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D.1

Gravitational fields

Theme D · Fields · SL and HL, with additional HL material marked

The old site split this across Unit 6 and Unit 10. It is now one topic, with the HL material sitting inside it rather than on a separate page — which is closer to how it is actually examined.

🎯What you need to be able to do

  • State Kepler’s three laws and say what each one is about.
  • Apply Newton’s law of gravitation, including the fact that the two forces are equal and opposite however unequal the masses are.
  • Use gravitational field strength, and sketch field lines around one mass and around two.
  • Analyse circular orbits by setting gravity equal to the centripetal force, and derive Kepler’s third law from it.
  • Explain why a higher orbit is a slower one, and why an astronaut in one appears weightless.
  • Describe the uniform field close to a planet’s surface, and say when the uniform approximation is safe.
  • HLUse gravitational potential and gravitational potential energy, including their signs.
  • HLAdd potentials as scalars, and use \( W = m\Delta V_g \) for the work done moving a mass.
  • HLUse the potential gradient \( g = -\Delta V_g/\Delta r \), and interpret equipotential surfaces.
  • HLCalculate escape speed and the total energy of an orbit.

🪐Kepler’s laws: what was known before anyone knew why

Copernicus proposed in 1543 that the Sun, not the Earth, is at the centre of the Solar System. In the 1590s Tycho Brahe spent years making planetary observations far more accurate than anyone had managed before. By 1618 Johannes Kepler had squeezed three laws out of Brahe’s numbers.

They are experimental laws. Kepler had no idea why they were true, and nor did anyone else for more than 250 years, until Newton produced a single hypothesis from which all three follow. That gap is the point of this section: the laws describe, Newton’s law explains. They apply to anything orbiting a much more massive body, not just planets round the Sun.

First law — the orbits are ellipses, with the Sun at one focus. An ellipse is a flattened circle with two focal points; the Sun sits at one of them and there is nothing at all at the other.

An ellipse with its two focal points marked. The Sun is drawn at the left focus and the other focus is marked as empty, with a note that nothing is there. The semi-major axis is marked as half the longest diameter, running from the centre of the ellipse to the far end. A planet is shown on the ellipse at its closest point to the Sun, labelled perihelion, and at its furthest, labelled aphelion. A note beneath states that the ellipse in the diagram is drawn far more flattened than any real planetary orbit: the Earth's orbit would be indistinguishable from a circle at this size, with the Sun slightly off-centre.
Real planetary orbits are much rounder than this. Earth’s is so nearly circular that you could not tell it from a circle by eye — what gives it away is that the Sun is not at the centre.

Second law — the line from the Sun to the planet sweeps out equal areas in equal times. A planet close in has a short line to sweep with, so it must travel further along its orbit in a given time to sweep the same area: it moves fastest when it is closest to the Sun and slowest when furthest away.

An elliptical orbit with the Sun at the left focus. Two shaded sectors are drawn. The first is near the Sun between points A and B, short and fat, with a long arc of orbit between A and B. The second is far from the Sun between points C and D, long and thin, with a much shorter arc between C and D. The two shaded areas are marked as equal, and the caption states that the time taken to go from A to B equals the time taken to go from C to D. Arrows on the orbit are drawn long near the Sun and short far away, showing that the planet moves fastest when closest.
Equal areas in equal times. The consequence students are asked for is the speed: fastest at the closest point, slowest at the furthest.

Third law — the ratio \( (\text{orbital period})^{2} / (\text{semi-major axis})^{3} \) is the same for every body orbiting the same central mass. For a circular orbit of radius \(r\) and period \(T\) that is

\[ \frac{T^{2}}{r^{3}} = \text{constant} \]
A graph of orbital period squared, in years squared, against semi-major axis cubed, in astronomical units cubed, for the four inner planets. Mercury, Venus, Earth and Mars are plotted as points and all four lie on a straight line through the origin of gradient one. A note states that Jupiter, at 140.9 and 140.7, and Saturn, at 867.4 and 867.3, lie on the same line far beyond the top right corner of the plot. A second note states that the constant is one only because the axes are measured in years and astronomical units; in SI units the same constant is four pi squared divided by G times the mass of the Sun.
Plotted this way the law is a straight line through the origin. That the four inner planets fall on it to within a fraction of a per cent is what convinced people the pattern was real, long before anyone could say why.
“The same” means the same central mass. The constant in \( T^{2}/r^{3} \) is \( 4\pi^{2}/GM \), so it depends on what you are orbiting. Planets round the Sun share one value; the Moon and every artificial satellite round the Earth share a completely different one. Comparing a planet with a satellite through Kepler’s third law is a guaranteed way to lose the marks.

🍎Newton’s law of universal gravitation

If you trip over, you fall towards the ground. Newton’s claim was that the same thing that does that to you holds the Moon in its orbit — that every mass in the Universe attracts every other one. That is what “universal” means here, and it is a much bigger claim than it looks.

\[ F = \frac{Gm_1m_2}{r^{2}} \qquad G = 6.67 \times 10^{-11}\ \text{N m}^{2}\ \text{kg}^{-2} \]
Two point masses, a small one labelled m one on the left and a much larger one labelled m two on the right, separated by a distance r measured centre to centre. An arrow on each mass points towards the other, and the two arrows are drawn exactly the same length, labelled as equal in magnitude and opposite in direction even though the masses are very unequal. Beneath, two spheres are drawn with their mass shown spread through the volume and then again concentrated at a point at each centre, with a note that the interaction between two spherical masses is the same as if all the mass sat at the centres, which is why r is measured centre to centre.
The two forces are a Newton’s third law pair, so they are always equal in size. The Earth pulls you down with exactly the force you pull the Earth up with; what differs is what that force does to each of you.

Five things about this law are worth having ready, because questions test them directly.

It is for point massesand for spheres, which behave as if all their mass were at the centre
There are two forcesone on each mass, equal in magnitude and opposite in direction — even when the masses are wildly unequal
It is always attractivethere is no gravitational repulsion
It acts between everythingthe force between two people is real, just far too small to notice
Orbits are taken as circulara good approximation for most planets, and the one the syllabus works in
\(r\) is measured centre to centre, and it is squared. For a satellite at an altitude of 800 km, \(r\) is 800 km plus the Earth’s radius of 6370 km — nearly nine times what the altitude alone would give, and since the force goes as \(1/r^{2}\) the error is a factor of eighty. And “inverse square” is a square: triple the separation and the force falls to a ninth, not a third.

🧮Gravitational field strength

A field is a region where a mass feels a force, and gravitational field strength is the force per unit mass:

\[ g = \frac{F}{m} = \frac{GM}{r^{2}} \]

It is a vector, pointing towards the mass, measured in N kg\(^{-1}\) — which is the same thing as m s\(^{-2}\). Notice what is not in that expression: the mass you place in the field. This is exactly why every object falls with the same acceleration, a fact that puzzled people for two thousand years and then dropped out of one line of algebra.

Two panels of gravitational field lines. In the first, a single spherical mass has straight field lines drawn radially inwards from all directions, with arrowheads pointing at the mass, and the lines shown crowded together close to the mass and spread apart far from it to indicate the field is stronger nearby. In the second panel, a large mass on the left and a smaller mass on the right have field lines that curve towards each one, and between them a point is marked where the two fields exactly cancel, drawn closer to the smaller mass. A note states that field lines never cross, because the field at a point has only one direction.
The spacing of the lines carries the strength. Between two masses there is a point where the two fields cancel exactly — and it lies nearer the smaller mass, because you have to get close to a small mass for its field to match a big one at a distance.

Outside a spherical mass the field falls off as \(1/r^{2}\) from the centre, so the surface value is not a special case of anything — it is just \(g\) evaluated at \(r = R\).

A graph of gravitational field strength against distance from the centre of the Earth. The curve begins at the Earth's surface at one Earth radius, where the value is 9.8 newtons per kilogram, and falls away as an inverse square, passing through 2.45 at two Earth radii, 1.09 at three, and 0.61 at four. Dashed guides mark those values. The region inside the Earth is shaded and labelled as not covered by this expression. A note states that the curve approaches zero but never reaches it, so there is nowhere that gravity is truly absent.
Doubling the distance from the centre quarters the field. At the height of the International Space Station, about 400 km up, \(g\) is still about 89% of its surface value — astronauts are not weightless because gravity is weak up there.

🌍Close to the surface, the field looks uniform

Stand on the ground and the field lines around you are, to any accuracy you could measure, parallel and evenly spaced, all pointing straight down. Nothing has changed about the physics — the field is still radial and still \( GM/r^{2} \) — but you are looking at such a tiny patch of a very large sphere that the convergence is invisible.

Two views of the same gravitational field. On the left, the whole Earth with field lines converging radially inwards from all directions, the lines further apart at greater distances. On the right, a small patch of the surface magnified, in which the same field lines are effectively parallel, evenly spaced and vertical, with the ground drawn as a flat horizontal line. A note states that the uniform picture is an approximation valid only over heights small compared with the radius of the Earth, and that it is the approximation behind the familiar formulas m g h and g equals 9.81.
The same field, twice. Over a few hundred metres the radial field is indistinguishable from a uniform one — and that approximation is exactly what \( \Delta E_p = mg\Delta h \) is built on.

This is why two different formulas for gravitational potential energy coexist, and it fixes when each is allowed. Over a cliff, a building or a laboratory bench, \(g\) is constant and \( \Delta E_p = mg\Delta h \) is exact enough. Over hundreds of kilometres it is not, and you must go back to \( E_p = -GMm/r \).

🛰️Circular orbits: gravity is the centripetal force

An orbiting satellite is not balanced between two forces. There is only one force on it — gravity — and that force is what makes it go round. Setting gravity equal to the centripetal force is the single move that opens almost every orbit question:

\[ \frac{GMm}{r^{2}} = \frac{mv^{2}}{r} \qquad\Longrightarrow\qquad v = \sqrt{\frac{GM}{r}} \]
A satellite on a circular orbit around the Earth, drawn at four positions round the circle. At each position a velocity arrow is drawn tangential to the circle, in the direction of travel, and a force arrow is drawn pointing straight at the Earth's centre. The two arrows are perpendicular at every position. Labels state that the velocity is always tangential and the gravitational force always points at the centre, that this force is the centripetal force rather than being balanced by one, and that because the force is always perpendicular to the velocity it does no work, which is why the speed stays constant.
The force is perpendicular to the velocity at every instant, so it does no work. That is why a circular orbit needs no fuel and why the speed never changes, only the direction.

Two things fall straight out. The satellite’s mass \(m\) cancels, so the orbit does not depend on what is orbiting — a bolt and a space station at the same radius travel at the same speed side by side. And because \(v \propto 1/\sqrt{r}\), a higher orbit is a slower one, which is the opposite of what most people guess.

🧑‍🚀Weightlessness is free fall, not the absence of gravity

What a bathroom scale reads is not your weight — it is the contact force the scale pushes back with. Those are equal only when you are not accelerating vertically. Put the scale in a lift and the reading changes with the lift’s acceleration; cut the cable, so that lift, scale and passenger all accelerate downwards at \(g\) together, and the reading falls to zero.

Two panels. On the left, a person standing on a bathroom scale inside a lift whose cable has snapped, so the lift is accelerating downwards at 9.8 metres per second squared; the weight arrow still acts on the person but the contact force from the scale is zero, so the scale reads zero. On the right, an astronaut inside an orbiting space station; gravity still acts on both the astronaut and the station and provides the centripetal force that keeps each in orbit, but because both accelerate together there is no contact force between them, so the astronaut floats. A note states that in both cases gravity is undiminished and it is the contact force, which is what we actually feel, that has gone.
Both panels show the same situation: everything in the picture is in free fall together, so nothing pushes on anything else. Gravity has not weakened in either — it is doing all the work.

An orbiting astronaut is in exactly that state permanently. Gravity is still pulling on the astronaut and on the station, and that pull is precisely the centripetal force keeping each of them in orbit. Because the two accelerate identically there is no contact force between them, and the astronaut floats. The correct term is apparent weightlessness: what has vanished is the contact force we normally feel, not the gravitational field, which at the ISS is still about 89% of its surface value.

✏️Worked example 1 — a satellite in low orbit

A satellite orbits Earth at an altitude of 800 km. Take \( M_E = 5.97 \times 10^{24} \) kg, \( R_E = 6.37 \times 10^{6} \) m, \( G = 6.67 \times 10^{-11} \).

Get \(r\) right first. \( r = 6.37 \times 10^{6} + 0.80 \times 10^{6} = 7.17 \times 10^{6} \) m. Using the altitude alone here is the error that wrecks the whole question.

(a) Orbital speed. \( v = \sqrt{GM/r} = \sqrt{(6.67 \times 10^{-11} \times 5.97 \times 10^{24})/(7.17 \times 10^{6})} = \sqrt{5.55 \times 10^{5}} = 7.45 \times 10^{3} \) m s\(^{-1}\), about 7.5 km s\(^{-1}\).

(b) Period. \( T = 2\pi r/v = (2\pi \times 7.17 \times 10^{6})/(7.45 \times 10^{3}) = 6045 \) s, about 101 minutes — a plausible low-Earth-orbit period, which is a useful sanity check.

(c) Field strength at that height. \( g = GM/r^{2} = 3.98 \times 10^{14} / (7.17 \times 10^{6})^{2} = 7.75 \) N kg\(^{-1}\) — still 79% of the surface value.

(d) HLEscape speed from that altitude. \( v_{\text{esc}} = \sqrt{2} \times v_{\text{orbit}} = 1.414 \times 7.45 = 10.5 \) km s\(^{-1}\).

Two things to carry away. Escape speed is always \(\sqrt{2}\) times the circular orbital speed at the same radius — useful as a shortcut and as a check. And part (c) is the answer to “why are astronauts weightless?”: they are not beyond gravity at all. They are in free fall, and so is everything around them.

✏️Worked example 2 — Kepler’s second law with numbers

The Earth is \( 1.471 \times 10^{11} \) m from the Sun at perihelion, where it travels at 30.29 km s\(^{-1}\), and \( 1.521 \times 10^{11} \) m away at aphelion. Find its speed at aphelion.

Sweeping equal areas in equal times means the quantity \( rv \) is the same at both ends of the orbit (each is twice the rate at which area is swept). So

\[ r_pv_p = r_av_a \qquad\Longrightarrow\qquad v_a = \frac{r_pv_p}{r_a} = \frac{1.471 \times 10^{11} \times 30.29 \times 10^{3}}{1.521 \times 10^{11}} \]

\( v_a = 2.93 \times 10^{4} \) m s\(^{-1} = 29.29 \) km s\(^{-1}\).

Check the direction of the change. Aphelion is the far end, so the Earth must be slower there — and 29.29 is less than 30.29. The orbit is only 3.4% wider at one end than the other, and the speed differs by the same 3.4%, which is how nearly circular Earth’s orbit really is.

✏️Worked example 3 — where the fields cancel

The Moon is \( 3.84 \times 10^{8} \) m from the Earth. Taking \( M_E = 5.97 \times 10^{24} \) kg and \( M_M = 7.35 \times 10^{22} \) kg, find the point on the line between them where the resultant gravitational field is zero.

Let the point be a distance \(x\) from the Earth’s centre, so it is \( d - x \) from the Moon’s. The two field strengths must be equal in magnitude:

\[ \frac{GM_E}{x^{2}} = \frac{GM_M}{(d-x)^{2}} \qquad\Longrightarrow\qquad \frac{x}{d-x} = \sqrt{\frac{M_E}{M_M}} = \sqrt{81.2} = 9.01 \]

So \( x = 9.01(d - x) \), giving \( x = 9.01d/10.01 = 0.900d = 3.46 \times 10^{8} \) m from the Earth’s centre.

Substitute back. \( GM_E/x^{2} = 3.33 \times 10^{-3} \) N kg\(^{-1}\) and \( GM_M/(d-x)^{2} = 3.33 \times 10^{-3} \) N kg\(^{-1}\). They match, so the point is right. Note that \(G\) cancelled and never had to be used, and that the point sits 90% of the way to the Moon — far off centre, because the Earth is 81 times the more massive.

🔄Kepler’s third law, derived rather than asserted

Substituting \( v = 2\pi r/T \) into \( v = \sqrt{GM/r} \) and squaring gives

\[ T^{2} = \frac{4\pi^{2}r^{3}}{GM} \qquad\Longrightarrow\qquad \frac{T^{2}}{r^{3}} = \frac{4\pi^{2}}{GM} \]

That is Kepler’s third law, and the constant is now identified: it depends only on the central mass. Three centuries of astronomy in two lines of algebra, which is roughly why Newton’s law was taken seriously.

It also runs backwards, which is how the Solar System was weighed. Measure a moon’s period and orbital radius and you can solve for the mass of the planet it goes round — the only way we have of finding the mass of anything we cannot put on a balance.

A geostationary satellite is just the case \( T = 24 \) hours. That fixes \( r \approx 4.2 \times 10^{7} \) m from the Earth’s centre, about 36 000 km up. It must also orbit over the equator and in the direction of the Earth’s rotation, or it will not stay above one point.

HLGravitational potential and potential energy

Gravitational potential \( V_g = -GM/r \) is the work done per unit mass in bringing a small mass from infinity to that point; gravitational potential energy is \( E_p = mV_g = -GMm/r \).

A graph of gravitational potential against distance from the centre of a mass. The curve is entirely below the horizontal axis. It is very deeply negative close to the mass, rises steeply at first and then more and more gradually, and approaches zero from below as the distance increases without ever reaching it. The zero of the axis is labelled as the value at infinity, which is the definition being used. Two points on the curve are marked, one close in and one far out, with a note that the closer point is more negative and therefore more tightly bound. A note beneath states that gravitational potential energy has exactly this shape, being the potential multiplied by the mass placed in the field, and that moving outwards always increases the potential even though every value stays negative.
Everything on this graph is negative, and that is a consequence of one choice: potential is defined as zero at infinity. Since gravity always attracts, you must do work to get out to infinity, so everywhere closer in sits below zero.

A more negative value means more tightly bound. Moving outwards always increases the potential — from −60 to −30 MJ kg\(^{-1}\) is an increase, and the arithmetic of negative numbers is where marks are quietly lost here.

\( E_p = -GMm/r \) and \( \Delta E_p = mg\Delta h \) are not two rival formulas. The second is the near-surface approximation of the first, valid only over a height change small enough that \(g\) has not noticeably changed. Use \( mg\Delta h \) for a ball thrown off a cliff; use \( -GMm/r \) for anything involving orbits, escape, or a change of altitude measured in hundreds of kilometres. Mixing them — especially setting \( mgh \) equal to \( GMm/r \) — produces confident nonsense.
A point P lying between two masses. The gravitational potential at P due to the first mass is minus 5.4 times ten to the fifth joules per kilogram and that due to the second is minus 3.2 times ten to the fifth. The two values are shown being added arithmetically, with no angles and no components, to give a total of minus 8.6 times ten to the fifth joules per kilogram. A note contrasts this with field strength, which is a vector and whose two contributions at the same point would have to be added as vectors, pointing in opposite directions and partly cancelling.
Potential is a scalar, so potentials simply add. Field strength is a vector and does not — which is why a point can have zero field and a very large negative potential at the same time.

That last point is worth pausing on, because it is a favourite of examiners. At the null point of Worked example 3 the two fields cancel exactly, so \( g = 0 \). The two potentials do not cancel — both are negative, so they add to something more negative than either. A place where nothing is pulled is not a place where nothing is bound.

Once you can find the potential anywhere, the work needed to move a mass between two points is immediate:

\[ W = m\,\Delta V_g \]

and because potential depends only on position, that work is independent of the path taken. Fields with this property are called conservative; gravitational and electric fields both are, and friction is the standard example of a force that is not.

✏️HLWorked example 5 — lifting a satellite, using potential alone

Find the work needed to raise a 1500 kg satellite from the Earth’s surface (\( r = R_E \)) to a height of one Earth radius (\( r = 2R_E \)). Take \( GM_E = 3.98\times10^{14} \) and \( R_E = 6.37\times10^{6} \) m.

Potential at each place. \( V_g(R_E) = -GM/R_E = -6.25\times10^{7} \) J kg\(^{-1}\), and \( V_g(2R_E) = -3.13\times10^{7} \) J kg\(^{-1}\).

\[ \Delta V_g = -3.13\times10^{7} - (-6.25\times10^{7}) = +3.13\times10^{7}\ \text{J kg}^{-1} \]
\[ W = m\,\Delta V_g = 1500 \times 3.13\times10^{7} = 4.7\times10^{10}\ \text{J} \]
The change in potential is POSITIVE even though both values are negative. Going from −62.5 to −31.3 MJ kg\(^{-1}\) is an increase, and it has to be: you are climbing, so work must be supplied. If your answer came out negative you subtracted the wrong way round. Note also how much easier this is than integrating a force that changes all the way up — that is the whole reason potential is worth defining.

HLEscape speed

To escape means to reach infinity, where the potential energy is zero, with nothing left over. So the kinetic energy at launch must be at least enough to cancel the negative potential energy:

\[ \tfrac{1}{2}mv^{2} = \frac{GMm}{r} \qquad\Longrightarrow\qquad v_{\text{esc}} = \sqrt{\frac{2GM}{r}} \]

The escaping mass cancels again, so escape speed is the same for a pebble and a spacecraft — 11.2 km s\(^{-1}\) from the Earth’s surface, and only 2.4 km s\(^{-1}\) from the Moon’s, which is why the Moon has no atmosphere: its gas molecules reach that speed and leave.

HLThe potential gradient

Field strength and potential are not two independent ideas: the field is the rate at which the potential changes with distance.

\[ g = -\frac{\Delta V_g}{\Delta r} \]
A graph of gravitational potential against distance from a mass, with the curve rising towards zero. A small step in distance, delta r, is marked on the horizontal axis and the corresponding rise in potential, delta V, on the vertical axis, forming a right-angled triangle on the curve. The gradient of the curve at that point is labelled as delta V over delta r, and the field strength is labelled as minus that gradient. Two positions are compared: close to the mass the curve is steep, so the gradient and therefore the field are large; far away the curve is nearly flat, so both are small. A note gives the units as joules per kilogram per metre, which are the same as newtons per kilogram.
The steepness of the potential curve is the field strength. Near the mass the curve plunges and the field is strong; far out it flattens and the field dies away.

The minus sign says the field points down the potential slope — towards the mass, where the potential is more negative — while the work you do to climb is done up it. The units, J kg\(^{-1}\) m\(^{-1}\), are the same as N kg\(^{-1}\), as they must be.

A quick way to feel the size of it: at the Earth’s surface \( g = 9.81 \) N kg\(^{-1}\), so the potential rises by 9.81 J kg\(^{-1}\) for every metre you go up — about 10 kJ kg\(^{-1}\) per kilometre. Equipotential surfaces drawn at equal steps of potential are therefore evenly spaced near the ground, which is the uniform-field picture from earlier arriving by a different route.

HLThe total energy of an orbit

For a circular orbit, \( \tfrac{1}{2}mv^{2} = GMm/2r \), so the kinetic energy is exactly half the size of the potential energy and opposite in sign:

\[ E_k = +\frac{GMm}{2r},\qquad E_p = -\frac{GMm}{r},\qquad E = E_k + E_p = -\frac{GMm}{2r} \]
A graph of energy against orbital radius showing three curves. The kinetic energy is positive and falls towards zero as the radius increases. The potential energy is negative, twice the size of the kinetic energy at every radius, and rises towards zero. The total energy is negative, exactly half the size of the potential energy, and also rises towards zero. All three are labelled directly on the curves. A note states that the total energy being negative is what it means for the satellite to be bound, and that a larger orbit has a larger, less negative, total energy, so raising a satellite requires energy to be put in.
The total energy is negative, which is precisely what “bound” means: you would have to add energy to reach zero and get away. A bigger orbit is a less negative, and therefore larger, total energy.
Drag makes a satellite go faster. Atmospheric drag removes energy, so \(E\) becomes more negative, so \(r\) must decrease. But \( v = \sqrt{GM/r} \) increases as \(r\) falls. So friction speeds the satellite up. There is no contradiction: as it spirals in it loses about twice as much potential energy as the kinetic energy it gains, and the difference is what the drag dissipates. This is a favourite of examiners precisely because the intuitive answer is wrong.

✏️HLWorked example 4 — the cost of a higher orbit

How much energy must be supplied to move a 1200 kg satellite from a 800 km orbit (\( r_1 = 7.17 \times 10^{6} \) m) to a geostationary orbit (\( r_2 = 4.22 \times 10^{7} \) m)? Take \( GM_E = 3.98 \times 10^{14} \).

Use the total energy at each radius and take the difference.

\[ E_1 = -\frac{GMm}{2r_1} = -\frac{3.98\times10^{14} \times 1200}{2 \times 7.17\times10^{6}} = -3.33 \times 10^{10}\ \text{J} \]
\[ E_2 = -\frac{GMm}{2r_2} = -\frac{3.98\times10^{14} \times 1200}{2 \times 4.22\times10^{7}} = -5.66 \times 10^{9}\ \text{J} \]

\( \Delta E = E_2 - E_1 = -5.66\times10^{9} - (-3.33\times10^{10}) = 2.77 \times 10^{10} \) J, about 28 GJ.

Positive, as it must be. Energy has to be supplied to climb, so a negative answer would mean a sign slip — almost always from dropping a minus somewhere in \( E_2 - E_1 \). Note also that the satellite is slower in the higher orbit (3.1 against 7.5 km s\(^{-1}\)): it has less kinetic energy and more total energy at the same time, which only makes sense once you accept that the potential energy term is twice the size and going the other way.

HLEquipotential surfaces

An equipotential is a surface on which the potential is the same everywhere. Around a point mass they are spheres, drawn as circles, and they are always perpendicular to the field lines. No work is done moving along one, because the potential has not changed — which is why a satellite in a circular orbit needs no fuel: it is running along an equipotential.

A spherical mass with radial field lines drawn pointing inwards towards it, and a set of circular equipotential surfaces drawn concentric with the mass, crossing the field lines at right angles everywhere. The equipotentials are labelled with equal steps of potential, and they are drawn close together near the mass and progressively further apart with distance, because equal steps of potential need more and more distance where the field is weaker. A note states that no work is done moving along an equipotential, and that the field lines always cross them at ninety degrees.
Drawn at equal steps of potential, the surfaces crowd together where the field is strong. The spacing is a picture of the field strength, just as the field-line spacing is: \( g = -\Delta V_g/\Delta r \).

🔭See it happen

PhET, Gravity and Orbits. Turn on the velocity and force vectors and watch gravity point permanently at the centre while the velocity stays tangential. Then reduce the orbital speed slightly and watch the orbit decay inwards — and the satellite speed up as it falls. Turning gravity off mid-orbit is also worth doing once: the satellite leaves along the tangent in a straight line, which is the clearest possible demonstration that the force is what bends the path.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. Find the gravitational field strength at the surface of Mars, given \( M = 6.42 \times 10^{23} \) kg and \( R = 3.39 \times 10^{6} \) m.
\( g = \dfrac{GM}{R^{2}} = \dfrac{6.67 \times 10^{-11} \times 6.42 \times 10^{23}}{(3.39 \times 10^{6})^{2}} = \dfrac{4.28 \times 10^{13}}{1.15 \times 10^{13}} = 3.7 \) m s\(^{-2}\), about 38% of Earth’s.
2. A minor planet orbits the Sun with a semi-major axis of 4.0 AU. Find its orbital period in years, without using \(G\) or the mass of the Sun.
Work in the units that make the constant 1. For anything orbiting the Sun, \( T^{2}/a^{3} \) is the same, and for the Earth \( T = 1 \) yr with \( a = 1 \) AU, so the constant is \( 1\,\text{yr}^{2}\,\text{AU}^{-3} \). Hence \( T^{2} = a^{3} = 4.0^{3} = 64 \), so \( T = 8.0 \) years. This is why astronomers use these units: Kepler’s third law becomes \( T^{2} = a^{3} \) exactly.
3. Find the orbital radius of a geostationary satellite. Take \( M_E = 5.97 \times 10^{24} \) kg and \( T = 24 \) hours.
From \( T^{2} = \dfrac{4\pi^{2}r^{3}}{GM} \), \( r^{3} = \dfrac{GMT^{2}}{4\pi^{2}} = \dfrac{3.98 \times 10^{14} \times (86400)^{2}}{39.5} = 7.53 \times 10^{22} \) m\(^{3}\), so \( r = 4.2 \times 10^{7} \) m from the Earth’s centre — about 36 000 km above the surface.
4. Find the orbital speed of that geostationary satellite, and compare it with the 7.5 km s\(^{-1}\) of the 800 km satellite.
\( v = \sqrt{GM/r} = \sqrt{\dfrac{3.98 \times 10^{14}}{4.22 \times 10^{7}}} = \sqrt{9.43 \times 10^{6}} = 3.1 \times 10^{3} \) m s\(^{-1}\). Check: \( 2\pi r/T \) gives the same, as it must. It is less than half the low-orbit speed, because \( v \propto 1/\sqrt{r} \) and the radius is nearly six times larger — the higher orbit is the slower one.
5. A comet on a long elliptical orbit is 30 times further from the Sun at aphelion than at perihelion. Find the ratio of its speeds at the two points, and name the law you used.
Kepler’s second law: equal areas in equal times means \( r_pv_p = r_av_a \), so \( v_p/v_a = r_a/r_p = 30 \). The comet moves 30 times faster at its closest approach. This is why a comet spends almost all of its period out in the cold and only a few weeks bright and visible near the Sun.
6. Find the escape speed from the surface of the Moon, given \( M = 7.35 \times 10^{22} \) kg and \( R = 1.74 \times 10^{6} \) m.
\( v_{\text{esc}} = \sqrt{\dfrac{2GM}{R}} = \sqrt{\dfrac{2 \times 6.67 \times 10^{-11} \times 7.35 \times 10^{22}}{1.74 \times 10^{6}}} = \sqrt{5.63 \times 10^{6}} = 2.4 \) km s\(^{-1}\) — low enough that gas molecules reach it, which is why the Moon has no atmosphere.
7. Two spheres of mass 60 kg and 6.0 kg are 0.50 m apart, centre to centre. Find the force on each, and state which is larger.
\( F = \dfrac{Gm_1m_2}{r^{2}} = \dfrac{6.67 \times 10^{-11} \times 60 \times 6.0}{0.25} = 9.6 \times 10^{-8} \) N. The forces on the two spheres are equal — neither is larger. They are a Newton’s third law pair, so the ten-fold difference in mass changes nothing about the size of the force; what it changes is the acceleration each one gets from it, which differs by a factor of ten.
8. Explain why astronauts on the International Space Station, about 400 km up, appear weightless, given that \(g\) there is still about 8.7 N kg\(^{-1}\).
They are not beyond gravity: at \( r = 6.77 \times 10^{6} \) m, \( g = GM/r^{2} = 8.7 \) N kg\(^{-1}\), about 89% of the surface value. They appear weightless because they are in free fall — the station and everything in it are accelerating towards the Earth at the same rate, so there is no contact force between an astronaut and the floor, and it is that contact force we normally feel as weight. The station keeps missing the Earth because of its sideways speed.
9. HLFind the gravitational potential at the Earth’s surface, taking \( GM = 3.98 \times 10^{14} \) and \( R = 6.37 \times 10^{6} \) m, and explain the sign.
\( V_g = -\dfrac{GM}{R} = -\dfrac{3.98 \times 10^{14}}{6.37 \times 10^{6}} = -6.3 \times 10^{7} \) J kg\(^{-1}\). It is negative because potential is defined as zero at infinity, and gravity is always attractive, so work must be done on a mass to move it out to infinity. Everything closer in than infinity therefore sits below zero. Multiplying by \(\sqrt{2}\)’s worth of algebra, \( v_{\text{esc}} = \sqrt{2 \times 6.3 \times 10^{7}} = 11.2 \) km s\(^{-1}\), which is a neat consistency check.
10. HLA satellite of mass \(m\) is moved from a circular orbit of radius \(r\) to one of radius \(2r\). Find the energy required, in terms of \(G\), \(M\), \(m\) and \(r\), and state what happens to its speed.
\( E = -\dfrac{GMm}{2r} \), so \( E_1 = -\dfrac{GMm}{2r} \) and \( E_2 = -\dfrac{GMm}{4r} \). The energy required is \[ \Delta E = E_2 - E_1 = -\frac{GMm}{4r} + \frac{GMm}{2r} = +\frac{GMm}{4r} \] Positive, so energy must be supplied — correct for climbing. Its speed falls, by a factor of \( \sqrt{2} \), since \( v = \sqrt{GM/r} \). The satellite ends up with more total energy and less kinetic energy, because the potential energy term is twice the size and moves the other way.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • The Physics Classroom — Circular Motion and Satellite Motion
  • HyperPhysics — gravity, orbits and gravitational potential