, the speed-up at perihelion, the constant $4\pi^2/GM$ common to all planets, and Newton's inverse-square reading of them, together form the definitive account a JEE student must carry of planetary and satellite motion. 🔉✕⇢

Gravitational Field 🔉✕⇢deep concept

Around any mass, at every point in the region surrounding it, a second point mass placed there experiences a gravitational force of attraction directed towards the first mass. It is convenient to describe this influence of a mass on the space around it not through the force on one particular particle, but through the force per unit mass. The gravitational field at a point is defined as the gravitational force experienced by a point mass of unit mass situated at that point. Thus if a point mass $m$ at some point feels a gravitational force $F$, the gravitational field there is $E = F/m$. 🔉✕⇢

Because the force on the particle is proportional to its mass, the ratio $F/m$ does not depend on the mass $m$ we happen to place at the point; it is a property of the point itself and of the masses that produce the force. The gravitational field is a vector: it has the same direction as the force, that is, directed towards the mass producing it. Its value tells us the gravitational force that would act on each kilogram of any point mass situated at that point. 🔉✕⇢

From the universal law of gravitation, the force of attraction between two point masses $m_1$ and $m_2$ separated by a distance $r$ has magnitude $F = G\,m_1 m_2 / r^2$, directed along the line joining them. Consider a single point mass $M$. The gravitational force on a point mass $m$ situated at a distance $r$ from $M$ is $F = GMm/r^2$. Dividing by $m$, the gravitational field of the point mass $M$ at distance $r$ has magnitude $E = GM/r^2$, directed towards $M$. The field falls off as the inverse square of the distance from the mass. 🔉✕⇢

This gravitational field is exactly the acceleration that a freely falling particle would have at that point. The acceleration experienced by a mass $m$, usually denoted by $g$, is related to the force by Newton's second law through $F = mg$, so that $g = F/m$. Comparing with the definition of the field, the gravitational field and the acceleration due to gravity are one and the same quantity, $E = g = GM/r^2$. For the earth, of mass $M_E$ and radius $R_E$, the field at the surface is $g = GM_E/R_E^2$. 🔉✕⇢

The direction of the gravitational field at any point is the direction of the force on a point mass placed there, and this force is always one of attraction, never of repulsion. For a single mass the field therefore points everywhere towards that mass, along the line joining the point to the mass. The magnitude of the field is the same at all points at the same distance $r$ from the mass, since the force of attraction depends only on the distance $r$ and not on the direction. 🔉✕⇢

When several point masses or particles are present together, each of them exerts its own gravitational force on a given point mass, and these forces will not all be in the same direction. From the principle of superposition and the law of vector addition, the resultant gravitational force on the point mass is the vector sum of the separate forces due to the individual masses. Dividing throughout by the mass of that point mass, the resultant gravitational field at a point is the vector sum of the fields produced separately by each of the several particles. 🔉✕⇢

In adding the fields we must add them vectorially, resolving each field into components and adding the components, not merely adding magnitudes. When the masses are arranged symmetrically about a point, the components of the several fields along one direction may cancel each other, leaving a smaller resultant or, in cases of complete symmetry, a resultant field of zero. For example, three equal masses placed at the vertices of a triangle produce fields at the centroid whose vector sum, by symmetry, vanishes; the resultant force on a mass placed there is zero. 🔉✕⇢

For the gravitational force between an extended object, like the earth, and a point mass, the simple formula $F = GMm/r^2$ is not directly applicable. Each point mass in the extended object will exert a force on the given point mass, and these forces will not all be in the same direction. We have to add up these forces vectorially for all the point masses in the extended object to obtain the total force, and hence the total field. This adding up is easily done using calculus. For two special cases a remarkably simple law results. 🔉✕⇢

The first special case is that of a hollow spherical shell of uniform density. The force of attraction between such a spherical shell and a point mass situated outside it is just as if the entire mass of the shell is concentrated at the centre of the shell. In terms of the field, a uniform spherical shell of mass $M$ produces, at any point outside it at a distance $r$ from the centre, a gravitational field of magnitude $E = GM/r^2$, directed towards the centre, exactly the field of a single point mass $M$ placed at the centre. 🔉✕⇢

Qualitatively this can be understood as follows. The gravitational forces caused by the various regions of the shell have components along the line joining the point mass to the centre as well as along a direction perpendicular to this line. When we sum over all the regions of the shell, the components perpendicular to this line cancel out, leaving only a resultant force along the line joining the point to the centre. The magnitude of this resultant force works out to be exactly that of the whole mass concentrated at the centre. 🔉✕⇢

The second special case concerns a point mass situated inside the hollow spherical shell. The force of attraction due to a spherical shell of uniform density on a point mass situated inside it is zero. Consequently the gravitational field at every point inside a uniform spherical shell is zero, no matter where inside the shell the point lies. A particle placed anywhere within the hollow shell feels no net gravitational force from the shell, and so the field there vanishes completely. 🔉✕⇢

Again this result can be understood qualitatively. The various regions of the spherical shell attract the point mass inside it in various directions. The nearer regions of the shell attract the point mass strongly but make up a smaller portion of the shell, while the farther regions attract it more weakly but are larger in extent. When the forces due to all the regions are added vectorially, these forces cancel each other completely, so that the resultant gravitational force, and hence the field, inside the shell is zero. 🔉✕⇢

It is worth noting that although the field inside the shell due to the shell itself is zero, the shell does not shield a particle inside it from the gravitational forces of other bodies outside. Unlike a metallic shell, which shields electrical forces, a spherical shell offers no gravitational shielding; masses outside continue to exert their gravitational force on a particle inside. Gravitational shielding is not possible, and the field inside is zero only when we account for the shell alone. 🔉✕⇢

A solid sphere of uniform density may be regarded as a collection of a large number of concentric spherical shells, with the smallest one at the centre and the largest one at its surface. The earth itself can be imagined to be such a sphere, made of a large number of concentric spherical shells. The field of the whole solid sphere at any point is then the superposition of the fields of all these concentric shells, and the two special results for a single shell let us find it easily both outside and inside the sphere. 🔉✕⇢

Consider first a point outside the earth. Such a point is obviously outside all the concentric shells that make up the earth. Hence each and every shell exerts a gravitational force at that outside point just as if its mass were concentrated at the common centre. The total mass of all the shells combined is just the mass of the earth. Therefore, at a point outside the earth, at a distance $r$ from the centre, the gravitational force, and hence the field, is just as if the entire mass $M_E$ of the earth were concentrated at its centre, giving $E = GM_E/r^2$. 🔉✕⇢

If the point mass is situated on the surface of the earth, then $r = R_E$, and the gravitational field there has magnitude $E = GM_E/R_E^2$. This is exactly the acceleration due to gravity $g$ at the surface, $g = GM_E/R_E^2$. Thus a spherically symmetric earth acts, for all points on or outside its surface, like a single point mass $M_E$ located at its centre, and its field is the familiar inverse square field of a point mass, directed towards the centre. 🔉✕⇢

For a point mass at a height $h$ above the surface of the earth, its distance from the centre is $(R_E + h)$, and since the point is outside the earth the field is $g(h) = GM_E/(R_E + h)^2$. This is clearly less than the value of $g$ on the surface. For small heights, when $h \ll R_E$, we may expand and, using the binomial expression, obtain $g(h) \approx g\left(1 - \dfrac{2h}{R_E}\right)$. Thus for small heights $h$ the field decreases from its surface value by a factor $(1 - 2h/R_E)$. 🔉✕⇢

Now consider a point inside the earth. Let a point mass $m$ be situated at a distance $r$ from the centre, so that the point $P$ lies at distance $r$. For all the shells of radius greater than $r$, the point $P$ lies inside them, and hence, by the result for a shell, they exert no gravitational force on the mass at $P$; they contribute nothing to the field there. Only the shells with radius less than or equal to $r$ ($\text{radius} \le r$) contribute, and these together make up a smaller sphere of radius $r$ on whose surface the point $P$ lies. 🔉✕⇢

This smaller sphere of radius $r$ therefore exerts a force on the mass $m$ at $P$ as if its mass $M_r$ were concentrated at the centre. Assuming the entire earth to be of uniform density $\rho$, the mass of the earth is $M_E = \dfrac{4\pi}{3} R_E^3 \rho$, while the mass of the smaller sphere of radius $r$ is $M_r = \dfrac{4\pi}{3}\rho\, r^3$. The field inside is then $E = \dfrac{GM_r}{r^2} = \dfrac{GM_E\, r}{R_E^3}$. So inside a uniform earth the gravitational field is directly proportional to the distance $r$ from the centre. 🔉✕⇢

The same reasoning gives the field at a depth $d$ below the surface. A point mass at depth $d$ is at a distance $(R_E - d)$ from the centre. The earth can be thought of as composed of a smaller sphere of radius $(R_E - d)$ and a spherical shell of thickness $d$. The force on the mass due to the outer shell of thickness $d$ is zero, by the result for a shell. The smaller sphere acts as if its entire mass were concentrated at the centre, and one finds $g(d) = g\left(1 - \dfrac{d}{R_E}\right)$. 🔉✕⇢

Putting these results together, the gravitational field of the earth has a simple behaviour with distance from the centre. Inside the earth, treated as a sphere of uniform density, the field grows in direct proportion to $r$, from zero at the centre to its maximum value $g = GM_E/R_E^2$ at the surface. Outside the earth, the field falls off as the inverse square of the distance, decreasing as $GM_E/r^2$ as we move farther away. The field is therefore largest at the surface, where the point is at the smallest distance while still outside all the mass. 🔉✕⇢

At the very centre of the earth the gravitational field is zero. This follows at once from the shell picture: a point at the centre lies inside every one of the concentric shells, and each shell exerts no force on a point mass inside it, so all the contributions cancel. Equivalently, in the expression $E = GM_E\, r / R_E^3$ for the field inside, putting $r = 0$ gives $E = 0$. A particle at the centre is pulled equally in all directions, and the resultant field vanishes. 🔉✕⇢

These field expressions answer familiar questions directly. A body that weighs a certain amount on the surface, where the field is $g$, would weigh less half way down to the centre. At $r = R_E/2$ the field inside is $E = GM_E (R_E/2)/R_E^3 = g/2$, exactly half the surface value, so the body weighs half as much there. Above the surface, at a height where the distance has doubled to $2R_E$, the inverse square dependence makes the field one quarter of its surface value. 🔉✕⇢

The principle of superposition also lets us locate points where the resultant field is zero between two masses. Consider two uniform spheres of masses $M$ and $4M$ whose centres are separated by a distance $6R$. Since each sphere acts as a point mass at its centre, the field of the first at a distance $r$ from its centre is $GM/r^2$ towards it, and that of the second is $4GM/(6R - r)^2$ towards it. At the neutral point these two opposing fields cancel, giving $(6R - r)^2 = 4r^2$, so $r = 2R$; there the resultant gravitational field is zero. 🔉✕⇢

An important general point underlies all of this. Although the gravitational force between two particles is central, that is, directed along the line joining them, the force between two extended bodies need not be along the line joining their centres. But for a spherically symmetric body the force on a particle external to the body is as if the entire mass is concentrated at the centre, and this force is therefore central. The gravitational field of such a body is radial everywhere, directed towards its centre, and this is what makes the earth's field so simple to treat. 🔉✕⇢

This same field of the earth, $E = GM_E/(R_E + h)^2$ at a distance $(R_E + h)$ from the centre, supplies the gravitational force $F = mE$ that provides the centripetal force keeping a satellite or the moon in its orbit around the earth. Whether we speak of the gravitational field, the force per unit mass, or the acceleration due to gravity $g$, we are describing the same quantity: $GM/r^2$ for a point mass or for any spherically symmetric mass, directed towards the centre, built up from the individual point masses of the body by superposition and simplified by the two shell results. 🔉✕⇢

Escape Speed 🔉✕⇢deep concept

If a stone is thrown by hand, we see it falls back to the earth. Using machines we can shoot an object with much greater speeds, and with greater and greater initial speed the object scales higher and higher heights before it falls back. A natural query then arises in our mind: can we throw an object with such a high initial speed that it does not fall back to the earth at all? In this section we fix the minimum speed with which an object must be projected from the surface of the earth so that it escapes the gravitational pull of the earth and reaches infinity. This special speed is called the escape speed. 🔉✕⇢

The principle of conservation of energy helps us to answer this question cleanly, without our having to follow the object along its path. The energy of an object is the sum of its potential and kinetic energy, and for an isolated object moving only under the gravitational force of the earth this total energy is conserved. So we may compare the energy of the object at the surface of the earth, where we project it, with the energy of the same object when it has reached infinity. Whatever the object does in between, these two energies must be equal. 🔉✕⇢

Suppose the object is thrown initially with a speed $V_i$ from a point at a distance $(h+R_E)$ from the centre of the earth, where $R_E$ is the radius of the earth and $h$ the height above its surface. Its potential energy there is the gravitational potential energy $-\,GmM_E/(h+R_E)$, where $m$ is the mass of the object and $M_E$ the mass of the earth. Its kinetic energy is $\tfrac{1}{2}mV_i^2$. Adding the constant $W_1$, the potential energy at infinity, the energy of the projectile initially is $E(h+R_E)=\tfrac{1}{2}mV_i^2-\dfrac{GmM_E}{(h+R_E)}+W_1$. 🔉✕⇢

Now suppose the object did reach infinity and that its speed there was $V_f$. As before, $W_1$ denotes the gravitational potential energy of the object at infinity. At infinity the distance from the centre of the earth is so large that the gravitational potential energy of the object due to the earth has fallen to $W_1$, and only the kinetic energy $\tfrac{1}{2}mV_f^2$ is left over. The total energy of the projectile at infinity is therefore $E(\infty)=W_1+\dfrac{mV_f^2}{2}$. One conventionally sets $W_1$ equal to zero, so that the potential energy at a point is just the work done in bringing the particle from infinity to that point. 🔉✕⇢

By the principle of energy conservation the energy at the surface point and the energy at infinity must be equal. Equating the two expressions and cancelling the common term $W_1$, we obtain $\dfrac{mV_i^2}{2}-\dfrac{GmM_E}{(h+R_E)}=\dfrac{mV_f^2}{2}$. This one equation contains all the physics of escape. It simply says that the kinetic energy we supply at the start, minus the gravitational potential energy the object must climb out of, ends up as the kinetic energy the object still carries at infinity. 🔉✕⇢

The right hand side $\tfrac{1}{2}mV_f^2$ is a positive quantity with a minimum value of zero, since a speed squared can never be negative. Hence the left hand side must also be a positive quantity; that is, an object can reach infinity only as long as $V_i$ is such that $\dfrac{mV_i^2}{2}-\dfrac{GmM_E}{(h+R_E)}\ge 0$. If the left hand side were negative, the object would not have enough energy to reach infinity: it would rise, slow down, stop at some finite distance, and then fall back to the earth, exactly as the thrown stone does. 🔉✕⇢

The minimum value of $V_i$ corresponds to the case when the left hand side of this inequality equals zero, that is, when the object arrives at infinity with just zero speed, $V_f=0$. Thus the minimum speed required for an object to reach infinity (i.e. to escape from the earth) is fixed by $\tfrac{1}{2}m(V_i)_{min}^2=\dfrac{GmM_E}{h+R_E}$. Notice that the mass $m$ of the object appears on both sides of this relation and therefore cancels out. This is our first important result about the escape speed. 🔉✕⇢

Solving for the minimum speed, we get $(V_i)_{min}=\sqrt{\dfrac{2GM_E}{h+R_E}}$. If the object is thrown from the surface of the earth, the height $h=0$, and we get the escape speed from the surface, $(V_i)_{min}=\sqrt{\dfrac{2GM_E}{R_E}}$. This is the minimum speed with which an object must be projected from the surface so that it escapes the gravitational pull of the earth and reaches infinity. Any smaller speed leaves the total energy negative and the object bound to the earth; this speed, or any greater speed, lets it escape. 🔉✕⇢

It is convenient to write this result using the acceleration due to gravity at the surface. Using the relation $g=GM_E/R_E^2$, we have $GM_E=gR_E^2$, so that $\sqrt{\dfrac{2GM_E}{R_E}}=\sqrt{\dfrac{2gR_E^2}{R_E}}=\sqrt{2gR_E}$. Hence the escape speed from the surface of the earth can be written in the two equivalent forms $(V_i)_{min}=\sqrt{\dfrac{2GM_E}{R_E}}=\sqrt{2gR_E}$. The second form is handy for numerical work because $g$ and the radius $R_E$ of the earth are both directly known quantities at the surface. 🔉✕⇢

Using the value of $g\approx 9.8\ \text{m/s}^2$ and the radius $R_E\approx 6.4\times 10^{6}\ \text{m}$ of the earth, we get numerically $(V_i)_{min}\approx\sqrt{2\times 9.8\times 6.4\times 10^{6}}\approx 1.12\times 10^{4}\ \text{m/s}$, that is about $11.2\ \text{km/s}$. This is called the escape speed, sometimes loosely called the escape velocity. So an object projected from the surface of the earth with a speed of about $11.2\ \text{km/s}$ or more will not fall back but will escape the gravitational pull of the earth and reach infinity. 🔉✕⇢

The first striking feature of this result is that the escape speed is independent of the mass $m$ of the object that is projected. We saw the mass $m$ cancel exactly when we equated the kinetic energy to the gravitational potential energy. A small stone and a massive projectile must both be given the same speed of about $11.2\ \text{km/s}$ at the surface of the earth in order to escape. What differs between them is the kinetic energy, and hence the work the machines must do, since a larger mass $m$ needs a larger $\tfrac{1}{2}mV_i^2$ even at the same speed. 🔉✕⇢

The second feature is that the escape speed does not contain the direction in which the object is projected. Whether the object is thrown straight up along the radius, at a slant, or nearly along the surface, the same minimum speed of about $11.2\ \text{km/s}$ suffices, because the conservation of energy argument used only the speed and the distance from the centre of the earth, never the direction. The escape speed is therefore the same for every direction of projection, and this is why we speak of a single escape speed for the earth rather than a set of values. 🔉✕⇢

It is worth being clear about what 'reach infinity' means physically. At the escape speed the object arrives at infinity with exactly zero speed, so both its kinetic energy and its gravitational potential energy tend to zero there, and its total energy is exactly zero. Its total energy at the surface was therefore also zero: the positive kinetic energy $\tfrac{1}{2}mV_i^2$ balanced the negative gravitational potential energy $-\,GmM_E/R_E$. This is the boundary case that separates an object that falls back from one that gets away. 🔉✕⇢

We can now read off the three cases directly from the sign of the total energy. If $V_i$ is less than the escape speed, the total energy is negative, the object is bound to the earth, and it rises to a greatest height and then falls back. If $V_i$ equals the escape speed, the total energy is zero and the object just reaches infinity with zero speed. If $V_i$ is greater than the escape speed, the total energy is positive and the object reaches infinity still carrying a finite speed $V_f$ given by $\tfrac{1}{2}mV_f^2=\tfrac{1}{2}mV_i^2-\dfrac{GmM_E}{R_E}$. 🔉✕⇢

The condition for escape, then, is simply that the total mechanical energy of the object, kinetic plus gravitational potential, must be greater than or equal to zero. Since the gravitational potential energy $-\,GmM_E/R_E$ at the surface is a fixed negative quantity for a given object, escape is a question of whether the kinetic energy supplied at the surface is large enough to make the sum non negative. This is the same statement as $V_i\ge (V_i)_{min}$, but phrased in terms of energy it makes the physics transparent. 🔉✕⇢

The general form $(V_i)_{min}=\sqrt{2GM_E/(h+R_E)}$ also tells us how the escape speed changes if the object is launched not from the surface but from a point at some height $h$ above it. The larger the distance $(h+R_E)$ from the centre of the earth, the smaller the escape speed, because the object then starts higher up in the gravitational potential and has less potential energy to climb out of. An object launched from a great height needs less speed to escape than one launched from the surface of the earth. 🔉✕⇢

For a quick check of the numerical value, use $(V_i)_{min}=\sqrt{2gR_E}$ with $g\approx 9.8\ \text{m/s}^2$ and $R_E\approx 6.4\times 10^{6}\ \text{m}$. The product $2gR_E\approx 2\times 9.8\times 6.4\times 10^{6}\approx 1.25\times 10^{8}\ \text{m}^2/\text{s}^2$, whose square root is about $1.12\times 10^{4}\ \text{m/s}$, i.e. $11.2\ \text{km/s}$. Because the escape speed depends only on $g$ and the radius $R_E$ through the combination $\sqrt{2gR_E}$, any body of the same $M$ and $R$ has the same escape speed, whatever the mass of the object thrown from it. 🔉✕⇢

Equation for the escape speed applies equally well to an object thrown from the surface of the moon, with $g$ replaced by the acceleration due to the moon's gravity on its surface and $R_E$ replaced by the radius of the moon. Both of these are smaller than their values on the earth, and the escape speed for the moon turns out to be about $2.3\ \text{km/s}$, about five times smaller than the escape speed of the earth. So a projectile leaves the moon far more easily than it leaves the earth, precisely because the moon's smaller mass and radius make its gravitational pull weaker at the surface. 🔉✕⇢

This small escape speed is the reason that the moon has no atmosphere. Gas molecules, if formed on the surface of the moon, are in ceaseless motion, and a good fraction of them have speeds larger than the moon's escape speed of about $2.3\ \text{km/s}$. Any gas molecule having a speed larger than this will escape the gravitational pull of the moon, exactly as our projectile did, and reach infinity, never to return to the surface. Over a long time practically all the gas molecules leak away, and the moon is left without an atmosphere. 🔉✕⇢

By contrast, the earth's escape speed of about $11.2\ \text{km/s}$ is large enough that the ordinary gas molecules of the air move far too slowly to reach it, so they stay bound to the earth and the earth retains its atmosphere. The comparison of the two escape speeds, $11.2\ \text{km/s}$ for the earth against $2.3\ \text{km/s}$ for the moon, thus explains at one stroke why the earth is wrapped in air while the moon is not. It is the same escape condition, applied to gas molecules instead of to a thrown stone. 🔉✕⇢

The escape idea also appears in the two sphere problem of the worked example, where a projectile of mass $m$ is projected from the surface of a uniform solid sphere of mass $M$ towards the centre of a second sphere of mass $4M$. There the projectile is acted upon by two mutually opposing gravitational forces, and there is a neutral point $N$ where the two forces cancel each other exactly. It is sufficient to project the particle with just the minimum speed that would enable it to reach $N$; thereafter the greater gravitational pull of the sphere of mass $4M$ would suffice to pull it in. 🔉✕⇢

To find that minimum speed we again use the conservation of energy, equating the mechanical energy at the surface of the sphere of mass $M$ to the mechanical energy at the neutral point $N$. The mechanical energy at the surface is the sum of the kinetic energy $\tfrac{1}{2}mv^2$ and the gravitational potential energy due to both spheres. Setting this equal to the energy at $N$, where the projectile arrives with zero speed, gives an equation from which the minimum speed $v$ follows. The method is identical to the escape calculation; only the potential energy now has two terms, one for each sphere. 🔉✕⇢

It is useful to compare the escape speed with the speed of a satellite in a circular orbit close to the surface of the earth. For such a satellite the gravitational force supplies the required centripetal force, giving an orbital speed $v_o=\sqrt{GM_E/R_E}=\sqrt{gR_E}$, close to about $7.9\ \text{km/s}$. Comparing this with the escape speed $\sqrt{2gR_E}$, we see at once that the escape speed is exactly $\sqrt{2}$ times the orbital speed, $(V_i)_{min}=\sqrt{2}\,v_o$. So a satellite already moving in a low orbit needs its speed increased only by the factor $\sqrt{2}$ to escape the earth entirely. 🔉✕⇢

In using these results one must remember the idealisations made. We treated the earth as a uniform sphere whose whole mass $M_E$ may be taken to act as if concentrated at its centre, so that at and above the surface the gravitational potential energy of the object is $-\,GmM_E/r$ with $r$ the distance from the centre. We also ignored the gravitational pull of other bodies and any slowing of the object as it climbs. Within these standard assumptions the escape speed from the surface of the earth is the clean result $\sqrt{2GM_E/R_E}=\sqrt{2gR_E}\approx 11.2\ \text{km/s}$. 🔉✕⇢

Summarising, the escape speed is the minimum speed with which an object must be projected from the surface of the earth so that it just escapes the gravitational pull of the earth and reaches infinity with zero speed. Obtained from energy conservation by setting the total energy at infinity to zero, it is $(V_i)_{min}=\sqrt{2GM_E/R_E}=\sqrt{2gR_E}\approx 11.2\ \text{km/s}$. It is independent of the mass $m$ of the object, since $m$ cancels, and independent of the direction of projection, since only the speed and the distance from the centre entered the argument. 🔉✕⇢

For problems at this level, three habits pay off. First, always start from the total energy, kinetic plus gravitational potential, and impose that it be greater than or equal to zero for escape; the algebra of $(V_i)_{min}$ then follows in one step. Second, remember the two equivalent forms $\sqrt{2GM/R}$ and $\sqrt{2gR}$ and use whichever data are given. Third, keep the physical picture in mind: an object thrown slower than the escape speed is bound and falls back to the earth, one thrown at the escape speed just reaches infinity, and gas molecules on the moon, moving faster than its small escape speed, leak away and leave the moon with no atmosphere. 🔉✕⇢

For the moon, the escape speed works out to be only about $2.3\,\text{km/s}$, far smaller than the escape speed from the surface of the earth. This is one reason why the moon has no atmosphere: gas molecules moving with speeds comparable to this escape speed are able to leave the moon, whereas on the earth the much larger escape speed of about $11.2\,\text{km/s}$ keeps the atmosphere bound to the planet. 🔉✕⇢

Energy of an Orbiting Satellite 🔉✕⇢deep concept

Earth satellites are objects which revolve around the earth, and the NCERT chapter is emphatic that their motion is very similar to the motion of planets around the Sun. For this reason Kepler's laws of planetary motion apply equally well to them, and in particular their orbits around the earth are circular or elliptic. The moon is the only natural satellite of the earth, moving in a near circular orbit with a time period of approximately $27.3$ days. Since 1957 many countries, including India, have launched artificial satellites for telecommunication, geophysics and meteorology. Before we can speak of the energy of such an orbiting satellite, we must first understand why it stays in orbit at all, and that reason lies entirely in the gravitational force of the earth. 🔉✕⇢

Consider a satellite of mass $m$ moving in a circular orbit of radius $(R_E + h)$ measured from the centre of the earth, where $R_E$ is the radius of the earth and $h$ is the height of the orbit above the surface. If $V$ is its speed, then to keep the satellite moving along a circle rather than along a straight line, a centripetal force directed towards the centre of the earth is required. Its magnitude is $F_{\text{centripetal}} = \dfrac{mV^2}{(R_E + h)}$. This is not a new force of nature; it is simply the name we give to whatever real force happens to point towards the centre and bend the path of the object into a circle. 🔉✕⇢

The crucial physical statement of this section is that this centripetal force is provided by the gravitational force of the earth. The earth, treated as a spherically symmetric body, attracts the satellite exactly as if its entire mass $M_E$ were concentrated at its centre, so the force of attraction on the satellite is $F_{\text{gravitation}} = \dfrac{G m M_E}{(R_E + h)^2}$. The satellite is thus perpetually falling towards the centre of the earth, yet its speed carries it sideways by just the right amount that it never gets closer to the surface. Gravity is the string; the orbit is the circle it traces. 🔉✕⇢

Equating the required centripetal force to the available gravitational force, the mass $m$ of the satellite cancels from both sides, and we obtain $V^2 = \dfrac{G M_E}{(R_E + h)}$. This is one of the most important results of the whole discussion: the orbital speed of a satellite does not depend on its own mass, only on the mass of the earth and on the radius of the orbit. A heavy communication satellite and a light one placed in the same orbit move with exactly the same speed and complete a revolution in exactly the same period. This mass-independence is the same feature that makes all objects fall with the same acceleration near the surface. 🔉✕⇢

A direct reading of $V^2 = \dfrac{G M_E}{(R_E + h)}$ is that the orbital speed $V$ decreases as the height $h$ increases. A satellite in a higher orbit, farther from the centre of the earth, moves more slowly than one in a lower orbit. This already foreshadows Kepler's law of areas, which tells us planets appear to move slower when they are farther from the Sun than when they are nearer. For the special case of a satellite skimming just above the surface, we set $h = 0$, giving $V^2(h=0) = \dfrac{G M_E}{R_E} = g R_E$, where we have used $g = \dfrac{G M_E}{R_E^{2}}$. This links orbital speed to the familiar acceleration due to gravity at the surface. 🔉✕⇢

We now build the energy of the orbiting satellite piece by piece, exactly as the NCERT text does. Using $V^2 = \dfrac{G M_E}{(R_E + h)}$, the kinetic energy of the satellite in a circular orbit with speed $V$ is $\text{K.E.} = \tfrac{1}{2} m V^2 = \dfrac{G m M_E}{2(R_E + h)}$. Notice that this quantity is positive, as any kinetic energy must be, and that it too is independent of the mass of the earth's rotation or the satellite's history — it is fixed entirely by where the orbit is. The higher the orbit, the larger $(R_E + h)$, and therefore the smaller the kinetic energy, consistent with the slower speed at greater distance. 🔉✕⇢

The potential energy is obtained by choosing, as the chapter does conventionally, the gravitational potential energy at infinity to be zero. The potential energy of a particle of mass $m$ at a point is then just the amount of work done in displacing the particle from infinity to that point. Since the gravitational force is attractive, this work is negative, and the potential energy at a distance $(R_E + h)$ from the centre of the earth is $\text{P.E.} = -\dfrac{G m M_E}{(R_E + h)}$. This is the same expression as the gravitational potential energy associated with two particles of masses separated by a distance $r$, namely $W(r) = -\dfrac{G m_1 m_2}{r}$, applied to the earth and the satellite. 🔉✕⇢

The total energy of the orbiting satellite is the sum of its kinetic and potential energies: $E = \text{K.E.} + \text{P.E.} = \dfrac{G m M_E}{2(R_E + h)} - \dfrac{G m M_E}{(R_E + h)}$. Combining these two terms gives the compact and central result $E = -\dfrac{G m M_E}{2(R_E + h)}$. The total energy of a circularly orbiting satellite is thus negative. The chapter phrases the internal structure of this result very precisely: the kinetic energy is positive whereas the potential energy is negative, but in magnitude the kinetic energy is exactly half the potential energy, so that the two combine into a negative total whose magnitude equals the kinetic energy. 🔉✕⇢

Why should the total energy be negative, and why does the sign matter so much? The negative sign is the mathematical signature of a bound system. The satellite is trapped by the gravitational force of the earth; it cannot wander off to infinity of its own accord. If the total energy were positive or zero, the object would escape to infinity, exactly as in the discussion of escape speed where an object can reach infinity only when its energy is not negative. Satellites are always at a finite distance from the earth, and hence their energies cannot be positive or zero. The negative total energy is therefore not an accident of our zero-reference choice but a direct statement that the satellite is captured. 🔉✕⇢

It is worth pausing on the relations among the three energies, because JEE problems lean on them constantly. For a circular orbit we have $\text{K.E.} = +\dfrac{G m M_E}{2(R_E + h)}$, $\text{P.E.} = -\dfrac{G m M_E}{(R_E + h)}$, and $E = -\dfrac{G m M_E}{2(R_E + h)}$. From these, $\text{P.E.} = 2E$ and $\text{K.E.} = -E = |E|$, while $\text{P.E.} = -2\,\text{K.E.}$. So if a problem hands you any one of these three numbers, the other two follow instantly. In particular the total energy equals the negative of the kinetic energy, and the kinetic energy equals half the magnitude of the potential energy. Committing this triangle of relations to memory removes most of the arithmetic from satellite-energy questions. 🔉✕⇢

The dependence of total energy on orbit radius is the next physically rich point. Since $E = -\dfrac{G m M_E}{2(R_E + h)}$, a higher orbit — larger $(R_E + h)$ — has a total energy that is smaller in magnitude, that is, less negative and therefore closer to zero. A student sometimes finds this confusing: a satellite lifted to a higher orbit has a greater (less negative) total energy, yet it moves more slowly. Both statements are correct simultaneously. The kinetic energy has dropped because the speed dropped, but the potential energy has risen even more steeply towards zero, and the net effect is a total energy that has increased towards zero. Raising a satellite therefore costs energy even though the satellite ends up moving slower. 🔉✕⇢

Turn now to the period of the orbiting satellite, which the chapter derives directly and then identifies with Kepler's third law. In every orbit the satellite traverses a distance $2\pi(R_E + h)$ with speed $V$, so its time period is $T = \dfrac{2\pi(R_E + h)}{V}$. Substituting the orbital speed $V = \sqrt{\dfrac{G M_E}{(R_E + h)}}$ gives $T = \dfrac{2\pi(R_E + h)^{3/2}}{\sqrt{G M_E}}$. This single expression already contains the whole content of Kepler's law of periods for satellites, because it says the period grows as the three-halves power of the orbital radius — larger orbits take disproportionately longer to complete. 🔉✕⇢

Squaring both sides makes the law explicit: $T^2 = k (R_E + h)^3$, where the constant is $k = \dfrac{4\pi^2}{G M_E}$. This is Kepler's law of periods, as applied to the motion of satellites around the earth, and it is the exact analogue of the planetary statement $T^2 = K_S R^3$ in which the constant $K_S$ is the same for all planets in circular orbits. The chapter's Points to Ponder stresses that this same relation applies to satellites orbiting the Earth. Note also that the relation holds for elliptical orbits if we replace $(R_E + h)$ by the semi-major axis of the ellipse, with the earth situated at one of the foci of that ellipse. 🔉✕⇢

For a satellite very close to the surface of the earth, $h$ can be neglected in comparison to $R_E$, and the period reduces to $T_0 = 2\pi\sqrt{\dfrac{R_E}{g}}$, where we used $g = \dfrac{G M_E}{R_E^{2}}$. Putting $g \simeq 9.8\ \text{m s}^{-2}$ and $R_E = 6400\ \text{km}$, the numerical value comes out to $T_0 = 2\pi\sqrt{\dfrac{6.4\times10^{6}}{9.8}}$, which is approximately $85$ minutes. This is the shortest possible period of any earth satellite; nothing can orbit the earth faster than a body just grazing its surface, because any real satellite must orbit above the surface where the speed is smaller and the period longer. 🔉✕⇢

A particularly important application is the geostationary, or geosynchronous, satellite. Such a satellite is placed so that its period of revolution around the earth is made equal to the rotational period of the earth about its own axis, namely one day. If in addition its orbit lies in the equatorial plane and it revolves in the same sense as the earth's rotation, then it appears to hang fixed over one point on the surface. This is exactly why such satellites are prized for telecommunication: a ground antenna can be aimed at a fixed direction in the sky. The moon itself displays a related synchrony, since its period of revolution of about $27.3$ days is roughly equal to the rotational period of the moon about its own axis. 🔉✕⇢

The radius of the geostationary orbit follows straight from Kepler's law of periods. Setting $T^2 = \dfrac{4\pi^2}{G M_E}(R_E + h)^3$ equal to the square of one day (about $24$ hours, or $86400$ seconds) and solving for $(R_E + h)$ gives an orbital radius of roughly $4.2\times10^4$ km from the centre of the earth, so the height $h$ above the surface is about $3.6\times10^4$ km — far higher than a near-surface satellite. Because the orbit is so large, the orbital speed $V = \sqrt{G M_E/(R_E + h)}$ is correspondingly small, and the total energy $E = -\dfrac{G m M_E}{2(R_E+h)}$ is very small in magnitude, that is, only weakly bound compared with a low satellite. 🔉✕⇢

The negative total energy leads naturally to the idea of binding energy. The binding energy of a satellite is the energy that must be supplied to it to just free it from the gravitational influence of the earth, that is, to raise its total energy from its negative orbital value up to zero, corresponding to being at rest at infinity. Since the total energy in orbit is $E = -\dfrac{G m M_E}{2(R_E + h)}$, the binding energy is $\left| E \right| = +\dfrac{G m M_E}{2(R_E + h)}$, which is numerically equal to the kinetic energy of the satellite in its orbit. To remove a satellite from orbit, one must supply an amount of energy equal to its kinetic energy. 🔉✕⇢

The chapter's exercise on a satellite at a height of $400$ km above the surface makes this concrete. There we must find how much energy must be expended to rocket the satellite out of the earth's gravitational influence. The satellite's total energy in orbit is $E_i = -\dfrac{G M_E m}{2(R_E + h)}$, and to send it out of the earth's gravitational influence we must raise its energy to $E_f = 0$. The energy that must be expended is therefore $E_f - E_i = +\dfrac{G M_E m}{2(R_E + h)}$. For the given data this evaluates to a definite positive number of joules; the key conceptual point is that the required energy is precisely the binding energy, the magnitude of the negative total energy the satellite already possesses. 🔉✕⇢

So far we have treated the circular orbit, but the chapter is careful to extend the energy picture to the elliptic case. When the orbit of a satellite becomes elliptic, both the kinetic energy and the potential energy vary from point to point along the orbit. As the satellite moves nearer to the earth its speed and kinetic energy increase while its potential energy becomes more negative; as it moves farther away the reverse happens. Yet the total energy, which remains constant throughout the motion, is again negative, just as in the circular orbit case. For an ellipse of semi-major axis $a$ the total energy takes the form $E = -\dfrac{G M m}{2a}$, and the kinetic energy relation $K = \dfrac{G M m}{2a}$ holds when suitably interpreted for the whole orbit. 🔉✕⇢

This constancy of total energy is a direct expression of the conservation laws highlighted in the Points to Ponder. In considering the motion of an object under the gravitational influence of another object, two quantities are conserved: the angular momentum and the total mechanical energy, whereas linear momentum is not conserved. Angular momentum conservation leads to Kepler's second law, the law of areas, and it is this law that governs how the varying speed of a satellite on an elliptic orbit distributes itself — faster near the earth, slower far away — even as the total energy stays fixed and negative. The energy accounting and the area law are two faces of the same underlying central-force motion. 🔉✕⇢

The energy of a satellite also clarifies the boundary between staying bound and escaping. Recall from the escape-speed discussion that an object thrown from the surface with speed $V_i$ can reach infinity only when $\dfrac{1}{2} m V_i^2 - \dfrac{G m M_E}{R_E} \ge 0$, and the minimum such speed is the escape speed $v_e = \sqrt{\dfrac{2 G M_E}{R_E}} = \sqrt{2 g R_E}$, numerically about $11.2$ km/s. Comparing with the orbital speed near the surface, $V = \sqrt{g R_E}$, we see the escape speed is exactly $\sqrt{2}$ times the near-surface orbital speed. A satellite is bound precisely because its total energy is negative; give it enough extra energy to reach zero and it escapes. 🔉✕⇢

It is illuminating to see escape as the limiting case of the binding-energy idea. The total energy of a low circular satellite is $E = -\dfrac{G m M_E}{2 R_E}$, whereas an object launched at exactly escape speed from the surface has total energy zero. The difference between them is the additional energy needed to unbind the orbiting satellite, again equal to its kinetic energy in orbit. This is why the same set of quantities — $G$, $M_E$, $R_E$ and the orbit radius — governs orbital speed, period, total energy, binding energy and escape speed alike. Every one of them is a rearrangement of the single gravitational relation $\dfrac{G m M_E}{r^2}$ acting as the centripetal force. 🔉✕⇢

The energy picture also explains the weightlessness of an astronaut in a satellite, a point the chapter is careful to correct a common misconception about. An astronaut experiences weightlessness in a space satellite not because the gravitational force is small at that location in space — at a few hundred kilometres the gravitational force is only slightly less than at the surface — but because both the astronaut and the satellite are in free fall towards the earth together. Both have the same orbital acceleration, supplied entirely by gravity acting as the centripetal force, so there is no contact force between the astronaut and the satellite floor, and the astronaut floats. The negative orbital energy and the free-fall weightlessness are two descriptions of the very same orbital motion. 🔉✕⇢

For the JEE candidate, the strategic content of this section can be reduced to a small, reliable toolkit. Start every satellite problem by writing the orbital condition $\dfrac{m V^2}{r} = \dfrac{G M_E m}{r^2}$ with $r = R_E + h$, which yields $V^2 = \dfrac{G M_E}{r}$. From there the kinetic energy is $\dfrac{G M_E m}{2r}$, the potential energy is $-\dfrac{G M_E m}{r}$, and the total energy is $-\dfrac{G M_E m}{2r}$. The period comes from $T = \dfrac{2\pi r}{V}$, i.e. $T^2 = \dfrac{4\pi^2}{G M_E} r^3$. With these five relations and the surface identity $g = \dfrac{G M_E}{R_E^{2}}$, almost any question on speed, period, energy, binding energy or transfer between orbits reduces to substitution. 🔉✕⇢

A frequent trap deserves explicit warning. Because the total energy is $-\dfrac{G M_E m}{2r}$ and the kinetic energy is $+\dfrac{G M_E m}{2r}$, students sometimes conclude that moving a satellite to a higher orbit lowers its energy since it slows down. The resolution is that total energy increases (becomes less negative) with $r$ even though kinetic energy decreases, because the potential energy climbs towards zero faster than the kinetic energy falls. When a satellite is boosted from a lower to a higher orbit, work must be done against gravity, its total energy rises, its kinetic energy and speed drop, and its period lengthens — all consistent, all encoded in the same two expressions. Keeping the signs straight is the single most important discipline here. 🔉✕⇢

In summary, the energy of an orbiting satellite is a small self-consistent world built from one force. The gravitational force of the earth supplies the centripetal force, fixing the orbital speed $V = \sqrt{G M_E / (R_E + h)}$ independent of the satellite's mass. This determines a positive kinetic energy $\dfrac{G m M_E}{2(R_E + h)}$, a negative potential energy $-\dfrac{G m M_E}{(R_E + h)}$ (with zero at infinity), and a negative total energy $-\dfrac{G m M_E}{2(R_E + h)}$ whose negativity marks the satellite as bound. The period obeys Kepler's law of periods $T^2 \propto (R_E + h)^3$, higher orbits mean slower speeds and smaller energy magnitudes, the binding energy equals the orbital kinetic energy, and escape corresponds to raising the total energy to zero. Every result is a rearrangement of the same inverse-square gravitation of the earth. 🔉✕⇢

✍️ Worked Examples Polya 5-move · JEE tier

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• Kepler's Three Laws of Planetary Motion — (JEE Main)
• Gravitational Field — (JEE Advanced)
• Escape Speed — (JEE Advanced)
• Energy of an Orbiting Satellite — (JEE Advanced)

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