Module 5 · Newtonian World and Astrophysics

Gravitational Fields

Specification: OCR A H556 | Section: 5.4 Gravitational fields | Teaching time: ~14 hours

By the end of this topic you should be able to…

define and use gravitational field strength \(g = F/m\)
apply Newton’s law of gravitation to point and spherical masses
derive and use \(g = GM/r^2\), \(V_g = -GM/r\) and \(E_p = -GMm/r\)
interpret \(g\)-against-\(r\), \(V_g\)-against-\(r\) and force–distance graphs
explain the neutral point between two masses and why potential does not become zero there
show Kepler’s third law and solve orbit and escape-speed questions

Part 1 · 5.4.1

Point and spherical masses

Every mass produces a gravitational field. A second mass placed in that field experiences a force directed towards the original mass.

Definition of field strength\[ g = \frac{F}{m} \]

Close to the Earth’s surface, the field is approximately uniform, so field lines are parallel and equally spaced. Further away, the field is radial, because a spherical mass can be modelled as a point mass at its centre.

Key point: \(g\) is a vector. It has a direction. In gravitational fields, that direction is always towards the mass.

Field lines show the direction of force on a small test mass.

Common misconception

“Gravity acts downwards.” More precisely: gravity acts towards the centre of the mass. “Downwards” is only a local description near the Earth’s surface.

Common misconception

Students often mix up little \(g\) and big \(G\). \(g\) is field strength. \(G\) is the universal gravitational constant.

Common misconception

For spherical masses, use the distance from centre to centre, not the height above the surface unless you first add the radius.

Part 2 · 5.4.2

Newton’s law of gravitation

Force between two point masses\[ F = \frac{GMm}{r^2} \]

The force is attractive, so when writing a final answer you should say it acts towards the other mass.

Field strength around a point mass\[ g = \frac{GM}{r^2} \]

This is an inverse-square law. If distance doubles, field strength falls by a factor of four.

Exam tip: sketch a quick centre–radius diagram before substituting. This avoids the classic mistake of using altitude instead of distance from the centre.

Use \(G = 6.67 \times 10^{-11}\,\mathrm{N\,m^2\,kg^{-2}}\), \(M_E = 5.97 \times 10^{24}\,\mathrm{kg}\) and \(R_E = 6.37 \times 10^6\,\mathrm{m}\).

\[ g = \frac{GM_E}{R_E^2} = \frac{(6.67\times10^{-11})(5.97\times10^{24})}{(6.37\times10^6)^2} = 9.81\,\mathrm{N\,kg^{-1}} \]

Answer: \(9.81\,\mathrm{N\,kg^{-1}}\), directed towards the centre of the Earth.

A satellite is \(3.00\times10^5\,\mathrm{m}\) above the Earth’s surface.

\[ r = R_E + h = 6.37\times10^6 + 3.00\times10^5 = 6.67\times10^6\,\mathrm{m} \] \[ g = \frac{GM_E}{r^2} = \frac{(6.67\times10^{-11})(5.97\times10^{24})}{(6.67\times10^6)^2} = 8.94\,\mathrm{N\,kg^{-1}} \]

Even 300 km up, the field is still large. That is why satellites are still very much in Earth’s gravity.

A probe moves from distance \(r\) to \(2r\) from a planet’s centre.

\[ g \propto \frac{1}{r^2} \] \[ g_{\text{new}} = g_{\text{old}}\times\frac{r^2}{(2r)^2} = g_{\text{old}}\times\frac{r^2}{4r^2} = \frac{g_{\text{old}}}{4} \]

Doubling distance makes the field strength one quarter as large.

Part 3 · 5.4.3

Kepler’s laws and showing Kepler’s third law

Kepler 1

Planets move in ellipses with the Sun at one focus.

Kepler 2

A line from the Sun to the planet sweeps out equal areas in equal times.

Kepler 3

For circular orbits, \(T^2 \propto r^3\).

Pupils need to be able to show Kepler’s third law from circular motion and Newtonian gravitation.

Start by equating forces\[ \frac{GMm}{r^2} = \frac{mv^2}{r} \]

Now replace orbital speed using \(v = 2\pi r/T\):

Substitute \(v = 2\pi r/T\)\[ \frac{GMm}{r^2} = \frac{m}{r}\left(\frac{2\pi r}{T}\right)^2 \]

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

What the examiner wants: start with gravitational force = centripetal force, substitute \(v=2\pi r/T\), rearrange cleanly, then state therefore \(T^2 \propto r^3\).

Show that the orbital speed in a circular orbit is \(v = \sqrt{GM/r}\).

\[ \frac{GMm}{r^2} = \frac{mv^2}{r} \] Cancel \(m\) and multiply by \(r\): \[ \frac{GM}{r} = v^2 \] \[ v = \sqrt{\frac{GM}{r}} \]

Notice that the satellite mass cancels. Orbital speed does not depend on the mass of the satellite.

For an orbit to be geostationary, it must:

• have period \(24\,\mathrm{h}\)
• lie above the equator
• move west-to-east, same direction as Earth’s rotation
• be circular so it stays above the same point

“24 hours” alone is not enough for full marks.

Part 4 · 5.4.4

Gravitational potential and potential energy

Field strength \(g\)

A vector. It tells you force per unit mass.

Potential \(V_g\)

A scalar. It tells you work done per unit mass in bringing mass from infinity.

Gravitational potential\[ V_g = -\frac{GM}{r} \]

Gravitational potential energy\[ E_p = mV_g = -\frac{GMm}{r} \]

Potential is defined as zero at infinity. That means the potential at every finite point in a gravitational field is negative.

Very common mistake: students see a point where resultant field strength is zero and assume the potential must also be zero. That is wrong. Fields are vectors and can cancel. Potentials are scalars and add.

A probe moves from \(r_1 = 2.0\times10^7\,\mathrm{m}\) to \(r_2 = 4.0\times10^7\,\mathrm{m}\) from a planet of mass \(6.0\times10^{24}\,\mathrm{kg}\).

\[ V_1 = -\frac{GM}{r_1} = -\frac{(6.67\times10^{-11})(6.0\times10^{24})}{2.0\times10^7} = -2.00\times10^7\,\mathrm{J\,kg^{-1}} \] \[ V_2 = -\frac{GM}{r_2} = -1.00\times10^7\,\mathrm{J\,kg^{-1}} \] \[ \Delta V = V_2 - V_1 = (-1.00\times10^7) - (-2.00\times10^7) = +1.00\times10^7\,\mathrm{J\,kg^{-1}} \]

The potential becomes less negative, so the change in potential is positive.

Two masses \(M\) and \(4M\) are separated by distance \(d\). Find the point between them where resultant field strength is zero.

Let the point be distance \(x\) from \(M\).
\[ \frac{GM}{x^2} = \frac{G(4M)}{(d-x)^2} \] \[ \frac{1}{x^2} = \frac{4}{(d-x)^2} \] \[ d-x = 2x \] \[ x = \frac{d}{3} \]

The zero-field point is closer to the smaller mass. The potential there is still negative, because both masses contribute negative potential.

Use energy conservation to show the escape-speed equation.

For escape, final kinetic energy at infinity is zero and final gravitational potential energy is zero.
\[ \frac{1}{2}mv^2 - \frac{GMm}{R} = 0 \] \[ \frac{1}{2}v^2 = \frac{GM}{R} \] \[ v = \sqrt{\frac{2GM}{R}} \]

Escape speed depends on the planet and launch radius, not on the mass of the object.

Graph focus

What the graphs are telling you

\(g\) against \(r\)

This falls as \(1/r^2\), so it drops quickly near the mass and then flattens out.

\(V_g\) against \(r\)

This follows \(-1/r\), so it always stays below zero and approaches zero from below.

Graph mistake

Students draw \(V_g\) crossing the axis. It should not: \(V_g\) tends to zero at infinity but remains negative at finite \(r\).

Graph mistake

Students sketch \(g\) and \(V_g\) with the same shape. They are different: \(g \propto 1/r^2\) and \(V_g \propto -1/r\).

Graph mistake

Students forget that the gradient of the \(V_g\)-graph is linked to field strength. Steeper graph means larger field strength magnitude.

Knowledge Check. Explain why the \(V_g\)-graph approaches zero from below as \(r\) becomes very large.

2 marks

Potential is defined as zero at infinity.
At any finite distance the potential is negative, but the magnitude decreases as distance increases.

Interactive

Interactive investigation: moving a test mass between two planets

Drag the probe along the line joining the two planets. The simulation reports real units:

field strengths in \(\mathrm{N\,kg^{-1}}\)
total potential in \(\mathrm{J\,kg^{-1}}\)
a prediction of the direction the probe would accelerate next

What to notice: the field can become zero between the masses, but the total potential stays negative.

\(g_L\): —

\(g_R\): —

\(g_{net}\): —

\(V_{total}\): —

What happens next?: —

\(M_L\) 6.0 \(M_R\) 9.0

Interactive

Force–distance graph: area = work done

Drag the two markers along the force–distance curve. The shaded area between them equals the work done in moving the chosen mass between those radii.

Investigation task: Set \(r_1 = 1\,R_\oplus\) and \(r_2 = 3\,R_\oplus\). Record the work done. Now double the mass \(m\) — predict what happens to the work done before you move the slider. Try halving the planet mass \(M\). Can you explain your results using \(W = \Delta E_p = -\frac{GMm}{r_2} + \frac{GMm}{r_1}\)?

\(r_1\): —

\(r_2\): —

Work done: —

\(M\) 6.0 × 10²⁴ kg \(m\) 5 × 10³ kg

Exam practice

Exam-Style Questions

Write an expression for the gravitational potential at the surface of a planet of mass \(M\) and radius \(R\). [Based on OCR H556 June 2017 Q23a]

1 mark

\(V_g = -GM/R\)

Explain why there can be a point between two masses where the resultant gravitational field strength is zero, but the resultant gravitational potential is not zero.

4 marks

Field strength is a vector.
Between the masses the two fields act in opposite directions.
So equal magnitudes can cancel.
Potential is a scalar.
Both gravitational potentials are negative.
So they add to a negative value rather than zero.

Show that for a circular orbit \(T^2 = \dfrac{4\pi^2}{GM}r^3\), and hence state Kepler’s third law.

5 marks

Set gravitational force equal to centripetal force.
\(GMm/r^2 = mv^2/r\)
Substitute \(v = 2\pi r/T\)
Rearrange to \(T^2 = 4\pi^2r^3/(GM)\)
State that \(T^2 \propto r^3\)

Two planets of masses \(3.0\times10^{24}\,\mathrm{kg}\) and \(7.0\times10^{24}\,\mathrm{kg}\) are separated by \(9.0\times10^7\,\mathrm{m}\). Calculate the point between them where the resultant gravitational field strength is zero.

5 marks

Let the distance from the smaller mass be \(x\).
\(G(3.0\times10^{24})/x^2 = G(7.0\times10^{24})/(9.0\times10^7-x)^2\)
Cancel \(G\), solve the equation.
\(x \approx 3.56\times10^7\,\mathrm{m}\)
Closer to the smaller mass.

Derive the escape-speed equation \(v = \sqrt{2GM/R}\) and state one reason why a real launch may require a greater speed than this ideal value.

4 marks

Use conservation of energy.
\(\tfrac12 mv^2 - GMm/R = 0\)
Rearrange to \(v = \sqrt{2GM/R}\)
Real-world reason: air resistance / atmosphere / non-instantaneous launch losses.

The gravitational field strength at the surface of a planet of radius \(R\) is \(g\). Write an expression for the gravitational field strength at a height \(h\) above the surface in terms of \(g\), \(R\) and \(h\).

2 marks

At height \(h\), distance from centre \(r = R + h\).
\(g' = GM/(R+h)^2 = g \times R^2/(R+h)^2\)

A satellite of mass \(m\) orbits a planet of mass \(M\) at radius \(r\) with period \(T\). A second satellite of mass \(2m\) orbits the same planet at the same radius. Compare the orbital periods of the two satellites.

2 marks

Both have the same period.
The satellite mass cancels in the derivation of \(T^2 = 4\pi^2r^3/(GM)\), so period depends only on \(r\) and \(M\).

A spacecraft travels from the surface of a planet (radius \(R\), mass \(M\)) to a point at distance \(3R\) from the planet’s centre. Calculate the change in gravitational potential energy per unit mass of the spacecraft. Give your answer in terms of \(G\), \(M\) and \(R\).

4 marks

\(\Delta V = V_2 - V_1\)
\(V_1 = -GM/R\) and \(V_2 = -GM/(3R)\)
\(\Delta V = -GM/(3R) - (-GM/R) = -GM/(3R) + GM/R = GM/R - GM/(3R) = 2GM/(3R)\)
So the change in potential energy per unit mass is \(+2GM/(3R)\).

A geostationary satellite orbits the Earth. State three conditions that must be satisfied for an orbit to be geostationary. Calculate the altitude of a geostationary orbit above the Earth’s surface. Use \(M_E = 5.97\times10^{24}\,\mathrm{kg}\), \(R_E = 6.37\times10^6\,\mathrm{m}\) and \(G = 6.67\times10^{-11}\,\mathrm{N\,m^2\,kg^{-2}}\).

7 marks

Conditions: period = 24 h; above the equator; circular orbit; same direction as Earth’s rotation. (any 3)
\(T = 86400\,\mathrm{s}\)
\(T^2 = 4\pi^2r^3/(GM_E)\)
\(r^3 = GM_ET^2/(4\pi^2) = (6.67\times10^{-11})(5.97\times10^{24})(86400)^2/(4\pi^2)\)
\(r^3 = 7.54\times10^{22}\), so \(r = 4.22\times10^7\,\mathrm{m}\)
Altitude \(h = r - R_E = 4.22\times10^7 - 6.37\times10^6 = 3.59\times10^7\,\mathrm{m}\)
\(\approx 3.6\times10^4\,\mathrm{km}\) above the surface.

Sketch a graph of gravitational potential \(V_g\) against distance \(r\) from the centre of a planet of radius \(R\) and mass \(M\). Mark the position \(r = R\) on your graph and label the value of \(V_g\) at \(r = R\). Explain why the graph does not extend below \(r = R\).

5 marks

Curve starts at \(r = R\) with \(V_g = -GM/R\) (negative, finite).
Curve rises (becomes less negative) approaching zero as \(r \to \infty\).
\(V_g\) never crosses zero — it approaches zero from below.
The formula \(V_g = -GM/r\) applies outside the sphere only.
Inside the sphere the field changes (not on OCR spec to calculate), so the graph stops at \(r = R\).

Topic Summary

Field strength

\(g\) is a vector. Around a point mass, \(g = GM/r^2\), directed towards the mass.

Potential

\(V_g\) is a scalar. \(V_g = -GM/r\). It stays negative at finite distance and tends to zero at infinity.

Most important distinction

Resultant field strength can be zero between two masses. Resultant potential does not become zero there, because negative potentials add.

\(g = F/m\)

\(F = GMm/r^2\)

\(g = GM/r^2\)

\(V_g = -GM/r\)

\(E_p = -GMm/r\)

\(T^2 = 4\pi^2r^3/(GM)\)

\(v = \sqrt{2GM/R}\)