Gravitation

We start from the definition of potential energy as the work needed to bring a mass in from infinity, and reach the -GMm/r formula and the potential that follows from it.

Setting up. A mass M sits at the origin. We bring a small mass m from infinity (where we agree U = 0) slowly in to a distance r. "Slowly" means the body has no kinetic energy at either end, so all the work done shows up as stored energy.

Step 1 — the force at a general distance x. By Newton's law of gravitation the attraction on m when it is a distance x from M has magnitude

and it points inwards, towards M.

Step 2 — fix a sign convention. Measure x outwards from M, so "outwards" is positive. Gravity pulls inwards, i.e. in the direction of decreasing x, so as a signed quantity the gravitational force is

Step 3 — work done by gravity over a tiny displacement dx. Work is force times displacement:

Step 4 — add up all the tiny bits from x = \infty to x = r:

Step 5 — take the constants outside the integral. G, M and m do not change during the journey:

Step 6 — integrate x^{-2}. Using \int x^{-2}dx = -x^{-1}:

Step 7 — the two minus signs cancel:

Step 8 — put in the limits (value at the top limit minus value at the bottom limit):

Step 9 — use 1/\infty = 0:

This is the work done by gravity as the mass falls in from infinity, and it is positive — gravity helped, as you would expect.

Step 10 — convert that into potential energy. Whenever a conservative force does work W, the potential energy falls by the same amount:

Step 11 — put in U(\infty) = 0 (our chosen zero) and W = GMm/r:

so

Step 12 — get the potential. The gravitational potential V is defined as the potential energy per unit mass, so divide by m:

Result.

Both are scalars — there is no direction to worry about, which is what makes potential so much easier to work with than field.

What it means. The sign is not a mistake. We chose U = 0 at infinity, and gravity is always attractive, so every real point is a place you would have to do work to climb out of. Hence U is negative everywhere, deepest (most negative) close in and rising towards zero far away — a "potential well".

Conditions used. M is treated as a point mass (or a uniform sphere, for which this is exactly true outside it), the mass is moved slowly so no kinetic energy is involved, and the zero of potential is taken at infinity. U belongs to the pair of masses, not to either one alone.

We start from g = GM/R^{2} at the surface and reach a formula for g at a height h above it.

Step 1 — g at the surface. A mass m on the surface is a distance R from the centre of the Earth, so its weight is GMm/R^{2}. But weight is also mg, so:

Step 2 — g at a height h. At height h the body is a distance R+h from the centre. The only change is the distance, so

Step 3 — form the ratio g_h/g. Divide Step 2 by Step 1. The GM cancels:

Step 4 — write it as a single squared bracket:

Step 5 — divide the top and bottom inside the bracket by R. Dividing top and bottom by the same thing does not change a fraction:

Step 6 — write the reciprocal as a negative power:

This formula is exact. The next two steps only simplify it for small heights.

Step 7 — use the binomial expansion. For small x, (1+x)^{n} \approx 1 + nx. Here x = h/R and n = -2:

Step 8 — tidy up:

Result.

What it means. g gets smaller as you climb, and for small heights the fractional drop is 2h/R — twice the fraction of the Earth's radius you have climbed. At the top of Sagarmatha (h \approx 8.8 km) this is about 0.28\%, so g falls from 9.800 to about 9.77 m s⁻². g only becomes zero at infinity.

Conditions used. The Earth is treated as a uniform sphere so that it acts like a point mass at its centre; rotation is ignored. Step 7 needs h \ll R — for a satellite at h = R you must use the exact form, which gives g/4, not g(1-2) = -g.

We start from the shell theorem and reach a formula for g at a depth d below the surface — and show that g becomes zero at the centre.

Setting up. Treat the Earth as a uniform sphere of radius R and density \rho. A body at depth d is at a distance (R - d) from the centre.

Step 1 — the shell theorem. A uniform spherical shell exerts no net gravitational force on anything inside it: the pulls from all sides cancel exactly. So the outer shell, of thickness d, can be ignored, and only the inner sphere of radius (R-d) attracts the body.

Step 2 — the mass of that inner sphere. Mass = density \times volume, and the volume of a sphere is \frac{4}{3}\pi(\text{radius})^{3}:

Step 3 — the field at the body's position. The inner sphere behaves like a point mass M' at the centre, a distance (R-d) away:

Step 4 — substitute M' from Step 2:

Step 5 — cancel (R-d)^{2} against two of the three factors in (R-d)^{3}, leaving one:

Step 6 — do the same thing at the surface. Putting d = 0 in Step 5 gives the surface value:

Step 7 — divide Step 5 by Step 6. All the constants \frac{4}{3}\pi G\rho cancel:

Step 8 — split the fraction on the right:

Step 9 — multiply both sides by g:

Step 10 — put d = R (the centre of the Earth):

Result.

What it means. Unlike the altitude case, g falls linearly with depth — and it reaches exactly zero at the centre, where a body would be weightless because the pulls from all directions cancel. Notice also that going up a small distance h costs 2h/R of g, while going down the same distance costs only d/R — twice as much is lost by climbing as by digging.

Conditions used. The Earth is assumed to be of uniform density, which is the only way Step 2 works. (In reality the core is far denser, so g actually rises slightly for the first 3000 km before falling.) Rotation and the non-spherical shape are ignored.

We start from the single physical idea that gravity is the centripetal force for a satellite, and reach expressions for its orbital speed and its period — and Kepler's third law as a bonus.

Setting up. A satellite of mass m moves in a circle of radius r = R + h about a planet of mass M and radius R, at a height h above the surface, with orbital speed v_o.

Step 1 — the gravitational force on the satellite. By Newton's law of gravitation, at a distance r from the centre:

Step 2 — the centripetal force it needs. To move in a circle of radius r at speed v_o, a mass m needs an inward force

Step 3 — equate them. Gravity is the only force acting, so it must be exactly the force required — no more and no less:

Step 4 — cancel m from both sides. It appears once on each side, so the answer will not depend on the satellite's mass:

Step 5 — multiply both sides by r:

Step 6 — take the square root:

Step 7 — rewrite using surface quantities. At the surface g = GM/R^{2}, so GM = gR^{2}. Substitute:

Step 8 — take R^{2} out of the root as R:

Step 9 — the period. The satellite travels one circumference 2\pi r in one period T at steady speed v_o, so time = distance \div speed:

Step 10 — substitute v_o from Step 6:

Step 11 — dividing by \sqrt{GM/r} is the same as multiplying by \sqrt{r/GM}:

Step 12 — bring the r outside the root inside it, where it becomes r^{2}:

Step 13 — square both sides:

Step 14 — group the constants. 4\pi^{2}, G and M are all fixed for a given planet:

Result.

That last line is Kepler's third law — here derived from Newton's laws rather than assumed from observation.

What it means. The satellite's mass cancelled in Step 4, so a tiny cubesat and a space station at the same height orbit at the same speed and with the same period. A higher orbit is a slower orbit (v_o \propto 1/\sqrt{r}) with a longer period — which is why a geostationary satellite has to sit 36 000 km up. For a satellite skimming the surface, h \approx 0 and v_o = \sqrt{gR} \approx 7.9 km s⁻¹.

Conditions used. The orbit is circular (for an ellipse, r becomes the mean distance), there is no air resistance, the planet is a uniform sphere, and the satellite's mass is far smaller than the planet's so the planet does not move.

We start from conservation of energy between the planet's surface and infinity, and reach the minimum launch speed needed to get away and never come back.

Setting up. A body of mass m is fired straight up from the surface of a planet of mass M and radius R with speed v_e. Take the gravitational potential energy to be zero at infinity, as usual.

Step 1 — energy at the surface. The body has kinetic energy \frac{1}{2}mv_e^{2} and potential energy -GMm/R (using U = -GMm/r with r = R). The total is

Step 2 — energy at infinity. "Just escapes" means it arrives at infinity with no speed left over, so its kinetic energy there is zero. Its potential energy there is also zero, by our choice of zero level. So:

Step 3 — apply conservation of energy. Gravity is a conservative force and we neglect air resistance, so the total energy does not change:

Step 4 — move the second term to the right:

Step 5 — cancel m from both sides. It appears once on each side, so the answer does not depend on the body's mass:

Step 6 — multiply both sides by 2:

Step 7 — take the square root:

Step 8 — rewrite in terms of g. At the surface g = GM/R^{2}, so GM = gR^{2}:

Step 9 — cancel one R:

Step 10 — a third form, in terms of density. The planet's mass is M = \frac{4}{3}\pi R^{3}\rho. Putting this into Step 7:

Result.

For the Earth, v_e = \sqrt{2 \times 9.8 \times 6.4\times10^{6}} = 11.2 km s⁻¹.

What it means. m cancelled, so a pebble and a rocket need the same escape speed. There is also no angle in the answer — energy depends only on distance, so the direction of projection does not matter either. Escape speed depends only on the planet: its mass (or density) and its radius.

Conditions used. Air resistance is neglected (a real rocket needs more), the body is given all its speed at once at the surface (not continuously pushed), the planet does not rotate, and there are no other bodies pulling on it.

We start from the two results just obtained and reach the ratio between them.

Step 1 — orbital speed for an orbit close to the surface. Put h = 0, i.e. r = R, into v_o = \sqrt{GM/r}:

Step 2 — write down the escape speed:

Step 3 — split the root. Since \sqrt{ab} = \sqrt{a}\sqrt{b}:

Step 4 — recognise the second factor as v_o from Step 1:

Result. v_e = \sqrt{2}\,v_o \approx 1.41\,v_o.

What it means. A satellite in low orbit is already travelling at 7.9 km s⁻¹; only 41 % more speed would let it leave the Earth altogether. In energy terms, escaping needs exactly twice the kinetic energy of orbiting, because v_e^{2} = 2v_o^{2}.

Condition used. The orbit must be close to the surface (h \ll R), so that the same R appears in both formulas.