Why is gravitational potential negative?

Zero is at infinity, and gravity attracts. Energy is released as a mass falls inwards, so every point closer has lower, negative potential.

How does this relate to mg h?

Over small height changes g is nearly constant and m V mg h. Over large ones only m V is accurate.

Does a rocket need to reach escape velocity to leave Earth?

Not if it keeps firing its engines. Escape velocity applies to an unpowered projectile given all its energy at launch.

How is gravitational potential like electric potential?

Both are work per unit (mass or charge) from infinity, both vary as 1/r, and both give field strength as the negative gradient.

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178FieldsStretch

Gravitational Potential

Near Earth’s surface, $mg\Delta h$ is good enough for potential energy. But launch a spacecraft thousands of kilometres up and $g$ is no longer constant. $\textbf{Gravitational potential}$ handles changing fields properly, and it explains why the energy needed to escape a planet entirely is finite.

35 min Topic 12 · Gravitational Fields

What you'll be able to do

Define gravitational potential and explain why it is negative
Use $V_{\text{grav}} = -GM/r$ and $\Delta E_{p} = m\Delta V$
Relate field strength to potential gradient, $g = -\Delta V/\Delta r$
Derive and use escape velocity, $v = \sqrt{2GM/r}$

Potential in a radial field

The $\textbf{gravitational potential}$ at a point is the work done per unit mass in bringing a small mass from infinity to that point. It is measured in $\mathrm{J\,kg^{-1}}$ .

At infinity the potential is zero. Because gravity attracts, the field does work on a mass as it moves inwards, so the work done $\textit{by an external agent}$ is negative, and potential at every finite distance is negative.

For a spherical mass $M$ , the potential at distance $r$ from the centre is $V = -GM/r$ . It rises towards zero as $r$ increases.

V_{\text{grav}} = -\dfrac{GM}{r}

Potential in a radial field.

Example 1Find the gravitational potential at Earth’s surface.

M = 5.97\times10^{24}

kg,

R = 6.37\times10^{6}

V = -\dfrac{6.67\times10^{-11} \times 5.97\times10^{24}}{6.37\times10^{6}}

Answer

-6.25\times10^{7}\ \mathrm{J\,kg^{-1}}

Potential energy and work

A mass $m$ at a point of potential $V$ has gravitational potential energy $E_{p} = mV = -GMm/r$ .

The work done moving a mass between two points is $\Delta E_{p} = m\Delta V$ . Moving outwards, $V$ becomes less negative, so $\Delta V$ is positive and work must be done.

For small height changes near the surface, $m\Delta V$ reduces to the familiar $mg\Delta h$ . For large changes, $mg\Delta h$ overestimates the energy because $g$ decreases with height.

\Delta E_{p} = m\Delta V

Work done moving a mass.

Example 1How much energy is needed to lift a 500 kg satellite from Earth’s surface to an orbit radius of

1.20\times10^{7}

m (ignoring its kinetic energy)?

V_{\text{orbit}} = -\dfrac{3.982\times10^{14}}{1.20\times10^{7}} = -3.318\times10^{7}\ \mathrm{J\,kg^{-1}}

\Delta V = -3.318\times10^{7} - (-6.251\times10^{7}) = 2.933\times10^{7}

\Delta E_{p} = 500 \times 2.933\times10^{7}

Answer

1.47\times10^{10}\ \mathrm{J}

Tip — Using $mg\Delta h$ with $g = 9.81$ here would give $2.8\times10^{10}$ J — nearly double. Large height changes always need $m\Delta V$ .

Field as potential gradient and equipotentials

Field strength is the negative of the potential gradient: $g = -\Delta V/\Delta r$ . On a graph of $V$ against $r$ , the gradient at any point gives the field strength there.

$\textbf{Equipotential surfaces}$ join points of equal potential. Around a planet they are concentric spheres, spaced further apart with increasing distance. No work is done moving along one.

Equipotentials are perpendicular to field lines, exactly as in electric fields.

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

Field from potential gradient.

Tip — For equal steps in potential, equipotentials get further apart moving outwards because the field weakens.

Escape velocity

To escape a planet’s gravity completely, a projectile must reach infinity, where $E_{p} = 0$ . It must therefore start with kinetic energy at least equal to the depth of the potential well.

Setting $\tfrac12 mv^{2} = GMm/r$ gives the $\textbf{escape velocity}$ . It does not depend on the projectile’s mass, and it ignores air resistance.

For Earth it is about $11.2\ \mathrm{km\,s^{-1}}$ . A black hole is an object whose escape velocity at its surface exceeds the speed of light.

v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}}

Escape velocity.

Example 1Find the escape velocity from the Moon.

M = 7.35\times10^{22}

kg,

R = 1.74\times10^{6}

v = \sqrt{\dfrac{2 \times 6.67\times10^{-11} \times 7.35\times10^{22}}{1.74\times10^{6}}}

Answer

2.37\times10^{3}\ \mathrm{m\,s^{-1}}

Equation recap

V = -GM/r

Radial field potential.

\Delta E_{p} = m\Delta V

Work done.

g = -\Delta V/\Delta r

Field from potential.

v_{\text{esc}} = \sqrt{2GM/r}

Escape velocity.

Common mistakes to avoid

Giving gravitational potential as positive.

It is negative everywhere at finite distance.

Using

mg\Delta h

for a satellite launch.

g

changes with height; use

m\Delta V

Writing

V = -GM/r^{2}

Potential goes as

1/r

; field strength as

1/r^{2}

Thinking escape velocity depends on mass launched.

m

cancels in the derivation.

Key takeaways

Gravitational potential is work per unit mass from infinity; it is always negative.
Radial field: $V = -GM/r$ ; potential energy $E_{p} = -GMm/r$ .
Work done $= m\Delta V$ ; field strength $g = -\Delta V/\Delta r$ .
Equipotentials are perpendicular to field lines.
Escape velocity $v = \sqrt{2GM/r}$ is independent of projectile mass.

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