Why a Smaller Planet Can Have Stronger Surface Gravity

A planet half the size of Earth can have twice Earth's surface gravity. That sounds backwards. Surely a smaller planet should have a weaker gravitational pull?

Not necessarily. Surface gravity depends on two things: how much mass a planet contains and how far its surface is from its centre. A smaller planet can have stronger surface gravity if its mass is concentrated into a sufficiently small space.

The key relationship is that surface gravitational field strength is proportional to the planet's mass divided by the square of its radius:

\[
g=\frac{GM}{R^2}
\]

The square on the radius is important. Halving a planet's radius, while keeping its mass unchanged, makes its surface gravitational field four times stronger. A planet does not need to be massive to have strong surface gravity. It needs enough mass relative to its radius.

Let's work through why this happens, how density changes the comparison, and why a stronger gravitational field does not automatically mean a greater gravitational force or gravitational potential energy.

## Why a smaller planet can have stronger surface gravity

Imagine two spherical planets.

- Planet A is large, with its mass spread throughout a relatively large volume.
- Planet B is smaller, but contains a substantial amount of mass in a much more compact volume.

You might predict that Planet A has stronger gravity because it contains more space, or perhaps because it looks more substantial. Neither prediction is useful. What matters is the amount of mass attracting you compared with your distance from the planet's centre.

Newton's law of universal gravitation gives the gravitational force between a planet and an object:

\[
F_g=\frac{GMm}{r^2}
\]

Here, \(F_g\) is gravitational force in newtons, \(G\) is the universal gravitational constant, \(M\) is the planet's mass in kilograms, \(m\) is the object's mass in kilograms, and \(r\) is the distance between their centres of mass in metres.

Gravitational field strength is the gravitational force per kilogram:

\[
g=\frac{F_g}{m}
\]

Substituting the force equation makes the object's mass cancel:

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

At the surface of a spherical planet, the distance from the planet's centre is its radius, \(R\). Therefore:

\[
\boxed{g_{\text{surface}}=\frac{GM}{R^2}}
\]

This equation assumes a spherically symmetric planet and ignores complications such as rotation and local variations in the planet's density. It is the standard model for comparing surface gravity.

Notice what the equation does not say: it does not say that a larger planet has stronger gravity. It says that a larger planetary mass increases the field, while a larger radius decreases it.

[IMAGE: Side-by-side cross-sections of two spherical planets. Show a large, relatively low-density planet and a smaller, compact, high-density planet. Mark each centre, surface radius, and inward gravitational field direction. Include a person or small test mass at each surface to establish the comparison. Do not assign numerical field strengths.]

### The radius has a surprisingly large effect

The radius is squared in the denominator. This means changes in radius can have a stronger effect than students initially expect.

Suppose a planet's mass stays exactly the same, but its radius becomes half as large.

The new surface field strength is:

\[
g_{\text{new}}=\frac{GM}{(R/2)^2}
\]

Since \((R/2)^2=R^2/4\),

\[
g_{\text{new}}=4\frac{GM}{R^2}
\]

Therefore:

\[
\boxed{g_{\text{new}}=4g_{\text{old}}}
\]

The planet's mass has not changed. Its surface has simply moved closer to the centre, where the gravitational field is stronger.

This is the central reason a smaller planet can have stronger surface gravity.

01Why the density of a planet matters

There is another way to understand the mass-radius relationship: consider the planet’s average density.

Density tells us how much mass is contained in a given volume:

\[ \rho=\frac{M}{V} \]

For a spherical planet, its volume is:

\[ V=\frac43\pi R^3 \]

Combining these relationships gives:

\[ M=\frac43\pi R^3\rho \]

Substitute this into the surface gravitational field equation:

\[ g=\frac{G\left(\frac43\pi R^3\rho\right)}{R^2} \]

Simplifying:

\[ \boxed{g=\frac43\pi G\rho R} \]

This gives us an important result:

\[ g\propto \rho R \]

Surface gravity is proportional to the product of average density and radius.

A useful contrasting case: two planets with equal density

Suppose two planets have exactly the same average density, but Planet A has half the radius of Planet B.

What happens to their surface gravity?

Since their densities are equal, we can compare their radii directly:

\[ \frac{g_A}{g_B}=\frac{R_A}{R_B} \]

Therefore:

\[ \frac{g_A}{g_B}=\frac12 \]

Planet A has half the surface gravitational field strength of Planet B.

This is the opposite of what you might expect if you only considered the effect of reducing radius while keeping mass constant.

The difference is that these are two different physical situations:

• Constant mass: Reducing radius increases surface gravity because the same mass is closer to the surface.
• Constant density: Reducing radius also reduces mass. The mass decreases faster than the radius, so surface gravity becomes weaker.

For a smaller planet to beat a larger planet’s surface gravity, its average density must be sufficiently high.

For example, if its radius is half that of the larger planet, its average density must be more than twice as high to produce stronger surface gravity.

This is a much more useful way to think about the title question than simply asking which planet is bigger.

02A historical check: Newton’s Moon test

Why does the gravitational field follow an inverse-square relationship in the first place?

An important part of the historical development came from Isaac Newton’s attempt to connect two observations:

1. Objects accelerate towards Earth’s surface when released.
2. The Moon continually accelerates towards Earth as it follows its orbit.

Newton wanted to know whether the same gravitational influence could explain both.

The Moon does not travel in a straight line. Its direction continually changes as it orbits Earth, which means it experiences centripetal acceleration towards Earth’s centre.

Newton investigated whether this acceleration was consistent with the gravitational attraction measured near Earth’s surface.

How the comparison worked

There was no single laboratory apparatus for this calculation. Newton combined two sets of measurements.

Terrestrial measurements

Measurements of falling objects and pendulums established the acceleration due to gravity near Earth’s surface.

Astronomical measurements

Observations of the Moon established its orbital period and its approximate distance from Earth’s centre.

The prediction from an inverse-square gravitational field was straightforward.

If the Moon is approximately 60 times Earth’s radius from its centre, its gravitational acceleration should be:

\[ g_{\text{Moon}}=\frac{g_E}{60^2} \]

Therefore:

\[ g_{\text{Moon}}=\frac{g_E}{3600} \]

The Moon should experience only about one 3600th of the gravitational acceleration measured at Earth’s surface.

But could its observed orbital acceleration actually support this prediction?

What happened historically?

Newton’s early attempt at this comparison did not produce a sufficiently close agreement. The estimate of Earth’s size available to him was inaccurate.

A better measurement of Earth by Jean Picard in the 1670s allowed Newton to revisit the calculation. The improved result was much closer to the inverse-square prediction. Newton’s work on universal gravitation was subsequently presented in the Principia in 1687.

A historical account of this calculation is available in Purdue University’s explanation of Newton’s Moon test.

The important connection to our question is that the distance-squared relationship is not just a mathematical feature. It describes how the gravitational field changes as you move away from a planet.

The planet’s mass determines the overall strength of its gravitational influence, while your distance from its centre determines how that influence changes with position.

03Radius is not altitude

There is another distinction that becomes important when comparing planets.

A planet’s radius and an object’s altitude are not interchangeable.

• Radius, \(R\), is the distance from the planet’s centre to its surface.
• Altitude, \(h\), is the distance above the planet’s surface.
• Radial distance, \(r\), is the distance from the planet’s centre to the object.

These quantities are related by:

\[ \boxed{r=R+h} \]

Why does this matter?

The gravitational field strength at altitude \(h\) is:

\[ \boxed{g=\frac{GM}{(R+h)^2}} \]

At the surface, \(h=0\), giving:

\[ g_{\text{surface}}=\frac{GM}{R^2} \]

Now suppose a spacecraft travels to an altitude equal to the planet’s radius.

Its altitude is \(h=R\), but its distance from the centre is:

\[ r=R+R=2R \]

Therefore:

\[ g=\frac{GM}{(2R)^2} \]

\[ \boxed{g=\frac14g_{\text{surface}}} \]

The gravitational field has fallen to one-quarter of its surface value, not one-half.

This also creates an interesting possibility when comparing two different planets.

A smaller planet can have stronger surface gravity, but a larger planet can have stronger gravity at the same altitude above their respective surfaces.

We’ll see exactly how this happens in a moment.

04Worked examples

The following examples use illustrative planetary data. They are not measurements of actual planets.

05Gravitational field, force, potential, and energy are different

These quantities are connected, but they describe different things.

Let’s look at the distinctions that matter most when comparing planets.

Gravitational field strength versus gravitational force

Gravitational field strength is a property of a location in a gravitational field.

Gravitational force depends on the mass of the object placed at that location.

The relationship is:

\[ \boxed{F_g=mg} \]

For example, suppose Planet S has a surface field strength of \(10.7\ \mathrm{N\\,kg^{-1}}\).

A \(2.0\ \mathrm{kg}\) object experiences:

\[ F_g=2.0\times10.7 \]

\[ F_g=21.4\ \mathrm{N} \]

A \(4.0\ \mathrm{kg}\) object at the same location experiences:

\[ F_g=4.0\times10.7 \]

\[ F_g=42.8\ \mathrm{N} \]

The gravitational field has not changed. The second object simply has twice the mass, so it experiences twice the gravitational force.

When comparing two planets, don’t confuse the planet producing the stronger field with the planet producing the greater force on two different test objects.

Gravitational field strength versus gravitational potential

Gravitational field strength tells you how much gravitational force acts per kilogram at a particular location.

Gravitational potential tells you the gravitational potential energy per kilogram at that location.

For a spherical planet:

\[ \boxed{\phi=-\frac{GM}{r}} \]

We use the convention that gravitational potential is zero at infinity.

This makes gravitational potential negative at any finite distance from an isolated planet.

Notice the difference between the equations:

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

\[ \phi=-\frac{GM}{r} \]

Field strength depends on the square of the distance, while the magnitude of gravitational potential depends on the first power of distance.

This is why the two quantities can give different comparisons.

A useful way to picture the distinction is to imagine a landscape.

• Gravitational field strength is related to the steepness of the landscape at your current position.
• Gravitational potential describes the energy level at that position relative to your chosen zero.

A steep section of a hill does not necessarily belong to the hill with the greatest total height.

The analogy is not exact: gravitational potential is energy per kilogram, not physical height. However, it helps explain why a planet with a stronger surface field does not automatically have a more negative surface potential.

Gravitational potential versus gravitational potential energy

Gravitational potential is measured per kilogram.

Gravitational potential energy depends on the mass of the object in the field.

Their relationship is:

\[ \boxed{U=m\phi} \]

For an object of mass \(m\) at the surface of a spherical planet:

\[ U=-\frac{GMm}{R} \]

Using zero potential energy at infinity, the work required to move an object slowly from rest at the surface to infinity is:

\[ \boxed{W_{\text{escape}}=\frac{GMm}{R}} \]

This assumes that the object starts and finishes at rest, and that energy losses and other forces are negligible.

A larger object requires more energy to escape than a smaller object in the same gravitational field.

More importantly, a planet with weaker surface gravity can still require more energy per kilogram to escape.

That is exactly what happened in our comparison between Planets S and L.

06A reliable decision process for comparing planets

When you encounter a gravitational comparison, first identify what the question is actually asking.

A few assumptions also matter:

• Treat the planets as spherical, with spherically symmetric mass distributions.
• Use the planet’s total mass and distance from its centre when calculating the external gravitational field.
• Ignore rotation, atmospheric effects, and local variations in gravity unless the question provides additional information.
• Keep distances in metres and masses in kilograms when using \(G\).
• State the reference convention when calculating gravitational potential or potential energy.

For a more detailed derivation of the surface-field relationship, see Gravitational Field Strength at a Planet’s Surface.

07Questions and solutions

The questions below build from straightforward surface-gravity comparisons towards situations where surface field strength, altitude, density, and gravitational energy interact.

Question 1

Planet P has a mass of \(0.30M_E\) and a radius of \(0.40R_E\), where \(M_E\) and \(R_E\) are Earth’s mass and radius.

Take Earth’s surface gravitational field strength as \(9.81\ \mathrm{N\\,kg^{-1}}\).

Calculate Planet P’s surface gravitational field strength and explain why it can be stronger than Earth’s despite having less mass.

Solution 1

Planet P has a surface gravitational field strength of \(18.4\ \mathrm{N\\,kg^{-1}}\), approximately 1.88 times Earth’s.

Use the comparison equation:

\[ \frac{g_P}{g_E} = \frac{M_P}{M_E} \left(\frac{R_E}{R_P}\right)^2 \]

Substitute:

\[ \frac{g_P}{g_E} = \frac{0.30}{(0.40)^2} \]

\[ =\frac{0.30}{0.16} \]

\[ =1.875 \]

Therefore:

\[ g_P=1.875\times9.81 \]

\[ \boxed{g_P=18.4\ \mathrm{N\\,kg^{-1}}} \]

Planet P has less mass than Earth, but its radius is substantially smaller.

The reduction in radius increases the surface gravitational field strength enough to outweigh the reduction in mass.

This is why comparing planetary masses alone gives an unreliable prediction.

Question 2

Planet A has a mass of \(0.50M_E\) and a radius of \(0.50R_E\).

Planet B has a mass of \(2.0M_E\) and a radius of \(1.50R_E\).

Two objects are placed on their respective surfaces:

• A \(2.0\ \mathrm{kg}\) object on Planet A.
• A \(5.0\ \mathrm{kg}\) object on Planet B.

Determine:

a) Which planet has the stronger surface gravitational field?

b) The gravitational force acting on each object.

c) Whether the object experiencing the greater force is on the planet with the stronger gravitational field.

Solution 2

Planet A has the stronger surface gravitational field, but the object on Planet B experiences the greater gravitational force.

Part a: Compare surface gravitational field strengths.

For Planet A:

\[ \frac{g_A}{g_E} = \frac{0.50}{(0.50)^2} \]

\[ \frac{g_A}{g_E}=2.0 \]

Therefore:

\[ g_A=2.0\times9.81 \]

\[ \boxed{g_A=19.62\ \mathrm{N\\,kg^{-1}}} \]

For Planet B:

\[ \frac{g_B}{g_E} = \frac{2.0}{(1.50)^2} \]

\[ \frac{g_B}{g_E} = \frac{2.0}{2.25} =0.8889 \]

Therefore:

\[ g_B=0.8889\times9.81 \]

\[ \boxed{g_B=8.72\ \mathrm{N\\,kg^{-1}}} \]

Planet A has the stronger surface gravitational field.

Part b: Calculate the gravitational forces.

Use:

\[ F_g=mg \]

For the object on Planet A:

\[ F_A=2.0\times19.62 \]

\[ \boxed{F_A=39.2\ \mathrm{N}} \]

For the object on Planet B:

\[ F_B=5.0\times8.72 \]

\[ \boxed{F_B=43.6\ \mathrm{N}} \]

Part c: Interpret the results.

The object on Planet B experiences the greater gravitational force.

However, Planet A has the stronger gravitational field.

The tempting mistake is to compare the two forces and assume that the larger force identifies the stronger planetary field.

That ignores the different masses of the objects.

Planet B’s object experiences a larger force because it has substantially more mass, not because Planet B has stronger surface gravity.

Question 3

Two spherical planets have the following properties:

A spacecraft travels to an altitude of \(4.0\times10^6\ \mathrm{m}\) above each planet’s surface.

a) Determine which planet has stronger surface gravity.

b) Calculate the gravitational field strength experienced by the spacecraft above each planet.

c) Explain why the planet with stronger surface gravity does not necessarily produce the stronger field at this altitude.

Solution 3

Planet P has stronger surface gravity, but Planet Q produces the stronger gravitational field at the specified altitude.

Part a: Compare the surface fields.

For Planet P:

\[ g_P=\frac{GM_P}{R_P^2} \]

\[ g_P= \frac{(6.67\times10^{-11})(2.0\times10^{24})} {(4.0\times10^6)^2} \]

\[ \boxed{g_P=8.34\ \mathrm{N\\,kg^{-1}}} \]

For Planet Q:

\[ g_Q= \frac{(6.67\times10^{-11})(6.0\times10^{24})} {(8.0\times10^6)^2} \]

\[ \boxed{g_Q=6.25\ \mathrm{N\\,kg^{-1}}} \]

Therefore, Planet P has stronger surface gravity.

Part b: Calculate the field strengths at altitude.

The spacecraft’s distance from the centre of each planet is:

\[ r=R+h \]

For Planet P:

\[ r_P=4.0\times10^6+4.0\times10^6 \]

\[ r_P=8.0\times10^6\ \mathrm{m} \]

Therefore:

\[ g_P= \frac{(6.67\times10^{-11})(2.0\times10^{24})} {(8.0\times10^6)^2} \]

\[ \boxed{g_P=2.08\ \mathrm{N\\,kg^{-1}}} \]

For Planet Q:

\[ r_Q=8.0\times10^6+4.0\times10^6 \]

\[ r_Q=1.2\times10^7\ \mathrm{m} \]

Therefore:

\[ g_Q= \frac{(6.67\times10^{-11})(6.0\times10^{24})} {(1.2\times10^7)^2} \]

\[ \boxed{g_Q=2.78\ \mathrm{N\\,kg^{-1}}} \]

Planet Q has the stronger field at the specified altitude.

Part c: Explain the change in ranking.

The spacecraft are at equal altitudes, but they are not at equal distances from their respective planet centres.

Their radial distances are:

• Planet P: \(8.0\times10^6\ \mathrm{m}\).
• Planet Q: \(1.2\times10^7\ \mathrm{m}\).

Planet Q also has three times the mass of Planet P.

At the surface, Planet P’s smaller radius gives it the stronger field. At the specified altitude, Planet Q’s greater mass produces the stronger field.

The mistake would be to assume that the surface-gravity ranking must remain unchanged at every altitude.

Question 4

Two spherical planets have the following average densities and radii:

Here, \(\rho\) and \(R\) are positive reference values.

a) Determine the ratio of their masses.

b) Determine the ratio of their surface gravitational field strengths.

c) Explain how the planet with the greater mass can have weaker surface gravity.

Solution 4

Planet Q has \(8/3\) times the mass of Planet P, but Planet P has 1.5 times the surface gravitational field strength of Planet Q.

Part a: Compare their masses.

Mass is density multiplied by volume:

\[ M=\rho V \]

For a spherical planet:

\[ V=\frac43\pi R^3 \]

Therefore, Planet P has mass:

\[ M_P=\frac43\pi(3\rho)R^3 \]

\[ M_P=4\pi\rho R^3 \]

Planet Q has mass:

\[ M_Q=\frac43\pi(\rho)(2R)^3 \]

\[ M_Q=\frac{32}{3}\pi\rho R^3 \]

Calculate the ratio:

\[ \frac{M_Q}{M_P} = \frac{\frac{32}{3}\pi\rho R^3} {4\pi\rho R^3} \]

\[ \boxed{\frac{M_Q}{M_P}=\frac83} \]

Planet Q has \(8/3\) times the mass of Planet P.

Part b: Compare their surface gravitational fields.

Since:

\[ g=\frac43\pi G\rho R \]

For Planet P:

\[ g_P=\frac43\pi G(3\rho)R \]

For Planet Q:

\[ g_Q=\frac43\pi G(\rho)(2R) \]

Divide:

\[ \frac{g_P}{g_Q} = \frac{3\rho R}{2\rho R} \]

\[ \boxed{\frac{g_P}{g_Q}=1.5} \]

Planet P has 1.5 times the surface gravitational field strength of Planet Q.

Part c: Explain the result.

Planet Q has more mass because its much larger volume more than compensates for its lower average density.

However, surface gravity depends on mass divided by radius squared.

Planet Q’s larger radius reduces its surface field strength enough to outweigh its greater mass.

This is why mass and density cannot be considered separately from radius.

Question 5

Two spherical, non-rotating planets have the following properties:

• Planet A has mass \(M\) and radius \(R\).
• Planet B has mass \(3M\) and radius \(2R\).

Both planets have spherically symmetric mass distributions.

A student makes the following claim:

“Planet A has stronger surface gravity, so it must also have the deeper gravitational potential and require more energy per kilogram to escape. At an altitude of \(R\) above each surface, Planet A must still have the stronger gravitational field.”

Test this claim.

Determine:

a) Which planet has stronger surface gravitational field strength, and by what factor?

b) The gravitational field strength at an altitude of \(R\) above each surface, expressed in terms of \(g_0=GM/R^2\).

c) Which planet has the more negative surface gravitational potential, and by what factor?

d) The ratio of the work required per kilogram to move an object slowly from each surface to infinity, assuming it starts and finishes at rest.

e) Explain why the student’s prediction fails, even though Planet A genuinely has stronger surface gravity.

Solution 5

The student’s claim is incorrect. Planet A has stronger surface gravity, but Planet B has a more negative surface potential, requires more escape energy per kilogram, and produces the stronger field at the specified altitude.

This question tests whether you can distinguish the local strength of a gravitational field from the overall gravitational potential of a planet.

Part a: Compare surface gravitational field strengths.

For Planet A:

\[ g_A=\frac{GM}{R^2} \]

Define:

\[ g_0=\frac{GM}{R^2} \]

Therefore:

\[ g_A=g_0 \]

For Planet B:

\[ g_B=\frac{G(3M)}{(2R)^2} \]

\[ g_B=\frac{3GM}{4R^2} \]

Therefore:

\[ g_B=\frac34g_0 \]

The ratio is:

\[ \frac{g_A}{g_B} = \frac{g_0}{0.75g_0} \]

\[ \boxed{\frac{g_A}{g_B}=\frac43} \]

Planet A has \(4/3\) times the surface gravitational field strength of Planet B.

Part b: Compare the fields at altitude \(h=R\).

The important step is to calculate the distance from each planet’s centre.

For Planet A:

\[ r_A=R+h \]

\[ r_A=R+R=2R \]

For Planet B:

\[ r_B=2R+h \]

\[ r_B=2R+R=3R \]

Now calculate the field strength at each position.

For Planet A:

\[ g_A’=\frac{GM}{(2R)^2} \]

\[ \boxed{g_A’=\frac14g_0} \]

For Planet B:

\[ g_B’=\frac{G(3M)}{(3R)^2} \]

\[ g_B’=\frac{3GM}{9R^2} \]

\[ \boxed{g_B’=\frac13g_0} \]

Compare:

\[ \frac{g_B’}{g_A’} = \frac{1/3}{1/4} \]

\[ \boxed{\frac{g_B’}{g_A’}=\frac43} \]

Planet B has the stronger gravitational field at this altitude.

The surface ranking has reversed because the two objects are at different radial distances, and Planet B has more mass.

Part c: Compare surface gravitational potentials.

With gravitational potential zero at infinity:

\[ \phi=-\frac{GM}{r} \]

At the surface of Planet A:

\[ \phi_A=-\frac{GM}{R} \]

At the surface of Planet B:

\[ \phi_B=-\frac{G(3M)}{2R} \]

Therefore:

\[ \phi_B=-\frac32\frac{GM}{R} \]

The ratio of their potential magnitudes is:

\[ \boxed{\frac{|\phi_B|}{|\phi_A|}=\frac32} \]

Planet B has a surface gravitational potential that is 1.5 times more negative than Planet A’s.

This is the key distinction.

Planet A has stronger surface gravity because its value of \(M/R^2\) is greater.

Planet B has a deeper gravitational potential because its value of \(M/R\) is greater.

Part d: Compare the escape work per kilogram.

The work required per kilogram to move an object slowly from the surface to infinity is:

\[ \frac{W_{\text{escape}}}{m}=\frac{GM}{R} \]

For Planet A:

\[ \frac{W_A}{m}=\frac{GM}{R} \]

For Planet B:

\[ \frac{W_B}{m}=\frac{3GM}{2R} \]

Therefore:

\[ \boxed{\frac{W_B}{W_A}=\frac32} \]

Planet B requires 1.5 times as much energy per kilogram to escape.

Part e: Explain the student’s misconception.

The student has assumed that stronger surface gravity automatically means a deeper gravitational potential.

However, the two quantities depend on different powers of radial distance:

\[ g\propto\frac{M}{R^2} \]

\[ |\phi|\propto\frac{M}{R} \]

Surface gravitational field strength describes the local gravitational pull per kilogram.

Gravitational potential describes the energy per kilogram relative to infinity.

Planet A produces a stronger gravitational field right at its surface, but Planet B’s greater mass creates a deeper gravitational potential.

The student’s mistake is treating a local measurement as though it describes the entire gravitational system.

The next useful step is to apply this changing gravitational field to objects in motion. Once you know how gravity depends on mass and distance, you can determine how it provides the centripetal force needed to keep a satellite in orbit.

Continue with Gravity as Centripetal Force in Circular Orbits for HSC Physics.