Gravitational Potential and Potential Energy

A2 · 10 min

At AS you used ΔEP=mgΔh\Delta E_P = mg\Delta h for gravitational potential energy. That formula assumes gg is constant, which fails for rockets, satellites and anything moving large distances from a planet. Gravitational potential fixes this: it describes the energy per unit mass at every point in a field, with zero at infinity. It gives the energy needed to launch satellites, the speed of meteors arriving at Earth and the energy of orbits, and its definition is one of the most frequently examined in Paper 4.

Why we need a new idea of potential energy

Lifting a mass through height hh near the surface takes work mghmgh, because the force needed is constant. But if the mass is lifted thousands of kilometres, the force needed gets smaller as gg falls off as 1/r21/r^2, so mghmgh overestimates the work. We need a way of describing gravitational energy that works at any distance.

The answer is to choose a natural zero: infinity, where the field is zero and the mass feels no force at all. Every other point is described by the work involved in bringing a unit mass there from infinity.

Gravitational potential

Definition

The gravitational potential ϕ\phi at a point is the work done per unit mass in bringing a small test mass from infinity to the point.

For a point mass MM (or outside a uniform sphere, measured from its centre):

Key result
ϕ=−GMr\phi = -\frac{GM}{r}

Unit: J kg−1\text{J kg}^{-1}. Gravitational potential is a scalar.

Why the potential is negative

Gravity is attractive. As a test mass moves in from infinity towards MM, the field pulls it inwards: the field does positive work on the mass, and an external agent bringing it in steadily would have to hold it back, doing negative work. So the work done per unit mass in bringing it from infinity is negative.

Equivalently: potential is zero at infinity and gravitational potential energy decreases as a mass falls towards a planet. Anything that decreases from zero becomes negative. The potential is most negative (lowest) at the planet's surface, and rises towards zero as r→∞r \to \infty.

y = -1/x + 0*sqrt(x - 1) (1, 0) -- (1, -1)

The graph shows ϕ\phi (in units of GM/RGM/R) against rr (in units of the planet radius RR), for points outside the planet. It is a −1/r-1/r curve: always negative, rising towards zero but never reaching it.

Gravitational potential energy of two point masses

If the potential at a point is ϕ\phi, then the work done in bringing a mass mm (rather than a unit mass) from infinity is mϕm\phi. That work is stored as the gravitational potential energy of the system.

Key result
EP=mϕ=−GMmrE_P = m\phi = -\frac{GMm}{r}

EPE_P is the gravitational potential energy of the two point masses MM and mm separated by rr. It is zero when they are infinitely far apart and negative otherwise.

The potential energy belongs to the pair of masses: it is the energy stored because they attract each other. In problems we usually say "the potential energy of the satellite" as shorthand.

Changes in potential energy

Only changes in energy can be measured. Moving a mass mm from r1r_1 to r2r_2:

ΔEP=mΔϕ=m(ϕ2−ϕ1)=GMm(1r1−1r2)\Delta E_P = m\Delta\phi = m(\phi_2 - \phi_1) = GMm\left(\frac{1}{r_1} - \frac{1}{r_2}\right)

If r2>r1r_2 > r_1 (moving away), ΔEP\Delta E_P is positive: work must be done on the mass.

Tip

For a small height change hh near the surface, GMm(1R−1R+h)=GMmhR(R+h)≈GMmR2h=mghGMm\left(\dfrac{1}{R} - \dfrac{1}{R + h}\right) = \dfrac{GMmh}{R(R + h)} \approx \dfrac{GMm}{R^2}h = mgh. The AS formula is the small-height approximation of the general one.

Potential and field strength

Field strength and potential are two descriptions of the same field.

  • Field strength is the negative of the potential gradient:
g=−ΔϕΔrg = -\frac{\Delta\phi}{\Delta r}

On a graph of ϕ\phi against rr, the gradient at any point gives the size of gg there. The minus sign says that the field points in the direction of decreasing potential (towards the mass). At the Earth's surface the gradient of the ϕ\phi–rr curve is GM/R2=9.81 J kg−1 m−1GM/R^2 = 9.81\ \text{J kg}^{-1}\ \text{m}^{-1}, which is 9.81 N kg−19.81\ \text{N kg}^{-1}.

  • The area under a graph of gg against rr between two distances equals the change in potential between them.

Surfaces of constant potential (equipotentials) around a planet are spheres centred on it. Moving along an equipotential needs no work, which is why a satellite in a circular orbit has constant potential energy.

Escape speed

An object launched from a planet's surface can escape completely (reach infinity with zero speed) if its kinetic energy is enough to raise its potential energy from −GMm/R-GMm/R to zero:

12mvesc2=GMmR⇒vesc=2GMR\tfrac{1}{2}mv_{\text{esc}}^{2} = \frac{GMm}{R} \quad\Rightarrow\quad v_{\text{esc}} = \sqrt{\frac{2GM}{R}}

For the Earth this is about 11.2 km s−111.2\ \text{km s}^{-1}. Escape speed is not a named syllabus formula, but questions often ask you to derive it from the energy argument above. It does not depend on the mass of the object or the direction of launch (ignoring air resistance).

Energy of a satellite in orbit

For a circular orbit, gravity provides the centripetal force: GMm/r2=mv2/rGMm/r^2 = mv^2/r, so mv2=GMm/rmv^2 = GMm/r.

EK=12mv2=GMm2r,EP=−GMmr,Etotal=EK+EP=−GMm2rE_K = \tfrac{1}{2}mv^{2} = \frac{GMm}{2r}, \qquad E_P = -\frac{GMm}{r}, \qquad E_{\text{total}} = E_K + E_P = -\frac{GMm}{2r}

A satellite in a higher orbit has less kinetic energy but more (less negative) total energy. Raising a satellite to a higher orbit requires energy input, even though it ends up moving more slowly.

Potential and energy problems
  1. Find each rr from the centre of the mass.
  2. Calculate the potential at each point with ϕ=−GM/r\phi = -GM/r (keep the minus sign).
  3. For several masses, add the potentials as scalars (no directions).
  4. Change in potential energy: ΔEP=m(ϕfinal−ϕinitial)\Delta E_P = m(\phi_{\text{final}} - \phi_{\text{initial}}).
  5. Use energy conservation: loss in EPE_P = gain in EKE_K (if no other forces act).

Worked examples

Potential at the Earth's surface

Calculate the gravitational potential at the Earth's surface. (M=5.97×1024 kgM = 5.97 \times 10^{24}\ \text{kg}, R=6.37×106 mR = 6.37 \times 10^{6}\ \text{m})

Solutionϕ=−GMR=−6.67×10−11×5.97×10246.37×106=−6.25×107 J kg−1\phi = -\frac{GM}{R} = -\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.37 \times 10^{6}} = -6.25 \times 10^{7}\ \text{J kg}^{-1}

Each kilogram at the surface needs 6.25×107 J6.25 \times 10^{7}\ \text{J} to be removed completely from the Earth's field.

Lifting a satellite: the general formula against mgh

Calculate the increase in gravitational potential energy when a 1000 kg1000\ \text{kg} satellite is raised from the Earth's surface to a height of 400 km400\ \text{km}. Compare with the value given by mghmgh.

SolutionΔEP=GMm(1R−1R+h)=6.67×10−11×5.97×1024×1000×(16.37×106−16.77×106)\Delta E_P = GMm\left(\frac{1}{R} - \frac{1}{R + h}\right) = 6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 1000 \times \left(\frac{1}{6.37 \times 10^{6}} - \frac{1}{6.77 \times 10^{6}}\right)ΔEP=3.69×109 J\Delta E_P = 3.69 \times 10^{9}\ \text{J}

Using mghmgh: 1000×9.81×4.00×105=3.92×109 J1000 \times 9.81 \times 4.00 \times 10^{5} = 3.92 \times 10^{9}\ \text{J}. This overestimates by about 6%6\% because gg decreases with height; mghmgh assumes it stays at 9.81 N kg−19.81\ \text{N kg}^{-1} all the way up.

Potential at a point between two masses

Calculate the gravitational potential at the point between the Earth and the Moon where the resultant field strength is zero, 3.46×108 m3.46 \times 10^{8}\ \text{m} from the Earth's centre. The Earth–Moon distance is 3.84×108 m3.84 \times 10^{8}\ \text{m} and MM=7.35×1022 kgM_M = 7.35 \times 10^{22}\ \text{kg}.

Solution

Potentials are scalars, so add them:

ϕ=−GMErE−GMMrM=−6.67×10−11×5.97×10243.46×108−6.67×10−11×7.35×10220.38×108\phi = -\frac{GM_E}{r_E} - \frac{GM_M}{r_M} = -\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{3.46 \times 10^{8}} - \frac{6.67 \times 10^{-11} \times 7.35 \times 10^{22}}{0.38 \times 10^{8}}ϕ=−1.151×106−1.29×105=−1.28×106 J kg−1\phi = -1.151 \times 10^{6} - 1.29 \times 10^{5} = -1.28 \times 10^{6}\ \text{J kg}^{-1}

The field strength there is zero but the potential is not. Field strength is the gradient of potential: zero field means the potential is at a maximum along the line, not that it is zero.

Escape speed from the Moon

Show that the escape speed from the surface of the Moon is about 2.4 km s−12.4\ \text{km s}^{-1}. (M=7.35×1022 kgM = 7.35 \times 10^{22}\ \text{kg}, R=1.74×106 mR = 1.74 \times 10^{6}\ \text{m})

Solution

To escape, the kinetic energy must at least equal the increase in potential energy from the surface to infinity:

12mv2=GMmR⇒v=2GMR=2×6.67×10−11×7.35×10221.74×106=2.37×103 m s−1\tfrac{1}{2}mv^{2} = \frac{GMm}{R} \quad\Rightarrow\quad v = \sqrt{\frac{2GM}{R}} = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 7.35 \times 10^{22}}{1.74 \times 10^{6}}} = 2.37 \times 10^{3}\ \text{m s}^{-1}

This is low enough that the Moon has been unable to hold on to a significant atmosphere (gas molecules at lunar daytime temperatures can reach it).

A meteoroid falling to Earth

A meteoroid is at rest relative to the Earth at a distance of 3R3R from the Earth's centre. Ignoring air resistance, calculate its speed when it reaches the Earth's surface.

Solution

Loss of potential energy = gain in kinetic energy:

12mv2=GMm(1R−13R)=2GMm3R\tfrac{1}{2}mv^{2} = GMm\left(\frac{1}{R} - \frac{1}{3R}\right) = \frac{2GMm}{3R}v=4GM3R=4×6.67×10−11×5.97×10243×6.37×106=9.13×103 m s−1v = \sqrt{\frac{4GM}{3R}} = \sqrt{\frac{4 \times 6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{3 \times 6.37 \times 10^{6}}} = 9.13 \times 10^{3}\ \text{m s}^{-1}
Energy to reach geostationary orbit

A 1000 kg1000\ \text{kg} satellite is to be placed in geostationary orbit at r=4.22×107 mr = 4.22 \times 10^{7}\ \text{m}. Ignoring the Earth's rotation and energy losses, calculate the minimum energy needed, starting from rest on the Earth's surface.

Solution

Final total energy in orbit:

Eorbit=−GMm2r=−6.67×10−11×5.97×1024×10002×4.22×107=−4.72×109 JE_{\text{orbit}} = -\frac{GMm}{2r} = -\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 1000}{2 \times 4.22 \times 10^{7}} = -4.72 \times 10^{9}\ \text{J}

Initial energy (potential energy only, at rest on the surface):

Esurface=−GMmR=−6.25×1010 JE_{\text{surface}} = -\frac{GMm}{R} = -6.25 \times 10^{10}\ \text{J}

Energy needed =Eorbit−Esurface=−4.72×109+6.25×1010=5.78×1010 J= E_{\text{orbit}} - E_{\text{surface}} = -4.72 \times 10^{9} + 6.25 \times 10^{10} = 5.78 \times 10^{10}\ \text{J}.

Watch out

Dropping the minus sign. ϕ=−GM/r\phi = -GM/r is negative everywhere. When calculating changes, write both values with their signs and subtract: Δϕ=ϕ2−ϕ1\Delta\phi = \phi_2 - \phi_1. Most errors in potential questions come from signs.

Watch out

Adding potentials as vectors. Potential is a scalar: add the values from each mass directly, with no directions or components. Field strengths are vectors and must be added with directions.

Watch out

Defining potential without "from infinity" or "per unit mass". "The work done in moving a mass to a point" scores nothing. The definition needs: work done per unit mass, bringing a small test mass from infinity to the point.

Exam tip
  • "Define gravitational potential at a point" (2 marks): work done per unit mass; in bringing a small test mass from infinity to the point.
  • "Explain why gravitational potential is negative": the potential at infinity is zero; gravitational forces are attractive, so work is done by the field (energy is released) as a mass moves from infinity towards the point; hence the work done on the mass is negative.
  • "Show that the change in potential energy is approximately mghmgh": use the binomial or R(R+h)≈R2R(R + h) \approx R^2 argument from the tip above.
  • The formula sheet gives ϕ=−GM/r\phi = -GM/r and EP=−GMm/rE_P = -GMm/r. You must still know what each symbol means and where rr is measured from.
Summary
  • Gravitational potential: work done per unit mass in bringing a small test mass from infinity to the point.
  • ϕ=−GM/r\phi = -GM/r (scalar, J kg−1\text{J kg}^{-1}); zero at infinity, negative everywhere else because gravity is attractive.
  • EP=mϕ=−GMm/rE_P = m\phi = -GMm/r for two point masses; ΔEP=GMm(1/r1−1/r2)\Delta E_P = GMm(1/r_1 - 1/r_2).
  • mghmgh is the approximation for small heights near the surface.
  • g=−Δϕ/Δrg = -\Delta\phi/\Delta r: field strength is minus the potential gradient; area under gg–rr gives Δϕ\Delta\phi.
  • Escape speed v=2GM/Rv = \sqrt{2GM/R} from energy conservation.
  • Orbit energies: EK=GMm/2rE_K = GMm/2r, EP=−GMm/rE_P = -GMm/r, total −GMm/2r-GMm/2r.

Practice questions

Question
  1. Define gravitational potential and explain why its value is negative near a planet.
  2. Calculate the gravitational potential at a height equal to the Earth's radius above the Earth's surface.
  3. Calculate the work done in moving a 50 kg50\ \text{kg} mass from the Earth's surface to a distance 2R2R from the Earth's centre.
  4. The gravitational potential at a distance of 1.0×107 m1.0 \times 10^{7}\ \text{m} from the centre of a planet is −4.0×107 J kg−1-4.0 \times 10^{7}\ \text{J kg}^{-1}. Calculate the mass of the planet.
  5. Calculate the escape speed from Mars (M=6.42×1023 kgM = 6.42 \times 10^{23}\ \text{kg}, R=3.39×106 mR = 3.39 \times 10^{6}\ \text{m}).
  6. A probe falls from rest at 1.0×107 m1.0 \times 10^{7}\ \text{m} from the Earth's centre. Ignoring air resistance, calculate its speed at the Earth's surface.
  7. A 200 kg200\ \text{kg} satellite orbits at r=2REr = 2R_E. Calculate its kinetic, potential and total energy.
  8. A 1000 kg1000\ \text{kg} satellite is moved from a circular orbit of radius 7.0×106 m7.0 \times 10^{6}\ \text{m} to a geostationary orbit of radius 4.22×107 m4.22 \times 10^{7}\ \text{m}. (a) Calculate the change in its total energy. (b) State and explain what happens to its kinetic energy.
Answers
  1. The work done per unit mass in bringing a small test mass from infinity to the point. Potential at infinity is zero; the attractive force means work is done by the field as the mass approaches, so the work done on the mass, and the potential, are negative.
  2. r=2Rr = 2R: ϕ=−GM/2R=−6.25×107/2=−3.13×107 J kg−1\phi = -GM/2R = -6.25 \times 10^{7}/2 = -3.13 \times 10^{7}\ \text{J kg}^{-1}.
  3. W=mΔϕ=50×(−3.13×107+6.25×107)=1.56×109 JW = m\Delta\phi = 50 \times (-3.13 \times 10^{7} + 6.25 \times 10^{7}) = 1.56 \times 10^{9}\ \text{J}.
  4. M=−ϕr/G=4.0×107×1.0×107/6.67×10−11=6.0×1024 kgM = -\phi r/G = 4.0 \times 10^{7} \times 1.0 \times 10^{7}/6.67 \times 10^{-11} = 6.0 \times 10^{24}\ \text{kg}.
  5. v=2GM/R=2×6.67×10−11×6.42×1023/3.39×106=5.0×103 m s−1v = \sqrt{2GM/R} = \sqrt{2 \times 6.67 \times 10^{-11} \times 6.42 \times 10^{23}/3.39 \times 10^{6}} = 5.0 \times 10^{3}\ \text{m s}^{-1}.
  6. 12v2=GM(1/R−1/r)=3.98×1014×(1/6.37×106−1/1.0×107)=2.27×107\tfrac{1}{2}v^2 = GM(1/R - 1/r) = 3.98 \times 10^{14} \times (1/6.37 \times 10^{6} - 1/1.0 \times 10^{7}) = 2.27 \times 10^{7}, so v=6.7×103 m s−1v = 6.7 \times 10^{3}\ \text{m s}^{-1}.
  7. EK=GMm/2r=3.98×1014×200/(2×1.274×107)=3.13×109 JE_K = GMm/2r = 3.98 \times 10^{14} \times 200/(2 \times 1.274 \times 10^{7}) = 3.13 \times 10^{9}\ \text{J}; EP=−6.25×109 JE_P = -6.25 \times 10^{9}\ \text{J}; total =−3.13×109 J= -3.13 \times 10^{9}\ \text{J}.
  8. (a) ΔE=GMm2(1r1−1r2)=3.98×1014×10002(17.0×106−14.22×107)=2.37×1010 J\Delta E = \dfrac{GMm}{2}\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right) = \dfrac{3.98 \times 10^{14} \times 1000}{2}\left(\dfrac{1}{7.0 \times 10^{6}} - \dfrac{1}{4.22 \times 10^{7}}\right) = 2.37 \times 10^{10}\ \text{J} (an increase). (b) EK=GMm/2rE_K = GMm/2r decreases because rr increases: the satellite moves more slowly in the higher orbit. Its potential energy increases by twice as much as its kinetic energy decreases, so the total increases.

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