Gravitation

Drop an apple and it falls. Look up at night and the Moon does not fall — yet both are obeying the very same law. That single insight, that the force pulling an apple to the ground and the force holding the Moon in its orbit are one and the same, is the achievement this chapter builds from the ground up. Newton did not invent gravity; he showed that Tycho Brahe's planetary data, distilled by Kepler into three elegant geometric laws, were themselves proof of a universal, inverse-square force acting between every pair of masses in the universe. This chapter walks that whole arc. It begins with Kepler's three laws of planetary motion, purely descriptive rules squeezed out of decades of naked-eye observation. It then shows how Newton reasoned from those laws to a single force law, F = Gm1m2/r squared, and how that law, tested with superposition and the geometry of spheres, explains why a whole planet can be treated as a point mass. From there the chapter turns practical: how G was measured in a laboratory by Cavendish, how g weakens as you rise above the surface or descend below it, how much energy it costs to separate two masses to infinity, how fast an object must be thrown to escape a planet forever, and how satellites — from the Moon to a communications satellite — trace out orbits governed by exactly the same equations. By the end you should move fluidly between Kepler's geometric language and Newton's algebraic one, derive g, escape speed, and orbital period from first principles rather than memorising them, and read a satellite's total energy as a single negative number that tells you, at a glance, whether it is bound or free. This is one of the most self-contained, derivation-rich chapters in the NCERT syllabus, and it rewards working through the algebra rather than skipping to the results.

Key takeaways

Kepler's three laws: elliptical orbits with the Sun at one focus, equal areas swept in equal times, and T squared proportional to a cubed for the semi-major axis a.
Newton's universal law of gravitation: F = Gm1m2/r squared, an attractive, central, inverse-square force between every pair of point masses, obeying superposition for many masses.
The shell theorem: a uniform shell attracts an external point mass as if concentrated at its centre, and exerts zero net force inside it.
G = 6.67 times ten to the minus 11 N m squared per kg squared, measured by Cavendish's torsion-balance experiment.
g = GM/R squared at a planet's surface; it falls with height as g(h) is approximately g(1 minus 2h/RE), and with depth as g(d) is approximately g(1 minus d/RE), reaching zero at the centre.
Gravitational potential energy W(r) = minus GMm/r, zero at infinite separation, so always negative for a bound pair; mgh is only its near-surface approximation.
Escape speed ve = the square root of 2GM/R = the square root of 2gR, about 11.2 km/s for Earth, independent of the object's mass or direction.
A circular satellite of radius (RE + h) has speed V = square root of GME/(RE+h), period T = 2 pi (RE+h) to the three-halves over square root of GME, and total energy E = minus GMEm/2(RE+h), always negative.
Weightlessness in orbit is free fall, not zero gravity — the satellite and everything in it accelerate together under the same pull.
Key idea: one inverse-square force, F = Gm1m2/r squared, generates every result in this chapter — Kepler's orbits, g and its variation, gravitational potential energy, escape speed, and every satellite's speed, period and negative total energy.

The core insight

A single attractive, central, inverse-square force between any two masses, F = Gm1m2/r squared, is enough to derive Kepler's planetary laws, the value and variation of g, gravitational potential energy, escape speed, and the motion and energy of every satellite.

What this chapter covers

This chapter moves in a line from observation to law to consequence. It opens with Kepler's three empirical laws of planetary motion — orbits, areas, and periods — squeezed purely from data, with no theory behind them. It then presents Newton's universal law of gravitation, the single force law that reproduces all three Kepler laws, together with the superposition principle and the shell theorem that lets a spherical planet be treated as a point mass. Next comes Cavendish's laboratory measurement of G, and how that lets you "weigh" the Earth. The chapter then examines g in detail — its surface value and how it falls off both above and below the surface. It closes with three ideas built on energy conservation: gravitational potential energy, escape speed, and the speed, period, and energy of an orbiting satellite. Every result traces back to the same inverse-square force introduced at the start.

The physical picture

Kepler's three laws

Working from Tycho Brahe's decades of naked-eye planetary observations, Johannes Kepler extracted three geometric regularities any planetary theory must reproduce. The law of orbits states that every planet moves in an elliptical orbit with the Sun at one focus, not the centre; the closest point is the perihelion and the farthest is the aphelion, and for a circle — a special case of the ellipse — the two foci merge and the semi-major axis becomes the radius. The law of areas states that the line joining a planet to the Sun sweeps out equal areas in equal times, which is why a planet moves fastest near perihelion and slowest near aphelion. The law of periods states that the square of a planet's orbital period T is proportional to the cube of the semi-major axis a of its orbit,

T^2 = K_S\, a^3, \qquad (1)

where: T is the orbital period, a is the semi-major axis of the elliptical orbit, and $K_S$ is a constant, the same for every planet orbiting the Sun. This single constant lets you compare any two planets' periods just from their orbit sizes, with no need to know either mass. The law of areas is really a disguised statement of angular momentum conservation, and — as the first derivation below shows — it is not special to gravity at all; it holds for any central force, one that always points along the line joining the two bodies.

Figure

An ellipse with the Sun at one focus and two shaded wedge shaped regions of equal area, one near the close point of the orbit and one near the far point, each wedge swept out in the same short time interval

Newton's law of gravitation and the superposition principle

Newton's leap was to propose one force law reproducing all three Kepler laws at once: every pair of point masses attracts every other with a force along the line joining them, proportional to each mass and inversely proportional to the square of the separation,

F = \frac{G\, m_1 m_2}{r^2}. \qquad (2)

where: F is the magnitude of the mutual attractive force, $m_1$ and $m_2$ are the two point masses, r is the distance between them, and G is the universal gravitational constant, the same everywhere regardless of the medium between the masses. The force on each mass points toward the other, so it is attractive and central. When more than two masses are present, the superposition principle applies exactly as for electric forces: the net force on any one mass is the vector sum of the individual pairwise forces due to every other mass — a fact NCERT demonstrates with three equal masses at the corners of a triangle and a fourth mass at the centroid, where symmetry alone shows the resultant is zero.

The shell theorem: gravity of spheres and shells

Real planets are not points, so applying equation (2) to the Earth and an object on its surface needs justification. NCERT states two results, provable with calculus, for a uniform hollow spherical shell. First, a shell attracts any point mass outside it exactly as if its entire mass were concentrated at its centre — force components perpendicular to the line joining the point to the centre cancel by symmetry, leaving only the radial component. Second, a shell exerts zero net force on any point mass placed inside it, since pulls from every region cancel completely. Building the Earth as nested shells, a point outside feels its whole mass as if concentrated at the centre, while a point at depth d feels only the smaller sphere of radius (RE − d) below it — the material above contributes nothing.

Measuring G: the Cavendish experiment

The constant G in equation (2) is extremely small and cannot be inferred from planetary data alone, since only the product GM shows up in orbital motion. Henry Cavendish measured G directly in 1798 using a torsion balance: a light bar with two small lead spheres at its ends, suspended by a fine wire, with two large lead spheres brought close to the small ones on opposite sides. The attraction between each large-small pair produces a torque that twists the wire until the wire's restoring torque balances it, and since restoring torque per unit twist angle is calibrated independently, the measured twist angle gives the force F, hence G, via F = GMm/d squared and FL = restoring constant times twist angle (d = sphere separation, L = bar length). The currently accepted value is

G = 6.67 \times 10^{-11}\ \text{N m}^2\text{kg}^{-2}. \qquad (3)

Once G is known, measuring g and RE at the Earth's surface lets you compute the Earth's mass — the popular claim that "Cavendish weighed the Earth."

g and how it varies with height and depth

At the Earth's surface, treating the whole planet as a point mass at its centre (by the shell theorem) gives the familiar expression for surface gravity,

g = \frac{G M_E}{R_E^2}. \qquad (4)

where: $M_E$ is the mass of the Earth and $R_E$ its radius. Above the surface, at height h, the object's distance from the centre becomes $(R_E + h)$ , so

g(h) = \frac{G M_E}{(R_E+h)^2} \approx g\left(1 - \frac{2h}{R_E}\right) \quad \text{for } h \ll R_E, \qquad (5)

using the binomial approximation. Below the surface, at depth d, only the smaller sphere of radius $(R_E - d)$ contributes (the shell theorem again), and since mass scales with the cube of radius for uniform density,

g(d) = g\left(1 - \frac{d}{R_E}\right). \qquad (6)

Together, equations (5) and (6) say g is maximum at the surface and falls off both ways: linearly to zero at the centre, and roughly as $1/(R_E+h)^2$ with height — for the same small distance, height costs g twice as fast as depth does, because of that extra factor of 2 in equation (5).

Figure

A graph of gravitational acceleration g plotted against distance from the centre of the Earth, rising in a straight line from zero at the centre to a maximum value at the surface, then curving downward above the surface following an inverse square fall off

Gravitational potential energy

Because gravity is attractive, separating two masses requires work done against the force, and that work is stored as gravitational potential energy. NCERT derives this by integrating the force from a reference point to a general separation r and choosing the potential energy to be zero at infinite separation (see the second derivation below for the full working), giving

W(r) = -\frac{G m_1 m_2}{r}, \qquad V = -\frac{G m}{r} \ \text{(potential of a mass } m\text{)}. \qquad (7)

where: W(r) is the potential energy of the two-particle system at separation r, and V is the gravitational potential of a single mass m, the potential energy per unit mass placed at that point. W(r) is always negative for finite r, growing less negative toward zero as r increases. For several particles the total potential energy is the sum over every distinct pair, the same superposition principle used for forces. NCERT's own example places four equal masses m at the corners of a square of side l: four edge pairs at distance l and two diagonal pairs at distance $\sqrt{2}\,l$ give total potential energy $W = -(4 + \sqrt{2})\,Gm^2/l \approx -5.41\,Gm^2/l$ , and potential at the square's centre $V = -4\sqrt{2}\,Gm/l$ .

Escape speed and orbiting satellites

If an object is thrown upward fast enough, does it ever come back? Energy conservation answers this directly: the object escapes to infinity only if its total mechanical energy at launch is zero or positive. The minimum launch speed for which this just barely happens is the escape speed,

v_e = \sqrt{\frac{2GM_E}{R_E}} = \sqrt{2gR_E} \approx 11.2\ \text{km/s for Earth}, \qquad (8)

derived in full below. A satellite, by contrast, never escapes — it is permanently bound, moving in a circular or elliptical orbit exactly as Kepler described for planets. For a satellite in a circular orbit of radius $(R_E+h)$ , gravity supplies exactly the centripetal force needed, $GM_Em/(R_E+h)^2 = mV^2/(R_E+h)$ , giving orbital speed and period

V = \sqrt{\frac{GM_E}{R_E+h}}, \qquad T = \frac{2\pi (R_E+h)^{3/2}}{\sqrt{GM_E}}. \qquad (9)

where: V is the constant orbital speed and T the orbital period. Squaring the period gives $T^2 = k(R_E+h)^3$ with $k = 4\pi^2/GM_E$ — Kepler's third law applied to Earth satellites rather than to planets around the Sun. For a satellite skimming just above the surface, h can be neglected next to $R_E$ , giving $T_0 = 2\pi\sqrt{R_E/g} \approx 85$ minutes, the shortest possible period for any Earth satellite. The kinetic, potential and total energy of a circular orbit follow from equation (9) and equation (7):

K = \frac{GM_Em}{2(R_E+h)}, \quad U = -\frac{GM_Em}{R_E+h}, \quad E = K+U = -\frac{GM_Em}{2(R_E+h)}. \qquad (10)

The total energy is negative, with the potential energy exactly twice the magnitude of the (positive) kinetic energy — a signature of every bound orbit, circular or elliptical, and the reason a satellite can never simply drift off to infinity unless extra energy is supplied.

Derivation ANewton's test of the inverse-square law using the Moon

The goal is to show, the way Newton himself reasoned, that the same force law governing a falling apple also governs the Moon's orbit, by comparing the Moon's centripetal acceleration, computed purely from its orbital radius and period with no assumption about the force law, against the value predicted by scaling down Earth's surface gravity g using the inverse-square law. If the two independently obtained numbers agree, the inverse-square hypothesis is confirmed, using only circular motion kinematics and the definition of g, with no calculus.

Step 1. Compute the Moon's centripetal acceleration from its orbit alone. For a body at constant speed V on a circle of radius $R_m$ with period T, $a_m = V^2/R_m = (2\pi R_m/T)^2/R_m = 4\pi^2 R_m/T^2$ . Using $R_m = 3.84\times10^8\ \text{m}$ and $T = 27.3$ days $= 2.36\times10^6\ \text{s}$ , this gives $a_m \approx 2.72\times10^{-3}\ \text{m/s}^2$ — a number that uses nothing about gravity, only the observed orbit.

Step 2. Predict the same acceleration from the inverse-square hypothesis. If gravity falls off as $1/r^2$ , then the acceleration at the Moon's distance should be smaller than surface gravity g by the square of the ratio of radii, since the Moon is much farther from Earth's centre than a point on the surface: $a_m(\text{predicted}) = g\,(R_E/R_m)^2$ .

Step 3. Substitute numbers. With $g = 9.8\ \text{m/s}^2$ , $R_E = 6.4\times10^6\ \text{m}$ and $R_m = 3.84\times10^8\ \text{m}$ , the ratio $R_E/R_m = 1.67\times10^{-2}$ , so $(R_E/R_m)^2 = 2.78\times10^{-4}$ , and $a_m(\text{predicted}) = 9.8 \times 2.78\times10^{-4} \approx 2.72\times10^{-3}\ \text{m/s}^2$ .

Step 4. Compare and conclude. Both routes — one kinematic, using only the Moon's orbit, the other dynamical, scaling down surface gravity — give the same number, $2.72\times10^{-3}\ \text{m/s}^2$ . This is Newton's own historic test: the force accelerating a falling apple and the force curving the Moon's path are the same inverse-square force, evaluated at two very different distances from Earth's centre.

Go deeper · Derivation B · 1 minGravitational potential energy from the work done against gravity

Derivation BGravitational potential energy from the work done against gravity

The goal is to obtain $W(r) = -GMm/r$ for the potential energy of two masses separated by distance r, by directly computing the work needed to move a mass m from a very large, effectively infinite, separation to a finite separation r against the attractive inverse-square force, fixing the arbitrary additive constant through the convention that potential energy vanishes at infinite separation. This is the calculus counterpart of the algebraic force law and explains why gravitational potential energy is always negative for bound systems.

Step 1. Set up the work integral for moving the mass from separation $r_1$ to separation $r_2$ (both measured from the centre of the fixed mass M). Since the gravitational force on m has magnitude $GMm/r^2$ directed toward M, the work associated with this displacement is $W_{12} = \int_{r_1}^{r_2} \dfrac{GMm}{r^2}\,dr$ .

Step 2. Evaluate the integral. Using $\int r^{-2}\,dr = -1/r$ , this becomes $W_{12} = GMm\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right)$ , a clean, finite expression that depends only on the two endpoints, not on the path taken between them — a hallmark of a conservative force.

Step 3. Send $r_1$ to infinity and define the potential energy there to be zero. As $r_1 \to \infty$ , the term $1/r_1 \to 0$ , so $W(\infty \to r_2) = -\dfrac{GMm}{r_2}$ . Calling this the potential energy $W(r_2)$ at separation $r_2$ , and dropping the now-redundant subscript, gives $W(r) = -GMm/r$ for any finite r. The constant of integration is fixed by convention, not derived — choosing it to vanish at infinity makes W(r) negative exactly when the two masses are bound.

Step 4. Read off the potential. Dividing by the test mass m gives the gravitational potential of the source mass M alone, $V(r) = -GM/r$ , a purely geometric property of M, independent of whatever second mass is placed there. For several source masses, superposition applies: total potential is the algebraic sum of the pairwise contributions, exactly as used for the four-particle square example above.

Go deeper · Derivation C · 1 minEscape speed from energy conservation

Derivation CEscape speed from energy conservation

The goal is to find the minimum launch speed from a planet's surface for which an object never falls back, using only conservation of total mechanical energy, kinetic plus gravitational potential, between the launch point and a hypothetical final point infinitely far away, with no calculus beyond the potential energy formula already derived above. The result, $v_e = \sqrt{2GM/R}$ , is one of the most quoted numbers in this chapter.

Step 1. Write the total energy at launch. An object of mass m launched with speed $v_i$ from the surface of a planet of mass M and radius R has kinetic energy $\tfrac{1}{2}mv_i^2$ and potential energy $-GMm/R$ (from equation 7 with $r=R$ ), so its total mechanical energy at launch is $E_i = \tfrac{1}{2}mv_i^2 - GMm/R$ .

Step 2. Write the total energy at infinity. There, potential energy is zero by the fixed convention, so if the object arrives with some final speed $v_f$ , its total energy is $E_f = \tfrac{1}{2}mv_f^2$ , never negative since kinetic energy cannot be negative.

Step 3. Apply conservation of energy and isolate the condition for barely escaping. Since $E_i = E_f$ for a body moving only under gravity, $\tfrac{1}{2}mv_i^2 - GMm/R = \tfrac{1}{2}mv_f^2 \geq 0$ . This is only possible if $\tfrac{1}{2}mv_i^2 \geq GMm/R$ , so the smallest launch speed for which escape is even possible is the one that makes this an equality, with $v_f \to 0$ at infinity: $\tfrac{1}{2}m v_e^2 = \dfrac{GMm}{R}$ .

Step 4. Solve for $v_e$ and evaluate for Earth. Rearranging gives $v_e = \sqrt{2GM/R}$ , and substituting $GM/R^2 = g$ rewrites it as $v_e = \sqrt{2gR}$ , needing only g and R. For Earth, $g = 9.8\ \text{m/s}^2$ and $R_E = 6.4\times10^6\ \text{m}$ give $v_e = \sqrt{2 \times 9.8 \times 6.4\times10^6} \approx 1.12\times10^4\ \text{m/s} = 11.2\ \text{km/s}$ . Because g and radius are both smaller on the Moon, its escape speed is only about 2.3 km/s — which is also why the Moon has essentially no atmosphere: gas molecules routinely reach that speed and leave.

Worked examples

Board level - weighing the Earth from g, RE and G

Given g = 9.81 m/s squared and RE = 6.37 times ten to the six metres, find the mass of the Earth ME using G = 6.67 times ten to the minus 11 N m squared per kg squared.

Step 1. Rearrange the surface-gravity relation, equation (4), to isolate the mass: $g = GM_E/R_E^2$ gives $M_E = gR_E^2/G$ .

Step 2. Substitute the given values. $R_E^2 = (6.37\times10^6)^2 = 4.058\times10^{13}\ \text{m}^2$ , so $gR_E^2 = 9.81 \times 4.058\times10^{13} = 3.981\times10^{14}$ .

Step 3. Divide by G to finish. $M_E = 3.981\times10^{14} / 6.67\times10^{-11} = 5.97\times10^{24}\ \text{kg}$ . Answer: $M_E \approx 5.97 \times 10^{24}$ kg. As a cross-check, NCERT also computes ME from the Moon's orbital data, $M_E = 4\pi^2R^3/GT^2$ , and finds $6.02\times10^{24}$ kg — the two methods agree to within one per cent.

Competitive level - mass and surface gravity of Mars from its moon Phobos (JEE and NEET pattern)

Phobos, a moon of Mars, has orbital period T = 7 hours 39 minutes and orbital radius R = 9.4 times ten cubed km. Find the mass of Mars, and hence its surface gravity, taking its radius as approximately 3.39 times ten to the six metres.

Step 1. Convert units and apply Kepler's third law, equation (9) squared: $T = 7\times3600 + 39\times60 = 27{,}540\ \text{s}$ , $R = 9.4\times10^6\ \text{m}$ , so $M_{\text{Mars}} = 4\pi^2R^3/(GT^2)$ .

Step 2. Compute numerator and denominator separately: $R^3 = (9.4\times10^6)^3 = 8.306\times10^{20}\ \text{m}^3$ , so $4\pi^2R^3 = 39.48 \times 8.306\times10^{20} = 3.279\times10^{22}$ ; and $GT^2 = 6.67\times10^{-11} \times (27{,}540)^2 = 5.06\times10^{-2}$ .

Step 3. Divide to get Mars's mass, then use it with the given radius via equation (4). $M_{\text{Mars}} = 3.279\times10^{22}/5.06\times10^{-2} = 6.48\times10^{23}\ \text{kg}$ , matching NCERT's value. Extending with the known Martian radius, $g_{\text{Mars}} = GM_{\text{Mars}}/R_{\text{Mars}}^2 \approx 3.76\ \text{m/s}^2$ . Answer: mass of Mars $\approx 6.48\times10^{23}$ kg, surface gravity $\approx 3.8$ m/s squared, about two-fifths of Earth's g. Timing note: about two and a half minutes. The trap is unit conversion of T to seconds before squaring — do this first, on scratch paper, since a slip here silently cubes or squares an error into the final answer.

Competitive level - neutral point and minimum launch speed between two spheres (JEE pattern)

Two uniform solid spheres of equal radius R but masses M and 4M have their centres 6R apart and are held fixed. A projectile of mass m is launched from the surface of the sphere of mass M, directly toward the centre of the sphere of mass 4M. Find the minimum launch speed v needed for the projectile to just reach the surface of the second sphere.

Step 1. Locate the neutral point N, where the two pulls exactly cancel. If ON = r (from the centre of M), the pulls balance when $GMm/r^2 = 4GMm/(6R-r)^2$ , giving $(6R-r)^2 = 4r^2$ , so $6R - r = \pm 2r$ . The physically relevant root is $r = 2R$ (the other, $r=-6R$ , lies outside the region between the spheres).

Step 2. It is enough to reach N with just zero speed — beyond N, the stronger pull of 4M takes over and pulls the projectile the rest of the way. Apply energy conservation between the launch point (surface of M, distance R from its centre and $5R$ from $4M$ 's centre) and N (distance $2R$ from M's centre and $4R$ from $4M$ 's centre), with speed at N taken as zero for the minimum case.

Step 3. Write the two energies and equate them. At launch, $E_i = \tfrac12 mv^2 - \dfrac{GMm}{R} - \dfrac{4GMm}{5R}$ . At N, $E_N = -\dfrac{GMm}{2R} - \dfrac{4GMm}{4R} = -\dfrac{GMm}{2R} - \dfrac{GMm}{R}$ . Setting $E_i = E_N$ and simplifying, the $-GMm/R$ term cancels from both sides, leaving $\tfrac12 v^2 = \dfrac{GM}{R}\left(\dfrac{4}{5} - \dfrac{1}{2}\right) = \dfrac{GM}{R}\cdot\dfrac{3}{10} = \dfrac{3GM}{5R}$ .

Step 4. Solve for v. $v^2 = \dfrac{6GM}{5R} \cdot \dfrac{1}{2}$ simplifies directly to $v = \sqrt{3GM/5R}$ . Answer: $v_{\min} = \sqrt{3GM/5R}$ , entirely in terms of the given masses and radius, with no numerical substitution needed since none was supplied. Timing note: about three minutes if the neutral-point equation is set up cleanly on the first attempt. The common slip is forgetting that beyond N, gravity from 4M assists the projectile, so reaching N with exactly zero speed is sufficient — chasing a nonzero arrival speed at the far surface wastes exam time.

Common traps

Treating orbital weightlessness as "no gravity." An astronaut several hundred kilometres up still feels nearly the same pull as at the surface; weightlessness there is free fall, not an absence of gravity.

Assuming linear momentum is conserved in a gravitational two-body problem. Angular momentum and total mechanical energy are conserved, not linear momentum; the individual momenta of the two bodies keep changing even though their sum is conserved.

Using mgh as the exact gravitational potential energy. mgh is only the near-surface, small-height approximation of $-GMm/r$ ; it silently assumes g is constant, which fails once h is a non-negligible fraction of RE.

Forgetting that potential energy is defined relative to a chosen zero. Only differences in potential energy are physically meaningful; NCERT's convention sets $W=0$ at infinite separation, which is why bound-state potential energies come out negative — a choice, not a law of nature.

Believing gravity can be shielded like an electric field. A hollow conductor blocks external electric fields, but no arrangement of mass can block gravity — a shell's own mass cancels to zero inside it purely by symmetry, and does nothing to stop an outside mass from pulling on a point inside the shell.

Mixing up the height and depth formulas for g. Equation (5), for height, carries a factor of 2; equation (6), for depth, does not — g reaches zero only at Earth's centre, never at any finite height.

Common trapWeightlessness in orbit does not mean gravity has switched off

The single biggest misconception in this chapter is treating an orbiting satellite, or an astronaut inside it, as being beyond the reach of Earth's gravity. In reality the gravitational force at typical low-orbit altitudes is only a few per cent weaker than at the surface — nowhere near zero. The satellite is in continuous free fall: gravity is precisely the centripetal force curving its path into a closed orbit rather than a straight line. Because the astronaut and the surrounding cabin fall together at the same rate, there is no contact force between them, which is what a bathroom scale would register as "weight." Remove gravity entirely and the satellite would fly off in a straight line, not stay in orbit — orbit is a story about gravity, not its absence.

You should be able to…

State Kepler's three laws and use the law of periods, T squared proportional to a cubed, to compare two orbiting bodies without needing either mass.
Write Newton's law of gravitation as F = Gm1m2/r squared and apply the superposition principle when more than two masses are involved.
Use the shell theorem to treat a spherically symmetric planet as a point mass, both outside it and, with the reduced mass, inside it.
Describe Cavendish's torsion-balance experiment and recall the numerical value of G.
Derive g = GM/R squared and its height and depth variation formulas.
Derive gravitational potential energy W(r) = minus GMm/r from the work integral, and explain why its zero at infinity is a convention.
Derive escape speed from energy conservation and compute it for Earth, the Moon, or any other body given g and R.
Derive the orbital speed, period and total energy of a circular satellite, explaining why the energy is always negative.
Explain, via free fall, why orbiting astronauts are weightless despite gravity still acting, and why gravitational shielding is impossible.

Ask the tutor: if you are unsure why a satellite's total energy must be negative, or want to walk through the Cavendish experiment's torque balance step by step, ask the tutor to work it through with you.

Why do astronauts orbiting Earth feel weightless when gravity is clearly still acting on them there?

Gravity at typical orbital altitudes is barely weaker than at the surface — an astronaut 400 km up still feels roughly 90 percent of surface gravity. What removes the sensation of weight is that the astronaut and spacecraft are both in continuous free fall together, following the same curved path. Since there is no relative acceleration between the astronaut's body and the surrounding cabin, there is no contact force pressing them against a floor or seat, which is exactly what a scale needs to register weight. It is the same feeling as the brief weightlessness at the top of a fast elevator drop, sustained indefinitely because the fall curves into a closed orbit instead of hitting the ground.

Why is gravitational potential energy always written as a negative number?

It is negative purely because of where the zero was chosen. NCERT's convention sets potential energy to zero when two masses are infinitely far apart, and since gravity is attractive, bringing masses closer together always releases energy — so any finite separation must have less energy than the zero at infinity, which means a negative number. This is not a statement about the force being "negative"; it reflects that a bound system, one needing extra energy to pull it apart to infinity, sits below the zero reference. If the zero were instead chosen at Earth's surface, as in the everyday formula mgh, potential energy would come out positive above the ground instead.

Why is the gravitational force zero inside a hollow shell but not zero inside a solid uniform sphere?

Inside a hollow shell, every bit of the surroundings is shell material, and pulls from all directions cancel exactly by the shell-theorem geometry — no mass is left over to pull it anywhere. Inside a solid sphere, imagine building it from nested shells: at depth d, every shell farther out cancels to zero as before, but shells closer to the centre have not yet been "passed," and their mass, concentrated as if at the centre, still pulls inward. That is why g decreases linearly with depth, reaching zero only at the very centre, not as soon as you step below the surface.

What is the actual difference between escape speed and a satellite's orbital speed?

Orbital speed is what keeps an object in a closed, bound path — fast enough that gravity provides exactly the centripetal force for a circle or ellipse, but not so fast it flies away. Escape speed is a larger target: the minimum speed for which total mechanical energy becomes zero or positive, so the object never returns even after travelling arbitrarily far. For a body skimming Earth's surface, orbital speed is about 7.9 km/s, while escape speed there is 11.2 km/s — almost exactly the square root of 2 times larger, following directly from comparing equations (8) and (9) at h=0.

Is Kepler's second law, about equal areas in equal times, special to gravity, or would it hold for any force?

It holds for any central force — one that always points along the line joining the two bodies, whatever its strength or distance dependence. The law of areas is really angular momentum conservation restated geometrically, and angular momentum is conserved whenever net torque about the centre is zero, which is automatic for a central force, since it has no component perpendicular to that line. Gravity happens to be central, which is why Kepler's second law applies to orbits, but the same reasoning applies equally to a charged particle orbiting a fixed charge under the also-central Coulomb force.