What a gravitational field is, and gravitational field strength g=F/m

Newton's law of gravitation, F=Gm₁m₂/r², and the inverse square law

Deriving and using g=Gm/r² for a radial field

Gravitational potential V_grav=-Gm/r, equipotentials, and the work done moving a mass between two points

Similarities and differences between gravitational and electric fields

Circular orbits: orbital speed, period, geostationary satellites and the derivation of T²∝r³

This whole topic grows from one force law. Before meeting it, we need a way to describe what a mass does to the space around it.

Field strength is a vector: its direction is the direction of the force on a mass. 1 N kg⁻¹ is the same as 1 m s⁻², so g is also the acceleration of a freely falling object.

The field is drawn with field lines. The arrow shows the direction of the force on a mass; the closer the lines, the stronger the field.

Around a spherical planet the field is radial: the lines point in towards the centre and spread out with distance, so the field weakens.

Over a region that is small compared with the planet, the lines are almost parallel and equally spaced: the field is uniform. That is why g=9.81 N kg⁻¹ can be used everywhere in a laboratory.

g=F/m defines the field at a point whatever is producing it; the test mass must be small enough not to disturb that field.

WORKED EXAMPLE · QUESTION

A lander of mass 420 kg rests on the surface of Titan, Saturn's largest moon. The gravitational force on it is 567 N. (a) Calculate the gravitational field strength at Titan's surface. (b) An astronaut with a spacesuit has a total mass of 115 kg. Calculate the gravitational force on the astronaut on Titan, and compare it with the force on Earth (g=9.81 N kg⁻¹).

SOLUTION

(a) g=F/m=567÷420= 1.35 N kg⁻¹

(b) Same field, different mass: F=mg=115×1.35= 155 N. On Earth, 115×9.81=1130 N, so the force on Titan is about 0.14 of the Earth value.

Newton's insight was that the force that makes an apple fall also holds the Moon in orbit, and that it weakens with distance in a precise way.

F is the force each mass exerts on the other (N), m₁ and m₂ are the two masses (kg), r is the distance between their centres (m) and G is the gravitational constant, 6.67×10⁻¹¹ N m² kg⁻². The law is in the formulae list. It is valid for point masses, and for uniform spheres if r is measured between their centres. The force is always attractive.

The force is proportional to each mass and to 1/r²: an inverse square law. Doubling r divides the force by 4; tripling it divides it by 9.

The forces on the two masses form a Newton's third law pair: equal in size and opposite in direction, even if one is a planet and the other a pebble.

The pebble accelerates far more because its mass is so much smaller.

A uniform sphere attracts as if all its mass were at its centre. So for a satellite, r = radius of the planet + height above the surface.

G is tiny, so gravitational forces between everyday objects are minute; gravity only becomes important when at least one mass is astronomical.

WORKED EXAMPLE · QUESTION

In a laboratory, a lead sphere of mass 12.0 kg and a small lead sphere of mass 0.800 kg are placed with their centres 0.150 m apart. (a) Calculate the gravitational force between them. (b) Compare it with the weight of the small sphere. (c) Calculate the gravitational field strength due to the large sphere at the centre of the small one.

SOLUTION

(a) F=(6.67×10⁻¹¹×12.0×0.800)/0.150²= 2.85 × 10⁻⁸ N

(b) W=0.800×9.81=7.85 N, so F/W= 3.6 × 10⁻⁹, far too small to notice without very sensitive apparatus.

(c) g=F/m=2.85×10⁻⁸÷0.800= 3.56 × 10⁻⁸ N kg⁻¹ (the same as GM/r²).

Place a small test mass m at distance r from the centre of a planet of mass M. By Newton's law the force on it is F=GMm/r². By definition g=F/m, so the test mass cancels:

(In the formulae list the producing mass is written m; here it is M to keep it apart from the test mass.) The equation is in the formulae list, and you must also be able to derive it in two lines, as above. It applies outside a spherical mass (r ≥ its radius), and the field points towards the centre.

g does not depend on the mass placed in the field: a feather and a hammer fall with the same acceleration in a vacuum.

Outside the planet, g∝1/r². The graph is a steep curve that never reaches zero.

Rearranged, M=gr²/G: measuring g at the surface and knowing the radius gives the planet's mass. For the Earth, 9.81×(6.37×10⁶)²÷6.67×10⁻¹¹=5.97×10²⁴ kg.

WORKED EXAMPLE · QUESTION

A space station orbits 420 km above the Earth's surface. The Earth has mass 5.97 × 10²⁴ kg and radius 6.37 × 10⁶ m. (a) Calculate the gravitational field strength at the station. (b) Express it as a percentage of the surface value, and explain why astronauts inside still seem weightless.

SOLUTION

(a) Distance from the centre, not from the surface: r=6.37×10⁶+0.42×10⁶=6.79×10⁶ m. Then g=(6.67×10⁻¹¹×5.97×10²⁴)/((6.79×10⁶)²)= 8.64 N kg⁻¹

(b) 8.64÷9.81= 88% of the surface value. Gravity is still strong; the astronauts and the station are both in free fall with acceleration g, so nothing presses them against the floor.

Field strength tells you the force. Potential tells you about energy: how much work it takes to move a mass around in the field.

Why negative? Zero is defined at infinity. Gravity attracts, so a mass moving in from infinity has work done on it by the field, and the work done by an external agent is negative. Every point near a planet is therefore at a potential below zero.

As r increases, V increases (becomes less negative) towards zero. Moving away from a planet means moving to a higher potential.

Work done moving a mass m between two points: W=mΔV=m(V_final-Vᵢₙᵢₜᵢₐₗ). This is the change in gravitational potential energy, and it depends only on the start and end points, not on the path.

Near the surface, where g is almost constant, mΔV reduces to the familiar mgΔh. Over large distances you must use mΔV, because g falls with height.

The field strength is the potential gradient: g=-ΔV/Δr. The tangent to a V–r graph gives the size of g; the minus sign means the field points towards decreasing potential.

Equipotentials

An equipotential is a surface on which every point has the same potential. Around a spherical planet they are concentric spheres (circles in a diagram).

ΔV=0 along an equipotential, so no work is done moving a mass along it. A satellite in a circular orbit stays on one equipotential.

Field lines always cross equipotentials at right angles.

Drawn at equal steps of potential, equipotentials get further apart with distance, because the field (the potential gradient) gets weaker. In a uniform field they are equally spaced, parallel planes.

WORKED EXAMPLE · QUESTION

Mars has mass 6.42 × 10²³ kg and radius 3.39 × 10⁶ m. A probe of mass 250 kg is lifted from the surface to a point 3.39 × 10⁶ m above the surface. (a) Calculate the gravitational potential at the surface and at the final position. (b) Calculate the minimum work done on the probe. (c) Explain why using W=mgΔh with the surface value of g gives too large an answer.

SOLUTION

(a) The final point is at r=2R. V_surface=-(6.67×10⁻¹¹×6.42×10²³)/(3.39×10⁶)= −1.263 × 10⁷ J kg⁻¹; V_2R=-1.263×10⁷÷2= −6.32 × 10⁶ J kg⁻¹

(b) W=mΔV=250×(-6.32×10⁶-(-1.263×10⁷))=250×6.32×10⁶= 1.58 × 10⁹ J

(c) g_surface=GM/R²=3.73 N kg⁻¹, so mgΔh=250×3.73×3.39×10⁶=3.16×10⁹ J, twice the true value, because g falls to one quarter of its surface value at 2R. Over large heights, use potential.

Taking the probe all the way to infinity (V=0) needs 250×1.263×10⁷=3.16×10⁹ J.

A probe launched from the surface with at least this much kinetic energy would never fall back: the idea of escape from a planet.

The field of a point mass and the field of a point charge have the same mathematical shape. Learn the pairs side by side:

Between individual particles the electric force is overwhelmingly stronger. Yet gravity dominates on the scale of planets and galaxies, because large bodies are almost exactly electrically neutral (their positive and negative charges cancel) while every bit of mass adds to the gravitational pull.

WORKED EXAMPLE · QUESTION

In an atom, an electron and a proton are 1.0 × 10⁻¹⁰ m apart. Calculate the electric force and the gravitational force between them, and the ratio of the two.

SOLUTION

Electric: F_E=(8.99×10⁹×(1.60×10⁻¹⁹)²)/((1.0×10⁻¹⁰)²)= 2.30 × 10⁻⁸ N

Gravitational: F_G=(6.67×10⁻¹¹×9.11×10⁻³¹×1.67×10⁻²⁷)/((1.0×10⁻¹⁰)²)= 1.01 × 10⁻⁴⁷ N

Ratio: F_E/F_G= 2.27 × 10³⁹. Both forces follow 1/r², so this ratio is the same at any separation: gravity is negligible inside atoms.

A satellite in a circular orbit moves at constant speed, but its velocity changes direction continuously, so it is accelerating towards the centre. By Newton's second law there must be a resultant force towards the centre: the centripetal force. The only force on it is gravity, which points towards the centre of the planet, at right angles to the velocity. Gravity therefore does no work, and the speed stays constant. Setting the gravitational force equal to the centripetal force gives everything else:

M is the mass of the central body, m the mass of the satellite, r the orbital radius from the centre, v the orbital speed and T the period. These results are not in the formulae list: derive them from F=Gm₁m₂/r² and F=mv²/r, both of which are in the formulae list.

The satellite's mass m cancels: every object at the same orbital radius has the same speed and period. Since 4π²/GM is a constant for one central body, T²∝r³.

Further out means slower and longer: v∝1/√r and T∝r^(3/2). A satellite at 4 times the radius moves at half the speed and takes 8 times as long to go round.

Rearranged, M=4π²r³/GT²: timing any moon or satellite gives the mass of the body it orbits.

Geostationary orbits

A geostationary satellite stays above the same point on the Earth's surface, so a fixed dish can point at it. It must:

have a period equal to the Earth's rotation period (about 24 hours);

orbit above the equator (its orbit must be centred on the Earth's centre and lie in the plane in which the surface rotates);

move in the same direction as the Earth rotates (west to east).

Because T fixes r through T²=(4π²/GM) r³, there is only one possible radius for a geostationary orbit.

WORKED EXAMPLE · QUESTION

The Earth rotates once every 8.62 × 10⁴ s. Its mass is 5.97 × 10²⁴ kg and its radius is 6.37 × 10⁶ m. (a) Show that the radius of a geostationary orbit is about 4.2 × 10⁷ m. (b) Calculate the height of the orbit above the surface and the orbital speed.

SOLUTION

(a) The gravitational force provides the centripetal force, so T²=4π²r³/GM. Rearranged: r³=GMT²/4π²=(6.67×10⁻¹¹×5.97×10²⁴×(8.62×10⁴)²)/4π²=7.49×10²² m³, so r= 4.22 × 10⁷ m ≈ 4.2 × 10⁷ m

(b) h=4.22×10⁷-6.37×10⁶= 3.58 × 10⁷ m; v=2π×4.22×10⁷÷8.62×10⁴= 3.07 × 10³ m s⁻¹. Check: √(GM/r) gives the same 3.07 × 10³ m s⁻¹.

Testing T² ∝ r³ with real orbits

If T=krⁿ, then lg T=nlg r+lg k, so a graph of lg T against lg r is a straight line of gradient n. For moons of one planet the gradient should be 1.5, confirming T²∝r³. Equally, a graph of T² against r³ is a straight line through the origin of gradient 4π²/GM.

WORKED EXAMPLE · QUESTION

Europa orbits Jupiter in a circle of radius 6.71 × 10⁸ m with a period of 3.07 × 10⁵ s (3.55 days). Calculate the mass of Jupiter.

SOLUTION

Gravity provides the centripetal force, so T²=4π²r³/GM; make M the subject: M=4π²r³/GT².

Substitute, with the period in seconds: M=(4π²×3.02×10²⁶)/(6.67×10⁻¹¹×9.42×10¹⁰)= 1.90 × 10²⁷ kg

The mass of Europa does not appear: it cancelled in the derivation. Any of the other three moons gives the same answer.

2 Circular motion and 4.3 Electric fields

F=mv²/r and v=ωr are what turn the law of gravitation into orbital speed and period. Coulomb's law, E=Q/4πε₀r² and electric potential have exactly the same structure as the equations here; learn the two topics as a pair.

Work and energy (Unit 1)

mgΔh is the special case of mΔV for a uniform field; this page explains when that approximation fails.

8 Stars and 5.9 The universe

Gravity holds stars and galaxies together; orbital speeds of stars in galaxies give evidence for dark matter, and the fate of the universe depends on its mean density.

1. Two asteroids attract each other with a force of 800 N. What is the force when the distance between their centres is doubled?

2. Explain why gravitational potential is negative near a planet.

3. At what height above the Earth's surface is the field strength one quarter of its surface value? (Earth radius 6.37 × 10⁶ m)

4. Satellite B orbits the Earth at four times the orbital radius of satellite A. How do their periods compare?

5. A satellite moves in a circular orbit. How much work does gravity do on it during one orbit, and why?

1. 200 N. The force is proportional to 1/r², so doubling r divides it by 4.

2. Potential is zero at infinity. Because gravity attracts, the field does work on a mass as it moves in from infinity, so the work done per unit mass by an external agent is negative; every nearby point is below zero.

3. g∝1/r², so one quarter needs r=2R. The height is 2R-R= 6.37 × 10⁶ m.

4. T²∝r³, so T∝r^(3/2): 4^(3/2)=8. Satellite B's period is 8 times A's.

5. None. The orbit lies on an equipotential (ΔV=0), and the gravitational force is always at right angles to the velocity, so the speed and kinetic energy stay constant.