9.1 Gravitational fields — Learn

Prerequisites: weight and Newton's laws from Section 3.2; circular motion from Section 3.5; work, energy and the conservation of energy from Section 3.3; vectors and resolving from Section 2.2; standard form, graphs and SI prefixes from Chapter 1.

By the end of this section you can: describe a gravitational field and its field lines; use Newton's law of gravitation; define and use gravitational field strength; define and use gravitational potential; find the energy needed to move a mass between two points; explain the link between field strength and the potential gradient; and find the speed, the period and the escape speed of an orbit.

1. Force fields and field lines

A force field is a region where a body feels a force without anything touching it. Gravity is a force field: the Earth pulls on the Moon across empty space. The electric force between two charges is a force field too, and so is the magnetic force between two magnets.

A field is drawn with field lines. For gravity, each line points in the direction of the force on a small mass placed at that point. The lines point towards the body that creates the field, and they are closer together where the field is stronger.

Two shapes of field matter in this chapter.

A radial field surrounds a point mass or a sphere. The lines point inwards and spread apart as you move away. The field is strongest at the surface.
A uniform field has parallel, equally spaced lines. The field strength is the same at every point. The region near the Earth's surface is nearly uniform.

The Earth is a sphere. Outside its surface, its field is the same as the field of a point mass at its centre. That result lets us treat the whole Earth as a single mass at the centre, at a distance $r$ from the point we are studying.

2. Newton's law of gravitation

Two point masses attract each other. The force is proportional to each mass and inversely proportional to the square of the distance between them:

F = \frac{GMm}{r^{2}}

Here $G$ is the gravitational constant, $G = 6.67 \times 10^{-11}\,\text{N}\,\text{m}^{2}\,\text{kg}^{-2}$ . The force acts along the line joining the two masses. The two masses pull each other with equal and opposite forces, which is Newton's third law.

The inverse-square law means that doubling the distance makes the force four times smaller. The force never reaches zero, however far apart the masses are.

Worked example 1. The Earth has a mass of $5.97 \times 10^{24}\,\text{kg}$ , and the Moon has a mass of $7.35 \times 10^{22}\,\text{kg}$ . Their centres are $3.84 \times 10^{8}\,\text{m}$ apart. Find the gravitational force between them.

Substitute the three values into the inverse-square law.

\begin{aligned} F &= \frac{GMm}{r^{2}} \\ &= \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 7.35 \times 10^{22}}{(3.84 \times 10^{8})^{2}} \\ &= \frac{2.927 \times 10^{37}}{1.475 \times 10^{17}} \\ &= 1.98 \times 10^{20}\,\text{N} \end{aligned}

The force between the Earth and the Moon is $1.98 \times 10^{20}\,\text{N}$ , and it pulls each body towards the other.

3. Gravitational field strength

The gravitational field strength $g$ at a point is the gravitational force per unit mass at that point:

g = \frac{F}{m}

Its unit is the newton per kilogram, $\text{N}\,\text{kg}^{-1}$ . At the Earth's surface, $g = 9.81\,\text{N}\,\text{kg}^{-1}$ , which is why a 1 kg mass has a weight of $9.81\,\text{N}$ .

Substituting $F = \dfrac{GMm}{r^{2}}$ into the definition gives the field strength of a point mass or a sphere:

g = \frac{GM}{r^{2}}

The field strength depends only on the mass $M$ that creates the field and the distance $r$ from its centre. It does not depend on the mass placed in the field. Notice that $g$ follows the same inverse-square law as the force.

Worked example 2. The spacecraft orbits at a height of 400 km above the Earth's surface. Find the gravitational field strength at the orbit.

Measure $r$ from the centre of the Earth, so add the height to the radius.

r = 6.37 \times 10^{6} + 4.00 \times 10^{5} = 6.77 \times 10^{6}\,\text{m}

Then use the field strength of a sphere.

\begin{aligned} g &= \frac{GM}{r^{2}} \\ &= \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{(6.77 \times 10^{6})^{2}} \\ &= \frac{3.98 \times 10^{14}}{4.58 \times 10^{13}} \\ &= 8.69\,\text{N}\,\text{kg}^{-1} \end{aligned}

The field strength at the orbit is $8.69\,\text{N}\,\text{kg}^{-1}$ . It is lower than the surface value, but an astronaut there still has a large weight. The apparent weightlessness of an astronaut in orbit comes from free fall, not from the absence of gravity.

4. Gravitational potential and potential energy

Field strength describes the force on a mass. Gravitational potential describes the energy of a mass. The gravitational potential $V$ at a point is the energy per unit mass needed to move a small mass from infinity to that point.

V = -\frac{GM}{r}

The unit is the joule per kilogram, $\text{J}\,\text{kg}^{-1}$ . The potential is defined as zero at infinity, so every real potential is negative. A mass falls towards the body, which releases energy, so the potential energy is lower than at infinity. The minus sign carries that meaning.

The gravitational potential energy of a mass $m$ at a point is

E_{p} = mV = -\frac{GMm}{r}

The potential difference between two points is the energy needed to move unit mass between them. To move a mass from the surface to infinity requires the energy $\dfrac{GMm}{R}$ , which is why it is positive.

Worked example 3. Find the gravitational potential at the Earth's surface and at the orbit, and the energy needed to lift an 800 kg spacecraft from the surface to the orbit.

Use $r = R$ for the surface and $r = 6.77 \times 10^{6}\,\text{m}$ for the orbit.

\begin{aligned} V_{\text{surface}} &= -\frac{GM}{R} = -\frac{3.98 \times 10^{14}}{6.37 \times 10^{6}} = -6.25 \times 10^{7}\,\text{J}\,\text{kg}^{-1} \\ V_{\text{orbit}} &= -\frac{GM}{r} = -\frac{3.98 \times 10^{14}}{6.77 \times 10^{6}} = -5.88 \times 10^{7}\,\text{J}\,\text{kg}^{-1} \end{aligned}

The potential difference is $\Delta V = (-5.88 \times 10^{7}) - (-6.25 \times 10^{7}) = 3.7 \times 10^{6}\,\text{J}\,\text{kg}^{-1}$ . Multiply by the mass.

E = m\,\Delta V = 800 \times 3.7 \times 10^{6} = 3.0 \times 10^{9}\,\text{J}

The potential is negative at both points, and less negative at the orbit. The energy needed to lift the spacecraft is $3.0 \times 10^{9}\,\text{J}$ .

5. Field strength and the potential gradient

The field strength and the potential are two views of the same field, and they are linked. The field strength is the negative gradient of the potential:

g = -\frac{\Delta V}{\Delta r}

The minus sign means that the field points in the direction in which the potential falls. A graph of $V$ against $r$ is a curve that rises towards zero as $r$ increases. The gradient of that curve is negative, so $g$ is positive.

For a uniform field the gradient is constant, so $g$ is constant. For a radial field the curve is steepest near the surface, so $g$ is largest there. This is the same link that you will meet again in Section 9.2 for electric fields.

Equipotential surfaces join points of equal potential. No work is done moving a mass along an equipotential surface, because the potential does not change. The field lines are always at right angles to the equipotentials.

6. Orbits

For a satellite in a circular orbit, gravity provides the centripetal force. Set the gravitational force equal to the centripetal force:

\frac{GMm}{r^{2}} = \frac{mv^{2}}{r}

The mass of the satellite $m$ cancels, which is why the orbital speed does not depend on the mass of the satellite.

v = \sqrt{\frac{GM}{r}}

The orbital period follows from the circumference and the speed:

T = \frac{2\pi r}{v}

Squaring the speed equation gives $r v^{2} = GM$ . Substituting $v = \dfrac{2\pi r}{T}$ gives

\frac{r^{3}}{T^{2}} = \frac{GM}{4\pi^{2}}

The right-hand side is constant for every satellite of the same planet, so $T^{2}$ is proportional to $r^{3}$ . This is Kepler's third law.

Worked example 4. Find the orbital speed and period of the spacecraft.

Substitute $r = 6.77 \times 10^{6}\,\text{m}$ into the speed equation.

\begin{aligned} v &= \sqrt{\frac{GM}{r}} = \sqrt{\frac{3.98 \times 10^{14}}{6.77 \times 10^{6}}} \\ &= \sqrt{5.88 \times 10^{7}} = 7.67 \times 10^{3}\,\text{m}\,\text{s}^{-1} \end{aligned}

Then use the circumference divided by the speed.

T = \frac{2\pi r}{v} = \frac{2\pi \times 6.77 \times 10^{6}}{7.67 \times 10^{3}} = 5.55 \times 10^{3}\,\text{s}

The speed is $7.67\,\text{km}\,\text{s}^{-1}$ , and the period is 5550 s, or about 92 minutes. This is the period of the International Space Station.

7. Escape speed

The escape speed is the smallest speed a body needs to leave a planet and never return. At the surface, the kinetic energy must equal the energy needed to move the mass to infinity:

\frac{1}{2}mv^{2} = \frac{GMm}{R}

Rearranging gives

v_{\text{esc}} = \sqrt{\frac{2GM}{R}}

For the Earth this is $1.12 \times 10^{4}\,\text{m}\,\text{s}^{-1}$ , or $11.2\,\text{km}\,\text{s}^{-1}$ . The escape speed does not depend on the mass of the escaping body, and it is $\sqrt{2}$ times the speed of a satellite in orbit at the surface.

Watch out. Two mistakes are common in this section. The first is to use $g = 9.81\,\text{N}\,\text{kg}^{-1}$ at a height of hundreds of kilometres. The field strength follows an inverse-square law, so it has fallen to $8.69\,\text{N}\,\text{kg}^{-1}$ at the orbit. Put the height into $r$ and recalculate. The second is to measure $r$ from the surface instead of from the centre. The law is about the distance between the centres, so a height of 400 km means $r = 6.37 \times 10^{6} + 4.00 \times 10^{5}\,\text{m}$ , not $4.00 \times 10^{5}\,\text{m}$ . Check the size: $r$ must be larger than the Earth's radius for a point outside the Earth.

8. Try it: the inverse-square law

The figure plots gravitational field strength against distance from the centre of the Earth. One slider moves a marker outwards. The readout gives the field strength, its ratio to the surface value, and the gravitational potential.

Predict first. Do not move the slider yet. Write down your answers.

At twice the Earth's radius from the centre, what fraction of the surface field strength is left?
Does the field strength ever reach zero?
Is the gravitational potential positive or negative at the orbit?

Test. Set the distance to $6.8\,\text{Mm}$ and read the field strength. Move the marker to twice the Earth's radius and check prediction 1. Move it far out and watch the curve.

Explain. Why does the curve fall steeply at first and then flatten? Write two or three sentences. Use the idea that the same number of field lines spreads over a larger and larger sphere.

Progress check

Return to the spacecraft. It orbits at a height of 400 km, where $r = 6.77 \times 10^{6}\,\text{m}$ .

Find the gravitational field strength at the orbit.
Find the orbital speed and the period.
Find the gravitational potential at the orbit, and the energy needed to lift the 800 kg spacecraft from the surface to the orbit.
Find the escape speed from the Earth's surface.
Explain why the orbital speed does not depend on the mass of the spacecraft.

If question 1 goes wrong, use $g = \dfrac{GM}{r^{2}}$ with $r$ measured from the centre. If question 2 goes wrong, set the gravitational force equal to the centripetal force and cancel the mass. If question 3 goes wrong, remember that the potential is negative and you need the difference between the two values. If question 4 goes wrong, set the kinetic energy equal to $\dfrac{GMm}{R}$ . If question 5 goes wrong, look at what cancels when you set the two forces equal.

You should now be able to

describe a force field and draw radial and uniform field lines;
use Newton's law of gravitation $F = \dfrac{GMm}{r^{2}}$ ;
define gravitational field strength and use $g = \dfrac{GM}{r^{2}}$ ;
explain the inverse-square variation of the field strength;
define gravitational potential and use $V = -\dfrac{GM}{r}$ ;
explain why the potential is negative and defined as zero at infinity;
find the energy needed to move a mass between two points;
relate field strength to the potential gradient, $g = -\dfrac{\Delta V}{\Delta r}$ ;
describe an equipotential surface;
find the orbital speed and period from $v = \sqrt{\dfrac{GM}{r}}$ and $T = \dfrac{2\pi r}{v}$ ;
use $\dfrac{r^{3}}{T^{2}} = \dfrac{GM}{4\pi^{2}}$ ;
find the escape speed from $\sqrt{\dfrac{2GM}{R}}$ .