Gravitation

1Newton's law of gravitation

Every particle attracts every other along the line joining them with F = Gm₁m₂/r², where G = 6.67 × 10⁻¹¹ N m² kg⁻². The force is always attractive, cannot be screened off, and acts equally on both bodies whatever their masses: the Earth pulls a falling apple exactly as hard as the apple pulls the Earth. Only the accelerations differ, and they differ by the mass ratio.

G is the smallest constant anybody measures, which is why gravitation is undetectable between laboratory objects and dominant between planets. Several masses acting at once add as a vector sum — each pull computed as though the others were not there.

Figure. The two nearest masses pull with equal force at right angles, so their resultant already lies along the diagonal; the far corner, at twice the distance squared, adds a third as much again in the same direction. The net pull is 1.91 Gm²/a² straight at the centre of the square, which is where symmetry says it must point.

Adding gravitational pulls

1. One pair at a timeCompute each Gm₁m₂/r² as if the other masses were absent. Superposition here is exact, not an approximation.
2. Draw the directionsEvery arrow points from the mass being pulled towards the mass doing the pulling. Attraction only, so there are no signs to guess.
3. Add as vectorsResolve along convenient axes, or exploit symmetry — equal masses placed symmetrically usually cancel in pairs.

2The field of a sphere

The gravitational field is force per unit mass, g = GM/r², directed inwards. What makes it usable is the shell theorem: a uniform spherical shell attracts an outside particle exactly as if all its mass sat at the centre, and exerts no force at all on a particle anywhere inside it. So outside the Earth g falls as 1/r² measured from the centre, and inside it only the mass in the sphere beneath you counts.

For a uniform sphere that enclosed mass is M(r/R)³, so g = GMr/R³ climbs linearly from zero at the centre to its maximum at the surface — which is the depth formula g_d = g(1 − d/R). Real Earth is not uniform; its core is far denser than its crust, so measured g actually rises a little for the first 2900 km down. Use g(1 − d/R) as the uniform-sphere result the exam asks for, and know why it is idealised.

Figure. Plotted for a sphere of uniform density: a straight rise from zero at the centre, a sharp corner at the surface, then a 1/r² tail. The peak is at r = R and nowhere else — g is smaller both above the surface and below it. At r = 2R the field is a quarter of surface g, and at r = 3R a ninth.

Why only the enclosed mass counts

1. Split the body into shellsAny spherically symmetric body is a stack of thin uniform shells, one inside the next.
2. Shells above you do nothingInside a shell the pulls from opposite sides cancel exactly: the far patch subtends the same cone, so its area grows as the square of its distance — and 1/r² cancels that square exactly.
3. Shells below you act from the centreEach of those pulls as a point mass at the centre, so only the mass within radius r survives — hence g ∝ r inside a uniform sphere.

3g above the surface

Above the surface there is nothing left to enclose, so g simply obeys the inverse square measured from the centre: g_h = g(R/(R + h))². Expanding that for h ≪ R gives the familiar g_h ≈ g(1 − 2h/R). The 2 comes from the square, and the expression is the first two terms of a series, not an identity — it has an expiry date.

Compare the two directions from the surface. Rise a height h and g falls by roughly 2h/R; descend the same distance and it falls by only d/R. Near the surface, going up costs exactly twice as much as going down.

Figure. The dashed line is the tangent to the true curve at h = 0, which is exactly what a first-order expansion is. The two agree while h is a percent or two of R and then separate fast: by h = R/2 the straight line has run into the axis while the real g is still 0.44 of its surface value. The approximation always under-reads, never over-reads.

4Rotation and the g you actually measure

A weighing scale on the ground sits in a rotating frame. At latitude λ the ground travels a circle of radius R cos λ, so part of the true gravitational pull is spent supplying that circular motion and the scale reads less: g_eff = g − Rω²cos²λ.

The two cosines come from different places, and that is the whole subtlety. One is the radius of the circle you are carried round; the other is the projection of the resulting centripetal direction onto your local vertical. Hence cos²λ, not cos λ — the correction is largest at the equator and exactly zero at the poles, where you are not going round anything.

Figure. Because it is cos² and not cos, the correction is almost flat across the tropics — Mumbai at 19° keeps 89% of the full equatorial reduction — and then falls away fastest at 45°, reaching exactly zero at the poles. Plotted for ω = 7.27 × 10⁻⁵ rad s⁻¹ and R = 6.4 × 10⁶ m, so the left-hand end is 0.034 m s⁻².

Where the two cosines come from

1. The radius of your circleAt latitude λ you are carried round a circle of radius R cos λ about the Earth's axis, not a circle of radius R.
2. What that circle demandsIt needs a centripetal acceleration ω²R cos λ directed at the axis — which is not the same direction as the pull towards the centre, except at the equator.
3. Project onto the verticalOnly the component along your local vertical shows up in the scale reading, and that costs a second cos λ. The leftover sideways part is real too: it tilts a plumb line by a few arcminutes at mid-latitudes.

5Potential and potential energy

Gravitational potential is the work done per unit mass in bringing a test mass in from infinity: V = −GM/r, with the zero deliberately placed at infinity rather than at the surface. It is negative everywhere because gravity does the work for you on the way in, so you must repay it to get back out. The energy of a pair of masses follows the same shape: U = −Gm₁m₂/r.

Field and potential are two views of the same thing, tied together by g = −dV/dr. Read that backwards and it says a region of constant V has zero field — which is exactly the inside of a spherical shell, where g = 0 but V holds its surface value −GM/R and does not drop to zero.

Figure. The whole curve lies below the dashed zero line and climbs towards it, never reaching it at any finite r. Its steepness is the field: the slope dV/dr at the surface is GM/R², which is g. A −1/r curve flattens far more slowly than the −1/r² field does, which is why potential still matters at distances where the field has become negligible.

Reading the minus signs

1. Zero is at infinityEvery value of V and U is measured against a body infinitely far away and at rest. That choice, not the physics, is what makes them negative.
2. Deeper means more negativeMoving in towards the mass lowers V; moving out raises it towards zero. 'Larger potential' means closer to zero, which is further out.
3. Work is a differenceThe work you must supply is ΔU = U_final − U_initial, and near the surface that reduces to mgh only while h ≪ R.

6Circular orbits: speed and period

A circular orbit is free fall that keeps missing. Gravity is the only force acting and it is exactly the centripetal force required, so GMm/r² = mv²/r, giving v_o = √(GM/r) and T = 2π√(r³/GM). Both depend on the orbital radius alone — the satellite's own mass cancels out, which is why a loose bolt and a space station in the same orbit drift along side by side.

Notice the direction of the dependence. A larger orbit is a slower orbit, v falling as 1/√r while T grows as √(r³). This is the opposite of the intuition that a higher orbit is a 'faster' one, and it is the single most reliable trap in the chapter.

Figure. Circular-orbit speed falls as 1/√r. The force is radial and the velocity tangential — perpendicular — so gravity turns the path without changing speed; the graph carries the r-dependence the undrawable circle would hide.

Setting up any orbit problem

1. One force, one jobGravity supplies the entire centripetal requirement: GMm/r² = mv²/r. There is no second force to balance and nothing 'flinging the satellite outwards'.
2. Measure r from the centrer = R + h, never h alone. This is the commonest arithmetic slip in the whole topic.
3. Trade GM for gR²Whenever the surface g of the planet is known, substituting GM = gR² removes both G and M and keeps the numbers small.

7Energy of a satellite

In a circular orbit K = ½mv_o² = GMm/2r while U = −GMm/r, so the total is E = −GMm/2r. Three statements fall out together and are best memorised as one line: E = −K, E = U/2, and E is negative — negative meaning bound. The binding energy, the extra energy needed to just barely escape from that orbit, is +GMm/2r.

Raising a satellite makes E less negative, which costs energy, and yet K falls while it happens. Fire the engine forwards and the satellite ends up in a higher and slower orbit. Total energy and speed move in opposite directions here, and examiners know it.

Figure. Three curves of the same 1/r shape. K sits above the dashed zero line, U twice as far below it, and E halfway between U and that line — so E is always the mirror image of K in it, which is the statement E = −K. All three climb towards zero as r grows: a distant orbit is a weakly bound one, and reaching E = 0 is escape.

8Escape velocity

Escaping means reaching infinity with nothing left over: ½mv_e² − GMm/R = 0, so v_e = √(2GM/R) = √(2gR). For the Earth that is 11.2 km s⁻¹, exactly √2 times the 7.9 km s⁻¹ needed to orbit at the surface.

Three things v_e does not depend on: the mass of the escaping body, the direction it is fired in (energy is a scalar — any direction that misses the ground will do), and the path it happens to take. It depends only on the mass and the radius of the body being left behind, which is why every planet and moon has its own fixed number.

Figure. At exactly v_e the speed decays as 1/√r and reaches zero only at infinity — the body slows forever and never turns round. Fired at 80% of v_e it stops dead at r = 2.8R, a height of just 1.8 R. Losing a fifth of the launch speed costs you the entire journey, which is the sense in which escape speed is a threshold rather than a target.

The energy argument in three lines

1. Write the total energy½mv² − GMm/r is conserved for a body moving freely under gravity alone.
2. Set the far end to zero'Just escapes' means v → 0 as r → ∞, and both terms vanish there — so the total energy is exactly zero throughout.
3. Solve at the launch point½mv_e² = GMm/R gives v_e = √(2GM/R), and m has cancelled before you ever put a number in.

9Kepler's laws

Kepler's three laws were the observational summary that Newton's inverse square law later explained. First: each planet moves on an ellipse with the Sun at one focus — at a focus, not at the centre. Second: the radius vector sweeps out equal areas in equal times, so a planet is fastest at perihelion and slowest at aphelion. Third: T² = 4π²a³/GM, with a the semi-major axis, which for a circle is just the radius.

The second law is angular momentum conservation wearing different clothes, and it gives the familiar v_p r_p = v_a r_a — but only at perihelion and aphelion, where the velocity happens to be perpendicular to the radius. Everywhere else the conserved quantity is vr sin θ and the shortcut does not apply.

Figure. Real data for Mercury, Venus, Earth and Mars, plotted as T² against a³ rather than T against a. The four points are collinear to three decimal places and the line passes through the origin — which is the whole content of the third law, and the reason plotting the raw T against a would have shown nothing but a curve. The slope is 1 in these units by construction.