Gravitational fields

Cambridge International AS & A Level Physics 9702 · A Level

All four subtopics here are printed under "A Level subject content" in the 9702 syllabus, and every objective carries the tier "A Level". None of it is AS. It is examined on Paper 4. Paper 1 is the AS multiple-choice paper, and the whole 9702 bank on this site comes from Paper 1, so no practice is tagged to this topic.

What a field is

A field of force is a region in which an object experiences a force without anything touching it. Gravitational, electric and magnetic fields are all of this kind, and the definitions in this topic have direct parallels in topic 18 on electric fields, which is worth noticing early because it halves the work.

Gravitational field strength is the force per unit mass acting on a small test mass placed at a point:

g = F / m

with unit N kg⁻¹. It is a vector, and its direction is the direction of the force on a mass, which is always towards the mass producing the field. Because F = mg is the weight of an object, g is also the free-fall acceleration, and 1 N kg⁻¹ is the same thing as 1 m s⁻².

Field lines

Field lines show the direction of the force on a mass placed in the field, and their spacing shows the strength.

• Around an isolated point mass or a uniform sphere, the lines are radial, pointing inwards, and they spread out with distance. The spreading is a picture of the inverse square law: the same number of lines is shared over a larger surface.
• Near the surface of the Earth, over a small region, the lines are parallel, equally spaced and vertically downwards. That is what a uniform field looks like, and it is why we can treat g as constant in projectile questions.

Gravitational field lines only ever converge on mass. There is no negative mass, so there is nothing for them to start from, and gravitational forces are only ever attractive.

Newton's law of gravitation

Every point mass attracts every other point mass with a force that is proportional to the product of their masses and inversely proportional to the square of their separation:

F = Gm₁m₂ / r²

where G is the gravitational constant, 6.67 × 10⁻¹¹ N m² kg⁻².

Three things must be right when using it:

• r is the separation of the centres, not the distance between the surfaces. For a point outside a uniform sphere, the whole mass may be treated as a point mass at the centre, which is what makes the law usable for planets at all. A satellite 300 km above the Earth's surface is 6.67 × 10⁶ m from the centre, not 3.00 × 10⁵ m.
• The force is attractive, and by Newton's third law the two forces form a pair: the Earth pulls the satellite with exactly the same magnitude of force as the satellite pulls the Earth. The accelerations differ enormously because the masses do.
• The inverse square relationship means doubling the separation gives a quarter of the force, and tripling it gives a ninth.

Worked example. The Earth has mass 5.98 × 10²⁴ kg and the Moon 7.35 × 10²² kg, with centres 3.84 × 10⁸ m apart.

F = 6.67 × 10⁻¹¹ × 5.98 × 10²⁴ × 7.35 × 10²² / (3.84 × 10⁸)² = 1.99 × 10²⁰

so the force is about 1.99 × 10²⁰ N.

Field strength of a point mass

The derivation is short and is asked for directly.

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Place a small test mass m at a distance r from a point mass M. The gravitational force on it is:

F = GMm / r²

By definition, the field strength is the force per unit mass:

g = F / m

Substituting and cancelling m:

g = GM / r²

The cancellation is the important step. The field strength depends on the mass producing the field and on the distance, and not at all on the mass placed in it. That is why all objects fall with the same acceleration.

Worked example. Take the Earth's mass as 5.98 × 10²⁴ kg and its radius as 6.37 × 10⁶ m.

g = 6.67 × 10⁻¹¹ × 5.98 × 10²⁴ / (6.37 × 10⁶)² = 9.83

so g at the surface is 9.83 N kg⁻¹, which is the value we round to 9.81.

Why g is nearly constant near the surface

Because r is measured from the centre of the Earth, climbing a mountain barely changes it.

Worked example. At the top of a 3000 m mountain, r becomes 6.373 × 10⁶ m.

g = 6.67 × 10⁻¹¹ × 5.98 × 10²⁴ / (6.373 × 10⁶)² = 9.82

The change is under 0.1 per cent, because 3000 m is a tiny fraction of 6370 km. For any change in height small compared with the Earth's radius, g is effectively constant, the field is uniform, and the familiar E_P = mgh applies. Once the height is comparable with the radius, that formula fails and the full expression must be used.

Circular orbits

A satellite in a circular orbit is in free fall. The gravitational force provides the centripetal force, and setting the two expressions equal is the whole method:

GMm / r² = mv² / r

The satellite's mass m cancels, giving

v² = GM / r

so a satellite's orbital speed depends only on the mass of the planet and the radius of the orbit, not on the satellite's own mass. A large satellite and a small one in the same orbit travel at the same speed, which is why an astronaut floats alongside their spacecraft.

Using v = 2πr/T instead gives the relationship between period and radius:

GMm / r² = 4π²mr / T², which rearranges to T² = 4π²r³ / GM

so T² is proportional to r³. That is Kepler's third law, and it comes straight out of Newton's law plus circular motion. A graph of T² against r³ is a straight line through the origin, and its gradient is 4π²/GM, which is a favourite way of asking you to find the mass of a planet from data about its moons.

Worked example. Find the radius of a geostationary orbit. Take GM for the Earth as 3.99 × 10¹⁴ N m² kg⁻¹ and the period as 24 hours.

T = 24 × 3600 = 86400

r³ = 3.99 × 10¹⁴ × 86400² / (4 × 9.870) = 7.544 × 10²²

Taking the cube root gives r = 4.23 × 10⁷ m, measured from the centre of the Earth, which is about 3.6 × 10⁷ m above the surface.

Geostationary orbits

A geostationary satellite stays above the same point on the Earth's surface. Three conditions are required, and all three are examined:

• Its period is 24 hours, matching the Earth's rotation.
• It orbits from west to east, the same direction as the Earth turns.
• It is directly above the Equator.

The last is the one most often missed, and the reason is worth understanding. The centre of any orbit must be the centre of the Earth, since that is where the gravitational force points. A circle centred on the Earth's centre and staying above a point in Britain is geometrically impossible: the satellite would have to orbit about an axis through Britain, and nothing pulls it that way. Only the equatorial plane contains the centre and keeps the satellite over a fixed latitude.

Because the radius is fixed by the period, every geostationary satellite sits in the same ring, which is why that orbit is a managed and crowded resource.

Gravitational potential

Gravitational potential φ at a point is the work done per unit mass in bringing a small test mass from infinity to that point. Its unit is J kg⁻¹, and it is a scalar.

For the field of a point mass:

φ = -GM / r

The minus sign is the part to understand rather than memorise. Gravity is attractive, so as a mass moves in from infinity the gravitational force does positive work on it, and an external agent must do negative work to bring it in slowly. Potential is therefore negative everywhere, and it becomes less negative as r increases, reaching zero at infinity, which is the defined reference point.

So potential increases as you move away from a mass, from a large negative value up towards zero. Students frequently invert this because "increasing" and "more negative" feel the same.

Field strength and potential are linked: g is the negative of the potential gradient. On a graph of φ against r, the gradient at a point gives -g. Notice that potential falls off as 1/r while field strength falls off as 1/r², so their graphs have different shapes and cannot be read as if they were the same curve.

Gravitational potential energy

The potential energy of a mass m at a point where the potential is φ is simply:

E_P = mφ

so for two point masses:

E_P = -GMm / r

This is the work done in bringing the mass from infinity to that point, and it is negative for the same reason as before. It reaches zero only at infinite separation, which is why the energy required to escape a field completely is exactly the magnitude of the potential energy at the start.

Worked example, escape speed. To escape from the surface of a planet, kinetic energy must be at least equal to the magnitude of the potential energy:

½mv² = GMm / r, so v² = 2GM / r

For the Earth, with GM = 3.99 × 10¹⁴ and r = 6.37 × 10⁶:

v² = 2 × 3.99 × 10¹⁴ / (6.37 × 10⁶) = 1.253 × 10⁸

so v = 1.12 × 10⁴ m s⁻¹, about 11.2 km s⁻¹. The mass of the escaping object cancels, so it is the same for a pebble and a rocket.

Worked example, moving a satellite. How much work is needed to lift a 500 kg satellite from an orbit of radius 7.0 × 10⁶ m to one of radius 1.4 × 10⁷ m? Take GM = 3.99 × 10¹⁴.

E_P at the first radius:

-3.99 × 10¹⁴ × 500 / (7.0 × 10⁶) = -2.850 × 10¹⁰

E_P at the second:

-3.99 × 10¹⁴ × 500 / (1.4 × 10⁷) = -1.425 × 10¹⁰

work done = -1.425 × 10¹⁰ - (-2.850 × 10¹⁰) = 1.425 × 10¹⁰

so 1.43 × 10¹⁰ J of work is required. Doubling the orbital radius halves the magnitude of the potential energy, which means the energy is only half way to escape, not most of the way. That is a useful sanity check on this kind of answer.

Common mistakes

• Measuring r from the surface of a planet rather than from its centre.
• Confusing the inverse square law for force and field with the inverse relationship for potential. Force and g go as 1/r²; φ and E_P go as 1/r.
• Dropping the minus sign in φ = -GM/r or E_P = -GMm/r, or treating the negative value as an error.
• Saying gravitational potential decreases as you move away from a mass. It increases, towards zero.
• Saying a heavier satellite must orbit faster at the same radius. Mass cancels.
• Forgetting that a geostationary orbit must be above the Equator, or giving a period of 12 hours.
• Using E_P = mgh over a distance comparable with the Earth's radius, where g is nowhere near constant.
• Treating gravitational field strength as a scalar, or potential as a vector. g is a vector; φ is a scalar.
• Writing that the Earth pulls a satellite harder than the satellite pulls the Earth.
• Confusing g at the surface, which is about 9.81 N kg⁻¹, with G, which is 6.67 × 10⁻¹¹ N m² kg⁻².

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